F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 2014 McGraw-Hill Education (Italy) srl

Size: px
Start display at page:

Download "F. P. Beer et al., Meccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn 9788838668579, 2014 McGraw-Hill Education (Italy) srl"

Transcription

1

2 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: Σ = 0: bp = 0 = Pb Σ = 0: ap = 0 = Pa From to B: 0 < x < a Σ Fy = 0: Pb V = 0 Pb Σ J = 0: x = 0 V = Pb = Pbx From B to : a < x < Pa Σ Fy = 0: V + = 0 V Pa Σ K = 0: + ( x) = 0 Pa = Pa( x) = Pab t section B: = PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

3 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: Σ B = 0: w + w = 0 = Σ = 0: w B w = 0 B = Free body diagram for determining reactions. Over whole beam, 0 < x < Place section at x. Replace distributed load by equivalent concentrated load. w Σ Fy = 0: wx V = 0 V w = x w x Σ J = 0: x + wx + = 0 w = ( x x ) aximum bending moment occurs at x =. w = x ( x) w max = 8 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

4 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl From to B (0 < x < a) : F = 0: wx V = 0 y x J = 0: ( wx) + = 0 From B to ( a < x < ) : V = wx wx = F = 0: wa V = 0 V = wa y a a J = 0: ( wa) x + = 0 = wax PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

5 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl wx 0 Σ Fy = 0: x V = 0 t x =, wx 0 x Σ J = 0: x + = 0 V wx 0 = wx 0 = w w V max = 0 0 V = 0 w = w 0 = max PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

6 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: = = wa From to B: 0 < x < a = 0: wa wx V = 0 F y x J = 0: wax + ( wx) + = 0 V = w( a x) = w ax x From B to : a < x < a = 0: wa wa V = 0 F y a J = 0: wax + wax + = 0 From to : a < x < F = 0: V w( x) + wa = 0 y V = 0 = wa V = w( x a) x J = 0: w( x) + wa( x) = 0 = wa[( x) ( x) ] PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

7 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl alculate reactions after replacing distributed load by an equivalent concentrated load. Reactions are = = w ( a ) From to B: 0 < x < a Σ Fy = 0: w( a) V = 0 V = w( a) Σ = 0: w( a) + = 0 From B to : a < x < a x a b = = w ( ax ) PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed, Place section cut at x. Replace distributed load by equivalent concentrated load. Σ Fy = 0: w( a) w( x a) V = 0 V w = x x a J = 0: w( a) x + w( x a) 0 + = [( ) ( ) ] = w a x x a

8 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5. (ontinu From to : a < x < t x =, Σ Fy = 0: V + w( a) = 0 Σ J = 0: + w( a)( x) = 0 w V = ( a) = w ( a )( x ) a max = w 8 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

9 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl t B, V = 00N, = 0 t E +, F = 0 : V 00 = 0 V = 00N y = 0 : (0.5)(00) = 0 t +, E = 45 N m F = 0 : V = 0 y (a) V = 00 N = 0 : + (0.)(500) (0.55)(00) = 0 = 45 N m t +, F = 0 : V = 0 V = 00N y = 0 : (0.5)(00) + (0.55)(500) (0.75)(00) = 0 (b) = 7.5 N m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

10 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5.8 (ontinued) t, F = 0 : V = 0 V = 00 N y = 0 : (0.)(00) (0.55)(00) + (0.85)(500) (.05)(00) = 0 = 7.5 N m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

11 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: = 0 : + ()(4) ()(40) = 0 = 8kN = 8kN = 0: ()(4) ()(40) = 0 = 7kN = 7kN to. 0 < x < m F = 0: 8 x V = 0 V = ( 8 x)kn y x J = 0: 8 x ( x) = 0 = ( 8x x )kn m to B. m < x < m F = 0: V 40= 0 y V = 40kN = 0: ( x)(40) = 0 K = (40x 0) kn m From the diagrams, (a) V = 40.0kN max (b) = 40.0kN m max PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

12 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reaction at : B = 0: R + (0.550)(75) + (0.00)(75) = 0 R = 85 N lso, R 5 N B = to : V = 85 N to : V = 0 N to B : V = 5 N t and B, = 0 Just to the left of, Σ = 0: (0.5)(85) + = 0 =.5 N m Just to the right of, Σ = 0: (0.5)(85) + (0.050)(75) + = 0 = 7.50 N m Just to the left of, = 0 N m Just to the right of, Σ = 0: (0.50)(85) + (0.00)(75) + = 0 Σ = 0: + (0.5)(5) = 0 =.5 kn (a) (b) V max = 85.0 N max =.5 N m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

13 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Over the whole beam, Σ F = 0:.5w.5.5 = 0 w = kn/m y to : 0 x < 0. m Σ F = 0: x V = 0 V = ( x) kn y x Σ J = 0: ( x) + = 0 = ( x ) kn m t, x = 0. m V = 0. kn, = kn m to : 0. m < x <. m = 90 N m Σ F = 0: x.5 V = 0 V = (x.5) kn y x Σ J = 0: ( x) + (.5)( x 0.) + = 0 t the center of the beam: x = 0.75 m V = 0 = 0.5 kn m =.5 N m = ( x.5x ) kn m +, t x = 0. m, V = 0.9 kn (a) aximum V = 0.9 kn = 900 N (b) aximum =.5 N m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

14 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reaction at : B = 0: (.0)() + (.5)() + (.8)(4.5)(.5) = 0 = 7.05 kn Use as free body. Σ = 0: (7.05)(.5) + (.8)(.5)(0.75) = 0 = 8.55 kn m = N m I c = bh = (80)(00) = 80 0 mm 4 4 = 80 0 m = (00) = 50 mm = 0.50 m c ( )(0.50) σ = = = Pa I 80 0 σ = 7. Pa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

15 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Use B as free body. = 0: (00)() = 0 = 00 lb ft = 4. 0 lb in For rectangular section, I = bh = (4)(8) = 70.7 in c = h = 4 in. c (4. 0 )(4) σ = = =.05 0 psi σ =.0 ksi I 70.7 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

16 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Use portion B as free body. = 0 : + ()(.)(.05) + (8)(.) = 0 =.45 kn m =.45 0 N m For W0 0 : S = mm = m Normal stress: σ =.45 0 = S = Pa σ = 7.7 Pa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

17 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: By symmetry, = B = 0 : = B = 80 kn F y Using left half of beam as free body, J = 0: (80)() + (0)(.) + (50)(0.4) + = 0 = 04 kn m = 04 0 N m For W0 5 : S = mm = m Normal stress: 04 0 σ = = = 9. 0 Pa S σ = 9. Pa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

18 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Use entire beam as free body. B = 0: (.)() + (.)(50) + (0.8)(50) = 0 = 7 kn Use portion as free body. = 0: (.4)(7) + (.)() = 0 = 09. kn m For W40, S = 90 0 mm Normal stress: σ = = S N m 90 0 m = Pa σ = 9.5 Pa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

19 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: = 0 : 4 4 (4)()() = 0 = 8 kn F = 0 : 8 + (4)() = 0 = 7 kn y to : 0 < x < m to : m < x < 4m to B: 4m < x < m Fy = 0: V 8= 0 V = 8 kn J = 0: + 8x = 0 = ( 8 x) kn m Fy = 0: V 8= 0 V = 8 kn J = 0: + 8x 4= 0 = ( 8x + 4)kN m F y = 0: V 4( x) = 0 V = ( 4x + 44) kn max = 5 kn m = 5 0 N m J = 0: x 4( x) = 0 = ( x) kn m For S50 5 section, S = 48 0 mm 5 0 N m Normal Stress: σ = = =. 0 Pa S 48 0 m σ =. Pa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

20 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Statics: onsider portion B and BE separately. = B = E = 0 Portion BE: E = 0: (9)(.) + (48)(.) () + (0)(.5) = 0 E = 48kN = 5kN t midpoint of B: Fy = 0: V = 0 = 0 : = (9)(.) (9)(0.) = 57. kn m Just to the left of : F = 0 : V = 9 48 = 44 kn y = 0 : = (9)(0.) (48)(0.) = 7 kn Just to the left of : Fy = 0 : V = 0 5 = + 04 kn = 0 : = (5)(.5) = + 84 kn m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

21 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5. (ontinued) From the diagram: = 84 kn m = 84 0 N m max For W0 0 rolled steel shape, S x = mm = m Stress: σ = m S max σ m 84 0 = = Pa σ = 99.5 Pa m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

22 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reaction at : = 0 : (5)(.)(.) = 0 B = 5 kn to : 0 < x <.m to B:.m < x < 4.8m F = 0: 5 V = 0 V = 5kN y = 0: x = 0 J = (0x 40) kn m Fy = 0 : 5 5( x.) V = 0 V = ( 5x + 75)kN K = 0: x x. + (5)( x.) = 0 = (.5x + 75x 7) kn m Normal stress: For W00., S = 98 0 mm σ = = = S 98 0 m N m Pa σ = 5.9 Pa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

23 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl By symmetry, = B Σ F = 0: B = 0 y = B = kn Shear: to : V = kn + to : V = kn + to E : V = kn + E to B: V = kn Bending moment: t, Σ = 0: ()() = 0 = kn m t, +Σ = 0: ()() + (8)() = 0 4kN m By symmetry, = kn m at E. = kn m max = 4 kn m occurs at E. E For W0.8, = 80 0 mm = 80 0 m S x Normal stress: σ max 4 0 max = = Sx 80 0 = Pa σ max = 4.9 Pa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

24 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl By symmetry, = B Σ F = 0: 8 + W 8 + B = 0 y = B = 8 0.5W Bending moment at : Σ = 0: (8 0.5 W)() + = 0 Bending moment at : = (8 0.5 W) kn m Σ = 0: (8 0.5 W)() + (8)() + = 0 = (8 W) kn m Equate: = W 8 = 8 0.5W (a) W = 0.7 kn =.7 kn m =.7 kn m =.7.0 N m =.7 kn m max For W0.8 rolled steel shape, W = 0.7 kn S x = 80 0 mm = 80 0 m (b) σ max.7 0 max Sx 80 0 = = = Pa σ = 9.5 Pa max PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

25 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl P = 480 N Q = 480 N Reaction at : Σ = 0: a + 480( a 0.5) 480( a) = 0 70 = 90 N a Bending moment at : Σ = 0: = 0 0 = 0.5 = 480 N m a Bending moment at : Σ = 0: 480( a) = 0 = 480( a) N m (a) (b) Equate: For rectangular section, 0 = 480( a) = 480 a a = 0.80 m a = 8 mm = 8. N = 4. N m = 4. N m S = bh S = ()() = 48 mm = 48 0 m σ 9 4. = = = Pa S.48 0 max max 9 σ = 99. Pa max PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

26 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl P = 480 N Q = 0 N Reaction at : Σ = 0: a + 480( a 0.5) 0( a) = 0 50 = 800 N a Bending moment at : Σ = 0: = 0 80 = 0.5 = 400 N m a Bending moment at : Σ = 0: 0( a) = 0 = ( a) N m (a) Equate: 80 = 0 0a = 400 a 0a + 80a 80 = 0 a = m,.087 m (b) Reject negative root. =.04 N = N m = N m For rectangular section, S = bh ()(8) 48 mm 48 0 S = = = m σ = = = Pa S 48 0 max max 9 max a = 89 mm σ = 89.5 Pa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

27 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Weight density: γ = ρg et = total length of beam. W = ρg = b ρg Reactions at and : = = W Bending moment at : W Σ = 0: + = 0 W = 8 Bending moment at center of beam: W W W Σ E = 0: = = 4 For a square section, Normal stress: W b ρ g max = = 8 8 S = b bρg/8 ρg σ = = = S b / b Solve for b: b = ρ g σ ata: =. m ρ = 780 kg/m g = 9.8 m/s (a) σ = 0 0 Pa (b) σ = 50 0 Pa (a) (b) (.) (780)(9.8) b = =. 0 m ()(0 0 ) (.) (780)(9.8) b = =. 0 m ()(50 0 ) b =. mm b =. mm PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

28 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Σ = 0: bp = 0 = Pb Σ = 0: ap = 0 = Pa t +, V = = Pb = 0 to B : 0 < x < a t +, B x 0 w = 0 wdx = 0 V V = 0 a apb Pba B = Vdx = dx = 0 0 Pb Pa V = P = P = B + to : a < x < B Pb V = Pba = x a w = 0 wdx = 0 V V = 0 B Pa Pab B = Vdx = ( a) = a Pab Pba Pab = B = = 0 Pa V = max Pab = PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

29 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl w Σ B = 0: + w = 0 = w Σ = 0: B w = 0 B = dv w dx = x 0 V V = wdx = wx V = V wx = wx d V dx = x xw = Vdx = wx dx 0 0 wx wx = V w = wx wx wx = + w = ( x x ) aximum occurs at x =, where V d = = 0 dx w max = 8 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

30 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Over B: V = 0 = 0 x 0 V = wdx = wx d = V = wx dx x wx = Vdx = 0 x 0 wx = t B: x = a VB = wa wa B = Over B: w = 0 dv 0 V constant VB dx = = = V = wa d V wa dx = = x x B a a = Vdx = wax = wa ( x a ) = wa( x a) wa = wax a PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

31 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl w = x w0 V V = 0, = 0 x wx 0 wx 0 V = = 0 dv W0x = w = dx d dx w0 x = V = V wx 0 = wx = V dx = dx x x wx 0 = PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

32 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: = = wa to B: 0 < x < a w = w V = = wa, = 0 x 0 V V = wdx = wx V = w( a x) d V wa wx dx = = x = Vdx = ( wa wx ) dx 0 0 VB = 0 B = wa x = wax wx B to : a < x < a V = 0 d V 0 dx = = x a = V dx = 0 B = B = wa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

33 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5.8 (ontinued) x a to : V V = wdx = w[ x ( a)] x x a a = V dx = [ wx ( a )] dx x = w ( a) x x a x ( a) = w ( a) x + ( a) x ( a) = w ( a) x + V = w[ x ( a)] x ( a) = wa w ( a) x + PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

34 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions. = = w ( a ) t. V = = w ( a ), = 0 to B. 0 < x < a w = 0 V V = wdx= 0 B VB = V = w ( a ) 0 a a a B = V dx = w ( a ) dx 0 0 B = w ( aa ) B to. a < x < a w = w x V V = wdx= w( x a) B a V = w( a) w( x a) = w( x) d V w( x) dx = = x x ( B a ) a = V dx = w x x ( wx x a a ) = + = w ( aa ) + wx ( x a+ a) ( ) = wx x a PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

35 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5.9 (ontinued) t. x = a V = w( a) = ( a) a to. V = V = w ( a ) = 0 t x =, a max = w 8 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

36 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl = 0 : (0.)(00) (0.55)(00) = 7.5 N m + (0.85)(500) (.05)(00) = 0 F = 0 : V = 0 y V = Shear: to : 00 N V = 00 N to : V = = 00 N to E: V = = 00 N E to B: V = = 00 N reas under shear diagram: to : Vdx = (00)(0.) = 0 N m to : Vdx = ( 00)(0.5) =.5 N m to E: Vdx = ( 00)(0.) = 90 N m Bending moments: = 7.5 N m E to B: Vdx = (00)(0.5) = 45 N m = + V dx = = 7.5 N m = + V dx = = 45 N m E E = + V dx = = 45 N m B E E = + V dx = = 0 (a) aximum V = 00 N PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed, (b) aximum = 7.5 N m

37 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: = 0 : + ()()() (40)() = 0 = 8kN = 0 : ()()() (40)() = 0 = 7kN Shear diagram: V = 8kN to : 0 < x < m w = kn/m V V = wdx= dx= 4 kn V 0 0 = 4 8 = kn to B: V B = + 7 = 40 kn reas of shear diagram: to : Vdx = ( 8 )() = 40 kn m to B: Vdx = ()(40) = 40 kn m Bending moments: = 0 = + Vdx = 0 40 = 40kN m = + Vdx = = 0 B (a) aximum V = 40.0 kn (b) aximum = 40.0 kn m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

38 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reaction at : Reaction at B: B =.75 kn Beam B and loading: (See sketch.) reas of load diagram: to : (.4)(.5) = 8.4 kn to B: (0.)(.5) =. kn Shear diagram: Σ B = 0:.0 + (.5)(.0)(.5) + (.5)() = 0 =.75 kn V V V + Over to, V =.75.5x V B =.75 kn = =.5 kn =.5 = 4.5 kn = 4.5. =.75 kn t G, V =.75.5x = 0 x =.98 m reas of shear diagram: G G to G: G to : to B: (.98)(.75) =.5089 kn m (0.474)(.5) = kn m (0.)( ) =.4 kn m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

39 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5.44 (ontinued) Bending moments: = 0 G + B = =.5089 kn m = =. kn m =..7 =.4 kn m =.4.4 = 0 (a) (b) V max =.75 kn max =.5 kn m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

40 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl = 0: B + ()(4) + (0.5)(4) = 0 = kn = 0: B ()(4) (.5)(4) = 0 B = kn Shear diagram: to : V = kn to : V = 4 = kn to B: V = 4 = kn reas of shear diagram: to : Vdx = ()() = kn m to : Vdx = ()( ) = kn m to E: Vdx = ()( ) = kn m Bending moments: = 0 = 0 + = kn m + + B = + 4 = kn m = = 4kN m = 4 + = kn m = = 0 (a) (b) V =.00 kn max max =.00 kn m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

41 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl By symmetry, = B. Σ F = 0: + B (4.5)(.8) = 0 y = B = 7.05 kn Shear diagram: to : V = 7.05 kn w =.8 kn/m t, V = 7.05 (.8)(.5) = 4.5 kn t +, V = 4.5 =.5 kn + to : w =.8 kn/m t, V =.5 (.5)(.8) =.5 kn t +, V =.5 = 4.5 kn + to B: w =.8 kn t B, V = 4.5 (.5)(.8) = 7.05 kn raw the shear diagram: V = 0 at point E, the midpoint of. reas of the shear diagram: to : to E: E to : to B: ( )(.5) = 8.55 kn m (.5)(0.75) = kn m (.5)(0.75) kn m = ( )(.5) 8.55 kn m = PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

42 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5.4 (ontinued) Bending moments: E B = 0 = = 8.55 kn m = = kn m = = 8.55 kn m = = 0 = N m For a rectangular section, (80)(00) S = bh = =. 0 mm =. 0 m aximum normal stress at : σ = = S. 0 = Pa σ = 7. Pa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

43 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: By symmetry, = B. = 0 : = B = 80 kn F y Shear diagram: to : V = 80 kn to : V = 80 0 = 50 kn to E: V = = 0 reas of shear diagram: to : Vdx = (80)(0.8) = 4 kn m to : Vdx = (50)(0.8) = 40 kn m to E: Vdx = 0 Bending moments: = 0 = = 4kN m = = 04 kn m = = 04 kn m E = 04 kn m = 04 0 N m max For W0 5, S = mm = m Normal stress: 04 0 σ = = = 9. 0 Pa S σ = 9.Pa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

44 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl (a) k =. aximum occurs at x =. wx 0 kw0( x) wx 0 w= = ( + k) kw. dv wx 0 = w= kw0 ( + k) dx wx 0 0 V = kw x ( + k) + V = 0 at x= 0 = 0 d wx 0 = V = kw 0 x ( + k ) dx kw0x w0x = ( + k) + = 0 at x = 0 = 0 kw0x ( + k) w0x = wx 0 = w0 x V wx 0 wx 0 = w 0 max = (b) k =. t V = 0 at x = ( ) ( ) w w w x =, = = = w t x =, = 0 V wx 0 wx 0 = 4 0 wx 0 wx = 4 4 w 0 = max 7 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

45 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl dv dx = w= w x d V = w0 + = dx x = w + x+ 0 x 0 = 0 at x = 0 = 0 = 0 at x= 0= w0 + = w0 (a) (b) x 0 0 V = w + w x 0 0 = w + w x d max occurs when V 0. xm 0 V = w0 ( x )/ = w0 ( x x / ) dx = = = xm = max = w 0 = 0.04w max 0 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

46 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl dv π x = w= w0 sin dx w 0 π x d V = cos + = π dx 0 w π x = sin + x+ π = 0at x = 0 = 0 = 0at x= 0= = 0 w 0 π x (a) V = cos π d V 0 at x dx = = = w 0 π x = sin π (b) w 0 max = π π sin w 0 max = π PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

47 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl = 0 : ()() ()(4)() + 4B = 0 B = 5.5 kn = 0 : (5)() + ()(4)() 4 = 0 B = 8.5 kn Shear: to : V = kn + : V = =.5kN B: V =.5 ()(4) = 5.5 kn ocate point where V = 0. reas of the shear diagram: d 4 d = d = d =.7 m 4 d =.8 m to : Vdx = (.0)() =.0kN m to : Vdx = (.7)(.5) = kn m to B: Bending moments: = 0 Vdx = (.8)( 5.5) = kn m = 0.0 =.0kN m = = kn m = = 0 B aximum = kn m = N m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

48 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5.55 (ontinued) For pipe: co = do = (0) = 80 mm, ci = di = (40) = 70 mm ( o i ) π π I = c c = (80) (70) =.5 0 mm S I.5 0 = = =.40 0 mm =.40 0 m c 80 o Normal stress: σ = = = 0. 0 Pa S.40 0 σ = 0. Pa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

49 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl w = 0 = 0: 4 R + ()(50) ()(50) = 0 = 0: R 4 R ()(50) ()(50) = 0 R = 50 kn = 50 kn Shear: to : V = 50 kn V = 50 kn to : V = = 00 kn to B: V = = 50 kn reas of shear diagram: to : Vdx = (50)() = 00 kn m to : Vdx = ( 00)() = 400 kn m to B: Vdx = (50)() = 00 kn m Bending moments: = 0 = + Vdx = = 00 kn m = + Vdx = = 00 kn m = + Vdx = = 0 B aximum = 00 kn m = 00 0 N m For W40 4 rolled steel section, S x = 00 0 mm = 00 0 m σ m max 00 0 = = =.4 0 Pa S 00 0 x σ =.4 Pa m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

50 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reaction: = 0 : (80)(.)() = 0 B = 7kN Shear: V = to : 7kN V = 7 kn V = 7 (80)(.) = 5 kn to : ocate point where V = 0: V( x) = 80x + 7 = 0 x = 0.95 m reas of shear diagram: V = 5 kn to : Vdx = (.)(7) = 9. kn m to E: E to : Vdx = (0.95)(7) =. kn m Vdx = (0.5)( 5) =.9 kn m to B: Vdx = (.)( 5) =.4 kn m Bending moments: = 0 kn m = =. kn m =. +. = 7. kn m E = 7..9 = 50.4 kn m = = kn m B For W50 80, S = 98 0 mm Normal stress: σ 7. 0 N m max = = = Pa S 98 0 m σ = 8.5 Pa m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

51 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl (a) Replace distributed load by equivalent concentrated load at the centroid of the area of the load diagram. For the triangular distribution, the centroid lies at x =. W = w0 (a) Fy = 0: R W = 0 R = w0 = 0:0= 0 equilibrium V = 0, = 0, at x = < x <, dv = w = w x dx 0 0 d = V = w x + = w x dx 0 wx 0 wx = + = Just to the left of, V w0( /) = = w0 9 5 Just to the right of, V = w0 + R = w0 9 8 Note sign change. aximum occurs at. 4 aximum = w For square cross section, I = a c = a w0 ( /) 4 = = w 0 8 (b) σ m c max 4 w 8 w w = = = = I 8 a 7 a a w 0 σ m = (.5 a) PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

52 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl (8)() mm 4 I = = c= d = 8 mm I S = = mm = m c t, = Sσ = ( )( 5.9) =.5 N m t, = Sσ = ( )( 9.9) =.5 N m Σ = 0:. (0.)(400) 0.P 0.5Q= 0 0.P+ 0.5Q = 8.5 () Σ = 0:.5 (0.05)(00) 0.P 0.5Q= 0 0.P+ 0.5Q= 0.5 () Solving () and () simultaneously, P = 500 N Q = 50 N Reaction force at : R = 0 R = 50 N m V = 50 N V = 50 =.5 N m =.5 N m =.5 N m V max = 50 N max = N m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

53 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl (a) t, (4)(0) 4 0 mm 4 0 mm I = = c= I S = = mm = m = Sσ c = (4.4 0 )(55 0 ) = 79 N m t F, F = (4.4 0 )(7.5 0 ) = 540 N m Using free body FB, Σ = 0: B= 0 F 540 B = = 800 N 0. Using free body EFB, Σ = 0: 79 Q+ (0.8)(800) = 0 Q = 0 N Using entire beam, Σ = 0: 0. P (0.7)(0) + (.)(800) = 0 P = 40 N = 00 N Shear diagram and its areas: to : Σ F = 0: = 0 V = 00 N = (0.)(00) = 70 N m + to E : V = = 0 N = (0.5)(0) = 80 N m E + to B: V = 0 0 = 800 N = (0.5)( 800) = 900 N m Bending moments: E EB y (b) E B = 0 = = 70 N m = = 900 N m = = 0 Normal stress. max = 900 N m σ 900 = = =.5 0 Pa S σ =.5 Pa max max max PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

54 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: onstruct shear and bending moment diagrams: Σ = 0:.4 + (.)(.8) + (0.8)(.) = 0 =.4 kn Σ = 0: (0.8)(.8) (.)(.) +.4 = 0 = kn max =.4 kn m=.4 0 N m σ all = Pa = 0 Pa max.4 0 Smin = = σ all 0 = 00 0 m = 00 0 mm (40) S = bh = h = 00 0 ()(00 0 ) h = 40 = 0 0 mm h = 7. mm PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

55 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: F = 0: (.)(8) = 0 y =. kn = 0: (.8)(.)(8) = 0 = 8.88 kn m Shear diagram: V = V =. kn B V =. (.)(8) = 0 reas of shear diagram: to B : (.)(.) = 5.9 kn m Bto : (.)(.) =.9 kn m Bending moments: = 8.88 kn m = =.9 kn m B = = 0 = 8.88 kn m = N m max σ = max S max S max N m = = = 40 0 m = 40 0 mm σ 0 Pa max For a rectangular section, S = bh S (40 0 ) h = = = 94mm b 5 h = 94 mm PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

56 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: By symmetry, =. Fy = 0 : ()(.5) ()(.5) ()(.5) = 0 Shear diagram: = =.75 kips V =.75 kips V B =.75 ()(.5) = 4.5 kips V = 4.5 ()(.5) = 4.5 kips V = 4.5 ()(.5) =.75 kips ocate point E where V = 0 : By symmetry, E is the midpoint of B. reas of the shear diagram: to B : ()(4.5) + ()(.5) = 8 kip ft B to E : ()(4.5) =.75 kip ft E to : ()( 4.5) =.75 kip ft to : By antisymmetry, 8 kip ft PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

57 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5. (ontinu) Bending moments: = 0 = = 8 kip ft B = = 4.75 kip ft E = = 8 kip ft = 8 8 = 0 σ max max max (4.75 kip ft)( in/ft) = S = = = 9.74 in S σ.750 ksi max For a rectangular section, S = bh S (9.74) h = = = 4.7 in. h = 4.7 in. b 5 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

58 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl By symmetry, B = Shear: to B: V =.5 kn V + = = 9 kn B V = 9 ()() = 9 kn to : V = =.5 kn reas of the shear diagram: Σ F = 0: B ()() = 0 B= =.5 kn to B: Vdx = (0.)(.5) =.5 kn m B to E: E to : to : y Vdx = (.5)(9) =.75 kn m Vdx =.75 kn m Vdx =.5 kn m Bending moments: = 0 B = =.5 kn m E = = 8.5 kn m = =.5 kn m =.5.5 = 0 aximum = 8.5 kn m = N m σ all = Pa = 0 Pa max Smin = = = σ 0 m = mm all For a rectangular section, S = bh = (00) h ()( ) h = = mm h = 0 mm 00 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

59 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl B = 0:.4 + (0.)(.)() = 0 =.7 kn = 0: (.8)(.)() +.4B = 0 B = 8. kn Shear: V V V B + B V =.7 kn =.7 (.4)() = 4.5 kn = =. kn =. (.)() = 0 ocate point where V = 0. d.4 d = d =.48 d = 0.9 m.4 d =.5 m reas of the shear diagram: to : to B: B to : Vdx = (0.9)(.7) =.5 kn m Vdx = (.5)( 4.5) =.75 kn m Vdx = (.)(.) =. kn m Bending moments: = 0 B = =.5 kn m =.5.75 =. kn m =. +. = 0 aximum =. kn m =. 0 N m σ all = Pa = 0 Pa S min σ all 0. 0 = = = 80 0 m = 80 0 mm For rectangular section, S = bh = b (50) = 80 0 ()(80 0 ) b = b = 48.0 mm 50 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

60 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Shape S, ( 0 mm ) W w = + x ( )kn/m = + x dv = w = x dx V = x x + V = 0 at x = 0, = 0 d V = = x x dx = x x + = 0 at x = 0, = 0 = x x occurs at x = m. max max = ()() () = 80 kn m = 80 0 N m σ all = 0 Pa = 0 0 Pa S 80 0 = = =.5 0 m = 5 0 mm min σ all 0 0 W ightest acceptable wide flange beam: W50 W W W W PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

61 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl S min Section modulus σ = 0 pa all max 8 kn m = = = m σ 0 Pa all = mm Shape S, ( 0 mm ) W W W40 90 W W0 00 W Use W50 9 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

62 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reaction: = 0 : 0 + (7.5)(0) + (5)(40) = 0 = 5kN Shear diagram: to B: V = 5 kn B to : V = 5 0 = 5 kn to : V = 5 40 = 5 kn reas of shear diagram: to B: (.5)(5) =.5 kn m B to : (.5)(5) =.5 kn m to : (5)( 5) = 75 kn m Bending moments: = 0 = =.5 kn m B = = 75 kn m = = 0 = 75 kn m = 75 0 N m max σ all = 0 Pa = 0 0 Pa Shape S x, ( 0 mm ) S S S S 75 0 = = = m min σ all 0 0 = mm ightest S-section: S PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

63 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: By symmetry, B =. F = 0 : 70 + B (9)(45) + 70 = 0 y B = = 7.5kN Shear: V = 70 kn V = = 70 kn B V + = = 0.5 kn B V = 0.5 (9)(45) = 0.5 kn V + = = 70 kn V = 70 kn raw shear diagram. ocate point E where V = 0. E is the midpoint of B. reas of the shear diagram: to B: Vdx = ()( 70) = 0 kn m B to E: Vdx = (4.5)(0.5) = kn m E to : Vdx = (4.5)( 0.5) = kn m to : Vdx = ()(70) = 0 kn m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

64 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5.7 (ontinu) Bending moments: = 0 = 0 0 = 0.5 kn m B = = 45.5 kn E = = 0 kn = = 0 = 45.5kN m = N m max σ all = 0Pa = 0 0 Pa Shape S( 0 mm ) S S S ightest S-shape S σ = S S = = =.55 0 m σ 0 0 = mm PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

65 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl = 0 : ()(.5) (.5)(.5) ()(.5) = 0 =.75 kn m max = =.75 kn m S S I max min σ all min min min c Smin c 4 4 ( c c ) min max max = c I min = (50) (.5 0 ) =.85 0 mm c c max.75 0 N m = = = 45 0 m = 45 0 mm 50 0 Pa I = I = = (50)(45 0 ) =.5 0 mm π = π π = mm t = c c = = 7.0 mm min max t = 9mm PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

66 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions. By symmetry, B = F = 0: B + 0.9w = 0 y Shear: V = 0 B = = 0.45w V = 0 0.w = 0.w B V = 0.w w = 0.5w B + V = 0.5w 0.5w = 0.5w V = 0.5w w = 0.w + V = 0.w 0.w = 0 reas: to B. (0.)( 0. w ) = 0.0 w entroid and moment of inertia: B to E (0.5)(0.5 w ) = 0.05 w Bending moments: = 0 B = 0 0.0w= 0.0w = 0.0w+ 0.05w= 0.05w E Part,mm y,mm y (0 mm ) d, mm. 4 4 d (0 mm ) I (0 mm ) Σ Y 0 0 = = 50 mm I = d + I = 0 0 mm PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

67 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5.8 (ontinued) Top: Bottom: Iy / = (0 0 )/0 = mm = m Iy / = (0 0 ) / ( 50) = 7. 0 mm = 7. 0 m Bending moment limits ( = σ Iy / ) and load limits w. Tension at B and : ompression at B and : Tension at E: ompression at E: 0.0 w = (80 0 )(45. 0 ) 0.0 w = ( 0 0 )(7. 0 ) 0.05 w = (80 0 )(7. 0 ) 0.05 w = ( 0 0)(45. 0 ) w = 8. 0 N/m w = N/m w = N/m w = N/m The smallest allowable load controls: w = N/m w = 7.8 kn/m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

68 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl Reactions: By symmetry, B = Σ F = 0: B+ 0.9w= 0 y B = = 0.45w Shear: V = 0 V = 0 0.w= 0.w B V + = 0. w+ 0.45w= 0.5w B V = 0.5w 0.5w= 0.5w V + = 0.5w+ 0.45w= 0. w V = 0. w 0. w= 0 reas: to B: (0.)( 0. w ) = 0.0 w B to E: (0.5)(0.5 w ) = 0.05 w Bending moments: = 0 B = 0 0.0w= 0.0w = 0.0 w+ 0.05w= 0.05w entroid and moment of inertia: Part,mm y,mm E y,(0 mm) d,mm 4 d (0 mm ) 4 I,(0 mm) Σ Y = = 0mm I =Σ d +Σ I = 0 0 mm 400 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

69 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5.8 (ontinued) Top: Bottom: Iy / = (0 0 ) / (50) = 7. 0 mm = 7. 0 m 8 Iy / = (0 0 ) / ( 0) = mm = m Bending moment limits ( = σ Iy / ) and load limits w. Tension at B and : ompression at B and : Tension at E: ompression at E: 0.0 w = (80 0 )(7. 0 ) 0.0 w = ( 0 0 )( ) 0.05 w = (80 0 )( ) 0.05 w = ( 0 0 )(7. 0 ) w = N/m w = N/m w =.4 0 N/m w = 4. 0 N/m The smallest allowable load controls: w = N/m w = 08.8 kn/m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

70 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl For W40 74, S = 40 0 mm = 40 0 m σ all = 40Pa = 40 0 Pa all = Sσ all = (40 0 )(40 0 ) = N m = 04.4 kn m Reactions: By symmetry, = B, = + Σ F = 0: + B ()() = 0 y = B = 98 kn = 98 0 N +Σ Fy = 0: + l = 0 = = ( l) kn () Shear and bending moment in beam B: 0 < x< a, V = 98 x kn t, x a. Set all. = = max = 98x x kn m = 98a a kn m = a a 98 = 04.4 a 98a = 0 a = 4.75 m,.487 m (a) By geometry, l = a=.5 m l =.5 m From (), = = 0.5 kn raw shear and bending moment diagrams for beam. V = 0 at point E, the midpoint of. PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

71 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5. (ontinued) rea from to E: Vdx = (0.50) l 9.0 kn m = S E = 9.0 kn m = N m E min σ all 40 0 = = =.44 0 m =.44 0 mm Shape S (0 mm ) W W W W W (b) Use W0 44. PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

72 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl B = = 0 VB = V = (7.) w=. w rea B to E of shear diagram: (.)(. w ) =.48 w = 0+.48w=.48w E entroid and moment of inertia: Part (mm ) y (mm) y (mm ) d (mm) Σ d (mm ) I (mm ) Y = =.4 mm I =Σ d +Σ I = 0. 0 mm ocation y (mm) Iy / (0 mm ) Top Bottom also (0 m ) PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

73 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5. (ontinued) Bending moment limits: = σ Iy / Tension at E: (0 0 )( ) = 9. 0 N m ompression at E: ( 50 0 )(58. 0 ) = N m Tension at and : (0 0 )(58. 0 ) = N m ompression at and : (a) llowable load w: ( 50 0 )( ) =. 0 N m.48w= 9. 0 w= N/m w =.485 kn/m Shear at : V = ( a+.) w rea to B of shear diagram: av ( ) + VB = aa ( + 7.) w Bending moment at (also ): (b) istance a: = a ( a+ 7.) w aa ( + 7.)( ) = a + a = a =.95 m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

74 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl B B = = 0 V = V = P rea B to E of shear diagram:.4 P E = P =.4 P = F entroid and moment of inertia: Part (mm ) y (mm) y (mm ) d (mm) Σ d (mm ) I (mm ) Y = =.4 mm I =Σ d +Σ I = 0. 0 mm ocation y (mm) Iy / (0 mm ) Top also (0 m ) Bottom PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

75 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5. (ontinued) Bending moment limits: = σ Iy / Tension at E and F: (0 0 )( ) = 9. 0 N m ompression at E and F: ( 50 0 )(58. 0 ) = N m Tension at and : (0 0 )(58. 0 ) = N m ompression at and : ( 50 0 )( ) =. 0 N m (a) llowable load P: Shear at :.4 P = 9. 0 P = N P = 4.0 kn V = P rea to B of shear diagram: Bending moment at : av = ap = ap = a (b) istance a: a =. 0 a =.7 m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

76 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl ead load: = 48 ft a = 4 ft P = 4 kips P = kips W = 0.75 kip/ft R rea to E of shear diagram: = RB = (48)(0.75) = 8 kips (8)(8) = kip ft max = kip ft = 59 kip in at point E. ap (4)() ive load: u = = =.4 ft ( P + P) ()(0) x= u = 4.4 =. ft x+ a =. + 4 =. ft x a = 48. =.4 ft Σ B = 0: 48 R + (5.4)(4) + (.4)() = 0 R = 4.5 kips PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

77 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl PROBE 5. (ontinued) Shear: to : V = 4.5 kips to : V = = kips to B: V = kips rea: to : (.)(4.5) = 9.5 kip ft Bending moment: = 9.5 kip ft = 8 kip in esign: S γ + γ = ϕ = ϕσ S min U γ + γ = ϕσ U (.5)(59) + (.75)(8) = (0.9)(0) = 84. in U min Shape S (in ) W W7 84 Use W7 84. W W 0 7 W PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

78 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl = 48 ft a = 4 ft P = 4 kips P = kips W = 0.75 kip/ft See solution to Prob for calculation of the following: = 59 kip in = 8 kip in For rolled steel section W7 84, S = in llowable live load moment * : * U * ϕσ U = γ + γ = ϕ = ϕσus S γ γ (0.9)(0)() (.5)(59) =.75 = 47 kip in Ratio: * = 8 = = + Increase.%. PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

79 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl = m, w = 50 N/m = 0.5 kn/m w = 00N/m= 0.kN/m, P = kn ead load: rea to of shear diagram: Bending moment at : ive load: R = ()(0.5) =.8 kn (8)(.8) =. kn m. kn m =. 0 N m R = [()(0.) + ] = 7.8 kn Shear at : V = 7.8 (8)(0.) = kn rea to of shear diagram: Bending moment at : esign: (8)(7.8 + ) = 4. kn m 4. kn m = 4. 0 N m γ + γ = ϕ = ϕσ S U U U γ + γ (.)(. 0 ) + (.)(4. 0 ) S = = ϕσ (0.9)(50 0 ) = m = mm For a rectangular section, S = bh S ()( ) h = = h = 8 mm b 75 PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

80 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl P R = R B = P Σ J = 0: x+ = 0 Px = 0 Px S = = σ σ all all < x < For a rectangular cross section, Equating, S = bh all Px Px bh = h = σ σ b all P x (a) t x =, h= h0 = h= h0, 0 < x< σ b For x >, replace x by x. all (b) Solving for P, allbh0 σ ()(7 0 )(0.05)(0.00) P = = = 0 0 N P = 0 kn ()(0.8) PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

81 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl By symmetry, = B For 0 x, Σ Fy = 0: w0+ B= 0 = B = w0 4 wx 0 dv wx 0 wx 0 w= = w= V = dx t x = 0, V = w0: = w0 4 4 wx 0 wx d = V = w = + w x dx 4 4 t x = 0, = 0: = 0 (a) t x =, For constant strength, For a rectangular section, w ( 4 ) 0 = x w 0 = = 4 w = 0, S S = S = = = = ( x 4 x ) σ σ σ 0 0 all all all S0 0 S h S = bh S0 = bh0 = S0 h0 (b) ata: = 800 mm h0 = 00 mm b= 5 mm σ all = 7 Pa x 4x h= h0 S (5)(00).7 0 mm.7 0 m 0 = bh0 = = = all 0 0 = σ S = (7 0 )(.7 0 ) = 0 N m ()( 0 ) w = = = 5 0 N/m w 0 = 5 kn/m (0.800) PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

82 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl dv wx 0 = w = dx wx 0 d V = = dx wx 0 wx 0 = = wx 0 S = = σ σ all all For a rectangular cross section, Equating, S = bh wx 0 wx 0 bh = h = σ all σ b all t x =, h= h = 0 w 0 all σ b (a) ata: = 750 mm = 0.75 m, b= 0 mm = 0.00 m w 0 = = σ all = = 00 kn/m 00 0 N/m, 00 Pa 00 0 Pa h x h = 0 / (b) h 0 (00 0 )(0.75) = = m (00 0 )(0.00) h 0 = 7.7 mm PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

83 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl dv π x = w= w0 sin dx w 0 π x V = cos + π w 0 V = 0 at x = 0 = π d w 0 π x = V = cos dx π w 0 πx w 0 πx = x sin = x sin π π π π w π x S = = 0 x sin σall πσ all π For a rectangular cross section, Equating, S = bh w π x 0 bh = x sin πσ all π w 0 π h x x = sin πσ allb π / t x =, (a) ata: / 0 w 0 w h= h0 = =.78 πσ allb π σ allb x π x h= h0 sin π π = 750 mm = 0.75 m, b= 0 mm = 0.00 m w 0 = = σ all = = 00 kn/m 00 0 N/m, 00 Pa 00 0 Pa / x h=.59 h0 sin π π x / (b) h 0 (00 0 )(0.75) =.78 = m (00 0 )(0.00) h 0 = 97. mm PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

84 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl R P = RB = 0 < x < P Σ J = 0: x+ = 0 Px max x = or =. Bending moment diagram is two straight lines. t, S = bh = et be the point where the thickness changes. t, max max x S = bh =. 00 mm h 00 mm 4. S h x = = = = = x = 0. m S l =. x = 0.9 l =.800 m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

85 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl 0.8 N R = RB = = 0.4 w Shear: to : V = 0.4 w to B: V = 0.4 w reas: to : (0.8)(0.4) w= 0. w to E: Bending moments: (0.4)(0.4) w= 0.08 w t, = 0.40 w to : = 0.40 wx t, S = bh = max = 0.40w et F be the point were the thickness changes. t F, S bh wx F = F F = 0.40 F F F 00 mm F 0.40 F = = = = = h 00 mm S h wx S w x F l = 0.5 m =. xf = 0.95 m l =.900 m PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

86 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl = = 0 RkN R B x Σ J = 0: 0 x+ (40 x) + = 0 = 0 x 0x kn m t center of beam, x = 4m = 0 kn m t, t center of beam, I = Ibeam + Iplate x = (8 l ) =.5 m = 95 kn m = 0 + (00)(7.5) + + (00)(7.5) 4 = mm 457 c = = mm I S = = mm = m c (a) Normal stress: 0 0 σ = = = 5. 0 Pa S σ = 5. Pa t, S = 40 0 mm = 40 0 m (b) Normal stress: 95 0 σ = = =. 0 Pa S 40 0 σ =. Pa PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

87 F. P. Beer et al., eccanica dei solidi, Elementi di scienza delle costruzioni, 5e - isbn , 04 cgraw-hill Education (Italy) srl R = R B = w x Σ J = 0: wx+ wx + = 0 w = ( x x ) w = x ( x ) where w = 0 kn/m and =. m. For the tapered beam, h= a+ kx a = 0 mm 00 0 k = = mm/m For a rectangular cross section, Bending stress: S = bh = b ( a+ kx ) w x x σ = = S b ( a+ kx) dσ To find location of maximum bending stress, set = 0. dx σ ( + ) ( ) ( )( + ) = = 4 d w d x x w a kx x x x a kx k dx b dx ( a+ kx) b ( a+ kx) w( a+ kx)( x) k( x x ) = b ( a+ kx) wa+ kx ax kx kx + kx = b ( a+ kx) wa ax+ kx = 0 = b ( a+ kx) PROPRIETRY TERI. 0 The cgraw-hill ompanies, Inc. ll rights reserved. No part of this anual may be displayed,

Problem 1: Computation of Reactions. Problem 2: Computation of Reactions. Problem 3: Computation of Reactions

Problem 1: Computation of Reactions. Problem 2: Computation of Reactions. Problem 3: Computation of Reactions Problem 1: Computation of Reactions Problem 2: Computation of Reactions Problem 3: Computation of Reactions Problem 4: Computation of forces and moments Problem 5: Bending Moment and Shear force Problem

More information

Stresses in Beam (Basic Topics)

Stresses in Beam (Basic Topics) Chapter 5 Stresses in Beam (Basic Topics) 5.1 Introduction Beam : loads acting transversely to the longitudinal axis the loads create shear forces and bending moments, stresses and strains due to V and

More information

Mechanics of Materials. Chapter 4 Shear and Moment In Beams

Mechanics of Materials. Chapter 4 Shear and Moment In Beams Mechanics of Materials Chapter 4 Shear and Moment In Beams 4.1 Introduction The term beam refers to a slender bar that carries transverse loading; that is, the applied force are perpendicular to the bar.

More information

Section 16: Neutral Axis and Parallel Axis Theorem 16-1

Section 16: Neutral Axis and Parallel Axis Theorem 16-1 Section 16: Neutral Axis and Parallel Axis Theorem 16-1 Geometry of deformation We will consider the deformation of an ideal, isotropic prismatic beam the cross section is symmetric about y-axis All parts

More information

MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS

MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS This is the second tutorial on bending of beams. You should judge your progress by completing the self assessment exercises.

More information

ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P

ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P This material is duplicated in the Mechanical Principles module H2 and those

More information

Shear Forces and Bending Moments

Shear Forces and Bending Moments Chapter 4 Shear Forces and Bending Moments 4.1 Introduction Consider a beam subjected to transverse loads as shown in figure, the deflections occur in the plane same as the loading plane, is called the

More information

BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL)

BEAMS: SHEAR AND MOMENT DIAGRAMS (GRAPHICAL) LECTURE Third Edition BES: SHER ND OENT DIGRS (GRPHICL). J. Clark School of Engineering Department of Civil and Environmental Engineering 3 Chapter 5.3 by Dr. Ibrahim. ssakkaf SPRING 003 ENES 0 echanics

More information

Introduction to Beam. Area Moments of Inertia, Deflection, and Volumes of Beams

Introduction to Beam. Area Moments of Inertia, Deflection, and Volumes of Beams Introduction to Beam Theory Area Moments of Inertia, Deflection, and Volumes of Beams Horizontal structural member used to support horizontal loads such as floors, roofs, and decks. Types of beam loads

More information

Structural Axial, Shear and Bending Moments

Structural Axial, Shear and Bending Moments Structural Axial, Shear and Bending Moments Positive Internal Forces Acting Recall from mechanics of materials that the internal forces P (generic axial), V (shear) and M (moment) represent resultants

More information

Mechanics of Materials. Chapter 5 Stresses In Beams

Mechanics of Materials. Chapter 5 Stresses In Beams Mechanics of Materials Chapter 5 Stresses In Beams 5.1 Introduction In previous chapters, the stresses in bars caused by axial loading and torsion. Here consider the third fundamental loading : bending.

More information

Method of Joints. Method of Joints. Method of Joints. Method of Joints. Method of Joints. Method of Joints. CIVL 3121 Trusses - Method of Joints 1/5

Method of Joints. Method of Joints. Method of Joints. Method of Joints. Method of Joints. Method of Joints. CIVL 3121 Trusses - Method of Joints 1/5 IVL 3121 Trusses - 1/5 If a truss is in equilibrium, then each of its joints must be in equilibrium. The method of joints consists of satisfying the equilibrium equations for forces acting on each joint.

More information

2. Axial Force, Shear Force, Torque and Bending Moment Diagrams

2. Axial Force, Shear Force, Torque and Bending Moment Diagrams 2. Axial Force, Shear Force, Torque and Bending Moment Diagrams In this section, we learn how to summarize the internal actions (shear force and bending moment) that occur throughout an axial member, shaft,

More information

8.2 Elastic Strain Energy

8.2 Elastic Strain Energy Section 8. 8. Elastic Strain Energy The strain energy stored in an elastic material upon deformation is calculated below for a number of different geometries and loading conditions. These expressions for

More information

Chapter 8. Flexural Analysis of T-Beams

Chapter 8. Flexural Analysis of T-Beams Chapter 8. Flexural Analysis of T-s 8.1. Reading Assignments Text Chapter 3.7; ACI 318, Section 8.10. 8.2. Occurrence and Configuration of T-s Common construction type.- used in conjunction with either

More information

Type of Force 1 Axial (tension / compression) Shear. 3 Bending 4 Torsion 5 Images 6 Symbol (+ -)

Type of Force 1 Axial (tension / compression) Shear. 3 Bending 4 Torsion 5 Images 6 Symbol (+ -) Cause: external force P Force vs. Stress Effect: internal stress f 05 Force vs. Stress Copyright G G Schierle, 2001-05 press Esc to end, for next, for previous slide 1 Type of Force 1 Axial (tension /

More information

MATERIALS AND MECHANICS OF BENDING

MATERIALS AND MECHANICS OF BENDING HAPTER Reinforced oncrete Design Fifth Edition MATERIALS AND MEHANIS OF BENDING A. J. lark School of Engineering Department of ivil and Environmental Engineering Part I oncrete Design and Analysis b FALL

More information

Chapter 1: Statics. A) Newtonian Mechanics B) Relativistic Mechanics

Chapter 1: Statics. A) Newtonian Mechanics B) Relativistic Mechanics Chapter 1: Statics 1. The subject of mechanics deals with what happens to a body when is / are applied to it. A) magnetic field B) heat C ) forces D) neutrons E) lasers 2. still remains the basis of most

More information

BEAMS: SHEAR FLOW, THIN WALLED MEMBERS

BEAMS: SHEAR FLOW, THIN WALLED MEMBERS LECTURE BEAMS: SHEAR FLOW, THN WALLED MEMBERS Third Edition A. J. Clark School of Engineering Department of Civil and Environmental Engineering 15 Chapter 6.6 6.7 by Dr. brahim A. Assakkaf SPRNG 200 ENES

More information

Solid Mechanics. Stress. What you ll learn: Motivation

Solid Mechanics. Stress. What you ll learn: Motivation Solid Mechanics Stress What you ll learn: What is stress? Why stress is important? What are normal and shear stresses? What is strain? Hooke s law (relationship between stress and strain) Stress strain

More information

5 G R A TINGS ENGINEERING DESIGN MANUAL. MBG Metal Bar Grating METAL BAR GRATING MANUAL MBG 534-12 METAL BAR GRATING NAAMM

5 G R A TINGS ENGINEERING DESIGN MANUAL. MBG Metal Bar Grating METAL BAR GRATING MANUAL MBG 534-12 METAL BAR GRATING NAAMM METAL BAR NAAMM GRATNG MANUAL MBG 534-12 5 G R A TNG NAAMM MBG 534-12 November 4, 2012 METAL BAR GRATNG ENGNEERNG DEGN MANUAL NAAMM MBG 534-12 November 4, 2012 5 G R A TNG MBG Metal Bar Grating A Division

More information

FOOTING DESIGN EXAMPLE

FOOTING DESIGN EXAMPLE County: Any Design: BRG Date: 10/007 Hwy: Any Ck Dsn: BRG Date: 10/007 FOOTING DESIGN EXAMPLE Design: Based on AASHTO LRFD 007 Specifications, TxDOT LRFD Bridge Design Manual, and TxDOT Project 0-4371

More information

International Nursing and Rehab Center Addition 4815 S. Western Blvd. Chicago, IL

International Nursing and Rehab Center Addition 4815 S. Western Blvd. Chicago, IL PROJECT International Nursing and Rehab Center Addition 4815 S. Western Blvd. Chicago, IL EXP. 11/30/2014 STRUCTURAL CALCULATIONS July 24, 2014 BOWMAN, BARRETT & ASSOCIATES INC. CONSULTING ENGINEERS 312.228.0100

More information

Lecture 8 Bending & Shear Stresses on Beams

Lecture 8 Bending & Shear Stresses on Beams Lecture 8 Bending & hear tresses on Beams Beams are almost always designed on the asis of ending stress and, to a lesser degree, shear stress. Each of these stresses will e discussed in detail as follows.

More information

MECHANICS OF SOLIDS - BEAMS TUTORIAL TUTORIAL 4 - COMPLEMENTARY SHEAR STRESS

MECHANICS OF SOLIDS - BEAMS TUTORIAL TUTORIAL 4 - COMPLEMENTARY SHEAR STRESS MECHANICS OF SOLIDS - BEAMS TUTORIAL TUTORIAL 4 - COMPLEMENTARY SHEAR STRESS This the fourth and final tutorial on bending of beams. You should judge our progress b completing the self assessment exercises.

More information

The following sketches show the plans of the two cases of one-way slabs. The spanning direction in each case is shown by the double headed arrow.

The following sketches show the plans of the two cases of one-way slabs. The spanning direction in each case is shown by the double headed arrow. 9.2 One-way Slabs This section covers the following topics. Introduction Analysis and Design 9.2.1 Introduction Slabs are an important structural component where prestressing is applied. With increase

More information

Shear and Moment Diagrams. Shear and Moment Diagrams. Shear and Moment Diagrams. Shear and Moment Diagrams. Shear and Moment Diagrams

Shear and Moment Diagrams. Shear and Moment Diagrams. Shear and Moment Diagrams. Shear and Moment Diagrams. Shear and Moment Diagrams CI 3 Shear Force and Bending oment Diagrams /8 If the variation of and are written as functions of position,, and plotted, the resulting graphs are called the shear diagram and the moment diagram. Developing

More information

REVIEW PROBLEMS & SOLUTIONS JAMES M. GERE BARRY J. GOODNO

REVIEW PROBLEMS & SOLUTIONS JAMES M. GERE BARRY J. GOODNO REVIEW ROEMS & SOUTIONS JMES M. GERE RRY J. GOODNO ppendix FE Exam Review roblems R-1.1: plane truss has downward applied load at joint and another load applied leftward at joint 5. The force in member

More information

Structural Design Calculation For Pergola

Structural Design Calculation For Pergola Structural Design Calculation For Pergola Revision :5 Prepared by :EC Date : 8/10/009 CONTENTS 1. Introduction... Design Code and Reference 3. Design Synopsis 4. Design Parameters 4.1 Design Load. 4. Design

More information

SOLUTION 6 6. Determine the force in each member of the truss, and state if the members are in tension or compression.

SOLUTION 6 6. Determine the force in each member of the truss, and state if the members are in tension or compression. 6 6. etermine the force in each member of the truss, and state if the members are in tension or compression. 600 N 4 m Method of Joints: We will begin by analyzing the equilibrium of joint, and then proceed

More information

DESIGN OF SLABS. 3) Based on support or boundary condition: Simply supported, Cantilever slab,

DESIGN OF SLABS. 3) Based on support or boundary condition: Simply supported, Cantilever slab, DESIGN OF SLABS Dr. G. P. Chandradhara Professor of Civil Engineering S. J. College of Engineering Mysore 1. GENERAL A slab is a flat two dimensional planar structural element having thickness small compared

More information

Module 5 (Lectures 17 to 19) MAT FOUNDATIONS

Module 5 (Lectures 17 to 19) MAT FOUNDATIONS Module 5 (Lectures 17 to 19) MAT FOUNDATIONS Topics 17.1 INTRODUCTION Rectangular Combined Footing: Trapezoidal Combined Footings: Cantilever Footing: Mat foundation: 17.2 COMMON TYPES OF MAT FOUNDATIONS

More information

Recitation #5. Understanding Shear Force and Bending Moment Diagrams

Recitation #5. Understanding Shear Force and Bending Moment Diagrams Recitation #5 Understanding Shear Force and Bending Moment Diagrams Shear force and bending moment are examples of interanl forces that are induced in a structure when loads are applied to that structure.

More information

MECHANICS OF SOLIDS - BEAMS TUTORIAL 1 STRESSES IN BEAMS DUE TO BENDING. On completion of this tutorial you should be able to do the following.

MECHANICS OF SOLIDS - BEAMS TUTORIAL 1 STRESSES IN BEAMS DUE TO BENDING. On completion of this tutorial you should be able to do the following. MECHANICS OF SOLIDS - BEAMS TUTOIAL 1 STESSES IN BEAMS DUE TO BENDING This is the first tutorial on bending of beams designed for anyone wishing to study it at a fairly advanced level. You should judge

More information

Chapter - 3 Design of Rectangular Beams and One-way Slabs

Chapter - 3 Design of Rectangular Beams and One-way Slabs Rectangular Beams and One-way Slabs Page 1 of 9 Chapter - 3 Design of Rectangular Beams and One-way Slabs 12 h A 12 strip in a simply supported one-way slab h b=12 L Rectangular Beams and One-way Slabs

More information

Chapter 8. Shear Force and Bending Moment Diagrams for Uniformly Distributed Loads.

Chapter 8. Shear Force and Bending Moment Diagrams for Uniformly Distributed Loads. hapter 8 Shear Force and ending Moment Diagrams for Uniformly Distributed Loads. 8.1 Introduction In Unit 4 we saw how to calculate moments for uniformly distributed loads. You might find it worthwhile

More information

Design of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar

Design of Steel Structures Prof. S.R.Satish Kumar and Prof. A.R.Santha Kumar Problem 1 Design a hand operated overhead crane, which is provided in a shed, whose details are: Capacity of crane = 50 kn Longitudinal spacing of column = 6m Center to center distance of gantry girder

More information

Page 1 of 18 28.4.2008 Sven Alexander Last revised 1.3.2010. SB-Produksjon STATICAL CALCULATIONS FOR BCC 250

Page 1 of 18 28.4.2008 Sven Alexander Last revised 1.3.2010. SB-Produksjon STATICAL CALCULATIONS FOR BCC 250 Page 1 of 18 CONTENT PART 1 BASIC ASSUMPTIONS PAGE 1.1 General 1. Standards 1.3 Loads 1. Qualities PART ANCHORAGE OF THE UNITS.1 Beam unit equilibrium 3. Beam unit anchorage in front..1 Check of capacity..

More information

m i: is the mass of each particle

m i: is the mass of each particle Center of Mass (CM): The center of mass is a point which locates the resultant mass of a system of particles or body. It can be within the object (like a human standing straight) or outside the object

More information

SOLUTIONS TO CONCEPTS CHAPTER 15

SOLUTIONS TO CONCEPTS CHAPTER 15 SOLUTIONS TO CONCEPTS CHAPTER 15 1. v = 40 cm/sec As velocity of a wave is constant location of maximum after 5 sec = 40 5 = 00 cm along negative x-axis. [(x / a) (t / T)]. Given y = Ae a) [A] = [M 0 L

More information

Two-Way Post-Tensioned Design

Two-Way Post-Tensioned Design Page 1 of 9 The following example illustrates the design methods presented in ACI 318-05 and IBC 2003. Unless otherwise noted, all referenced table, figure, and equation numbers are from these books. The

More information

Lab for Deflection and Moment of Inertia

Lab for Deflection and Moment of Inertia Deflection and Moment of Inertia Subject Area(s) Associated Unit Lesson Title Physics Wind Effects on Model Building Lab for Deflection and Moment of Inertia Grade Level (11-12) Part # 2 of 3 Lesson #

More information

HØGSKOLEN I GJØVIK Avdeling for teknologi, økonomi og ledelse. Løsningsforslag for kontinuasjonseksamen i Mekanikk 4/1-10

HØGSKOLEN I GJØVIK Avdeling for teknologi, økonomi og ledelse. Løsningsforslag for kontinuasjonseksamen i Mekanikk 4/1-10 Løsningsforslag for kontinuasjonseksamen i 4/1-10 Oppgave 1 (T betyr tension, dvs. strekk, og C betyr compression, dvs. trykk.) Side 1 av 9 Leif Erik Storm Oppgave 2 Løsning (fra http://www.public.iastate.edu/~statics/examples/vmdiags/vmdiaga.html

More information

A Case Study Comparing Two Approaches for Applying Area Loads: Tributary Area Loads vs Shell Pressure Loads

A Case Study Comparing Two Approaches for Applying Area Loads: Tributary Area Loads vs Shell Pressure Loads 1 A Case Study Comparing Two Approaches for Applying Area Loads: Tributary Area Loads vs Shell Pressure Loads By Dr. Siriwut Sasibut (Application Engineer) S-FRAME Software Inc. #1158 13351 Commerce Parkway

More information

Analysis of Stresses and Strains

Analysis of Stresses and Strains Chapter 7 Analysis of Stresses and Strains 7.1 Introduction axial load = P / A torsional load in circular shaft = T / I p bending moment and shear force in beam = M y / I = V Q / I b in this chapter, we

More information

Technical Notes 3B - Brick Masonry Section Properties May 1993

Technical Notes 3B - Brick Masonry Section Properties May 1993 Technical Notes 3B - Brick Masonry Section Properties May 1993 Abstract: This Technical Notes is a design aid for the Building Code Requirements for Masonry Structures (ACI 530/ASCE 5/TMS 402-92) and Specifications

More information

CLASSICAL STRUCTURAL ANALYSIS

CLASSICAL STRUCTURAL ANALYSIS Table of Contents CASSCA STRUCTURA ANAYSS... Conjugate beam method... External work and internal work... 3 Method of virtual force (unit load method)... 5 Castigliano s second theorem... Method of consistent

More information

Design Of Reinforced Concrete Structures ii Two-Way Slabs

Design Of Reinforced Concrete Structures ii Two-Way Slabs 1. Inroduction When the ratio (L/S) is less than 2.0, slab is called two-way slab, as shown in the fig. below. Bending will take place in the two directions in a dish-like form. Accordingly, main reinforcement

More information

Today s Objective COMPOSITE BODIES

Today s Objective COMPOSITE BODIES Today s Objective: Students will be able to determine: a) The location of the center of gravity, b) The location of the center of mass, c) And, the location of the centroid using the method of composite

More information

p atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh

p atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh IVE1400: n Introduction to Fluid Mechanics Statics : Pressure : Statics r P Sleigh: P..Sleigh@leeds.ac.uk r J Noakes:.J.Noakes@leeds.ac.uk January 008 Module web site: www.efm.leeds.ac.uk/ive/fluidslevel1

More information

OUTCOME 1 STATIC FLUID SYSTEMS TUTORIAL 1 - HYDROSTATICS

OUTCOME 1 STATIC FLUID SYSTEMS TUTORIAL 1 - HYDROSTATICS Unit 41: Fluid Mechanics Unit code: T/601/1445 QCF Level: 4 Credit value: 15 OUTCOME 1 STATIC FLUID SYSTEMS TUTORIAL 1 - HYDROSTATICS 1. Be able to determine the behavioural characteristics and parameters

More information

STRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION

STRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION Chapter 11 STRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION Figure 11.1: In Chapter10, the equilibrium, kinematic and constitutive equations for a general three-dimensional solid deformable

More information

4.2 Free Body Diagrams

4.2 Free Body Diagrams CE297-FA09-Ch4 Page 1 Friday, September 18, 2009 12:11 AM Chapter 4: Equilibrium of Rigid Bodies A (rigid) body is said to in equilibrium if the vector sum of ALL forces and all their moments taken about

More information

When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid.

When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Fluid Statics When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Consider a small wedge of fluid at rest of size Δx, Δz, Δs

More information

CENTER OF GRAVITY, CENTER OF MASS AND CENTROID OF A BODY

CENTER OF GRAVITY, CENTER OF MASS AND CENTROID OF A BODY CENTER OF GRAVITY, CENTER OF MASS AND CENTROID OF A BODY Dr. Amilcar Rincon-Charris, MSME Mechanical Engineering Department MECN 3005 - STATICS Objective : Students will: a) Understand the concepts of

More information

Deflections. Question: What are Structural Deflections?

Deflections. Question: What are Structural Deflections? Question: What are Structural Deflections? Answer: The deformations or movements of a structure and its components, such as beams and trusses, from their original positions. It is as important for the

More information

Statics of Structural Supports

Statics of Structural Supports Statics of Structural Supports TYPES OF FORCES External Forces actions of other bodies on the structure under consideration. Internal Forces forces and couples exerted on a member or portion of the structure

More information

The elements used in commercial codes can be classified in two basic categories:

The elements used in commercial codes can be classified in two basic categories: CHAPTER 3 Truss Element 3.1 Introduction The single most important concept in understanding FEA, is the basic understanding of various finite elements that we employ in an analysis. Elements are used for

More information

Design of reinforced concrete columns. Type of columns. Failure of reinforced concrete columns. Short column. Long column

Design of reinforced concrete columns. Type of columns. Failure of reinforced concrete columns. Short column. Long column Design of reinforced concrete columns Type of columns Failure of reinforced concrete columns Short column Column fails in concrete crushed and bursting. Outward pressure break horizontal ties and bend

More information

B.TECH. (AEROSPACE ENGINEERING) PROGRAMME (BTAE) Term-End Examination December, 2011 BAS-010 : MACHINE DESIGN

B.TECH. (AEROSPACE ENGINEERING) PROGRAMME (BTAE) Term-End Examination December, 2011 BAS-010 : MACHINE DESIGN No. of Printed Pages : 7 BAS-01.0 B.TECH. (AEROSPACE ENGINEERING) PROGRAMME (BTAE) CV CA CV C:) O Term-End Examination December, 2011 BAS-010 : MACHINE DESIGN Time : 3 hours Maximum Marks : 70 Note : (1)

More information

New approaches in Eurocode 3 efficient global structural design

New approaches in Eurocode 3 efficient global structural design New approaches in Eurocode 3 efficient global structural design Part 1: 3D model based analysis using general beam-column FEM Ferenc Papp* and József Szalai ** * Associate Professor, Department of Structural

More information

SECTION 5 ANALYSIS OF CONTINUOUS SPANS DEVELOPED BY THE PTI EDC-130 EDUCATION COMMITTEE LEAD AUTHOR: BRYAN ALLRED

SECTION 5 ANALYSIS OF CONTINUOUS SPANS DEVELOPED BY THE PTI EDC-130 EDUCATION COMMITTEE LEAD AUTHOR: BRYAN ALLRED SECTION 5 ANALYSIS OF CONTINUOUS SPANS DEVELOPED BY THE PTI EDC-130 EDUCATION COMMITTEE LEAD AUTHOR: BRYAN ALLRED NOTE: MOMENT DIAGRAM CONVENTION In PT design, it is preferable to draw moment diagrams

More information

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS

EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS EDEXCEL NATIONAL CERTIICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQ LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS 1. Be able to determine the effects of loading in static engineering

More information

A transverse strip of the deck is assumed to support the truck axle loads. Shear and fatigue of the reinforcement need not be investigated.

A transverse strip of the deck is assumed to support the truck axle loads. Shear and fatigue of the reinforcement need not be investigated. Design Step 4 Design Step 4.1 DECK SLAB DESIGN In addition to designing the deck for dead and live loads at the strength limit state, the AASHTO-LRFD specifications require checking the deck for vehicular

More information

COMPLEX STRESS TUTORIAL 3 COMPLEX STRESS AND STRAIN

COMPLEX STRESS TUTORIAL 3 COMPLEX STRESS AND STRAIN COMPLX STRSS TUTORIAL COMPLX STRSS AND STRAIN This tutorial is not part of the decel unit mechanical Principles but covers elements of the following sllabi. o Parts of the ngineering Council eam subject

More information

COMPUTATIONAL ENGINEERING OF FINITE ELEMENT MODELLING FOR AUTOMOTIVE APPLICATION USING ABAQUS

COMPUTATIONAL ENGINEERING OF FINITE ELEMENT MODELLING FOR AUTOMOTIVE APPLICATION USING ABAQUS International Journal of Advanced Research in Engineering and Technology (IJARET) Volume 7, Issue 2, March-April 2016, pp. 30 52, Article ID: IJARET_07_02_004 Available online at http://www.iaeme.com/ijaret/issues.asp?jtype=ijaret&vtype=7&itype=2

More information

MECHANICS OF MATERIALS

MECHANICS OF MATERIALS T dition CHTR MCHNICS OF MTRIS Ferdinand. Beer. Russell Johnston, Jr. John T. DeWolf ecture Notes: J. Walt Oler Texas Tech University Stress and Strain xial oading - Contents Stress & Strain: xial oading

More information

L4. Pipes Materials and Loads

L4. Pipes Materials and Loads The Islamic University of Gaza- Civil Engineering Department SanitaryEngineering- ECIV 4325 L4. Pipes Materials an Loas Base on Dr. Fahi Rabah lecture notes Pipes Materials an loas Materials use for pipes:

More information

PROBLEM 4.1 SOLUTION. Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A, (b) point B.

PROBLEM 4.1 SOLUTION. Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A, (b) point B. PROBLEM.1 Knowing tht the couple shown cts in verticl plne, determine the stress t () point A, (b) point B. SOLUTON () (b) For rectngle: For cross sectionl re: 1 = bh 1 1 = 1 + + = ()(1.5) + ()(5.5) +

More information

GUARDRAIL INNOVATION FLOWS FROM HERE

GUARDRAIL INNOVATION FLOWS FROM HERE INNOVATION FLOWS FROM HERE ENSURING SAFETY ON CANADIAN HIGHWAYS The economics, versatility, and performance of Guardrail surpass all other roadside barrier systems. Guardrail provides highly-visible protection

More information

Activity 2.3b Engineering Problem Solving Answer Key

Activity 2.3b Engineering Problem Solving Answer Key Activity.3b Engineering roblem Solving Answer Key 1. A force of 00 lbs pushes against a rectangular plate that is 1 ft. by ft. Determine the lb lb pressure in and that the plate exerts on the ground due

More information

Thank you for joining our live webinar today. We will begin shortly. Please standby. Thank you. Need Help? Call ReadyTalk Support: 800.843.

Thank you for joining our live webinar today. We will begin shortly. Please standby. Thank you. Need Help? Call ReadyTalk Support: 800.843. Thank you for joining our live webinar today. We will begin shortly. Please standby. Thank you. Need Help? Call ReadyTalk Support: 800.843.9166 Today s audio will be broadcast through the internet. Alternatively,

More information

ABSTRACT 1. INTRODUCTION 2. DESCRIPTION OF THE SEGMENTAL BEAM

ABSTRACT 1. INTRODUCTION 2. DESCRIPTION OF THE SEGMENTAL BEAM Ninth LACCEI Latin American and Caribbean Conference (LACCEI 11), Engineering for a Smart Planet, Innovation, Information Technology and Computational Tools for Sustainable Development, August 3-, 11,

More information

Rigid and Braced Frames

Rigid and Braced Frames Rigid Frames Rigid and raced Frames Rigid frames are identified b the lack of pinned joints within the frame. The joints are rigid and resist rotation. The ma be supported b pins or fied supports. The

More information

PLQE. The base of the economy gearboxes with square output flange. Economy Line. Formerly PLE

PLQE. The base of the economy gearboxes with square output flange. Economy Line. Formerly PLE 20 ormerly PL The base of the economy gearboxes with square output flange The PL series with square output flange. A powerful alternative for additional higher radial and axial loads. Low backlash High

More information

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur

Module 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur Module Analysis of Statically Indeterminate Structures by the Matrix Force Method esson 11 The Force Method of Analysis: Frames Instructional Objectives After reading this chapter the student will be able

More information

Copyright 2015 Pecivilexam.com all rights reserved- Breadth Exam 120 solved Problems. www.pecivilexam.com

Copyright 2015 Pecivilexam.com all rights reserved- Breadth Exam 120 solved Problems. www.pecivilexam.com www.pecivilexam.com 2015 PE Exam Standard Specification PE Civil Exam E-Book120-Mix Questions & Answers (pdf Format) For Breath Exam (Morning Session) Breadth Exam (morning session): These practice exams

More information

INTRODUCTION TO BEAMS

INTRODUCTION TO BEAMS CHAPTER Structural Steel Design LRFD Method INTRODUCTION TO BEAMS Third Edition A. J. Clark School of Engineering Department of Civil and Environmental Engineering Part II Structural Steel Design and Analysis

More information

INTRODUCTION TO BEAMS

INTRODUCTION TO BEAMS CHAPTER Structural Steel Design LRFD etho INTRODUCTION TO BEAS Thir Eition A. J. Clark School of Engineering Department of Civil an Environmental Engineering Part II Structural Steel Design an Analsis

More information

Shear Center in Thin-Walled Beams Lab

Shear Center in Thin-Walled Beams Lab Shear Center in Thin-Walled Beams Lab Shear flow is developed in beams with thin-walled cross sections shear flow (q sx ): shear force per unit length along cross section q sx =τ sx t behaves much like

More information

DESIGN OF SLABS. Department of Structures and Materials Engineering Faculty of Civil and Environmental Engineering University Tun Hussein Onn Malaysia

DESIGN OF SLABS. Department of Structures and Materials Engineering Faculty of Civil and Environmental Engineering University Tun Hussein Onn Malaysia DESIGN OF SLABS Department of Structures and Materials Engineering Faculty of Civil and Environmental Engineering University Tun Hussein Onn Malaysia Introduction Types of Slab Slabs are plate elements

More information

Introduction to Railroad Track Structural Design

Introduction to Railroad Track Structural Design BCR2A 09 Railroad Track Design Including Asphalt Trackbeds Pre-Conference Workshop Introduction to Railroad Track Structural Design Don Uzarski, Ph.D., P.E. uzarski@illinois.edu Interaction, Vertical Load

More information

Bending Stress in Beams

Bending Stress in Beams 936-73-600 Bending Stress in Beams Derive a relationship for bending stress in a beam: Basic Assumptions:. Deflections are very small with respect to the depth of the beam. Plane sections before bending

More information

Introduction to Mechanical Behavior of Biological Materials

Introduction to Mechanical Behavior of Biological Materials Introduction to Mechanical Behavior of Biological Materials Ozkaya and Nordin Chapter 7, pages 127-151 Chapter 8, pages 173-194 Outline Modes of loading Internal forces and moments Stiffness of a structure

More information

EUROPEAN ORGANISATION FOR TECHNICAL APPROVALS

EUROPEAN ORGANISATION FOR TECHNICAL APPROVALS E TA TECHNICAL REPORT Design of Bonded Anchors TR 29 Edition June 27 EUROPEAN ORGANISATION FOR TECHNICAL APPROVALS TABLE OF CONTENTS Design method for bonded anchors Introduction..4 1 Scope...2 1.1 Type

More information

Modeling Beams on Elastic Foundations Using Plate Elements in Finite Element Method

Modeling Beams on Elastic Foundations Using Plate Elements in Finite Element Method Modeling Beams on Elastic Foundations Using Plate Elements in Finite Element Method Yun-gang Zhan School of Naval Architecture and Ocean Engineering, Jiangsu University of Science and Technology, Zhenjiang,

More information

Incorporating Innovative Materials for Seismic Resilient Bridge Columns

Incorporating Innovative Materials for Seismic Resilient Bridge Columns Incorporating Innovative Materials for Seismic Resilient Bridge Columns WSDOT Including Contributions from: Dr. M. Saiid Saiidi University Nevada, Reno Brain Nakashoji University Nevada, Reno Presentation

More information

DYNAMIC ANALYSIS OF THICK PLATES SUBJECTED TO EARTQUAKE

DYNAMIC ANALYSIS OF THICK PLATES SUBJECTED TO EARTQUAKE DYNAMIC ANALYSIS OF THICK PLATES SUBJECTED TO EARTQUAKE ÖZDEMİR Y. I, AYVAZ Y. Posta Adresi: Department of Civil Engineering, Karadeniz Technical University, 68 Trabzon, TURKEY E-posta: yaprakozdemir@hotmail.com

More information

DIRECT SHEAR TEST SOIL MECHANICS SOIL MECHANICS LABORATORY DEPARTMENT OF CIVIL ENGINEERING UNIVERSITY OF MORATUWA SRI LANKA

DIRECT SHEAR TEST SOIL MECHANICS SOIL MECHANICS LABORATORY DEPARTMENT OF CIVIL ENGINEERING UNIVERSITY OF MORATUWA SRI LANKA DIRECT SHEAR TEST SOIL MECHANICS SOIL MECHANICS LABORATORY DEPARTMENT OF CIVIL ENGINEERING UNIVERSITY OF MORATUWA SRI LANKA DIRECT SHEAR TEST OBJEVTIVES To determine the shear strength parameters for a

More information

CHAPTER 13 CONCRETE COLUMNS

CHAPTER 13 CONCRETE COLUMNS CHAER 13 CONCREE COUMNS ABE OF CONENS 13.1 INRODUCION... 13-1 13.2 YES OF COUMNS... 13-1 13.3 DESIGN OADS... 13-1 13.4 DESIGN CRIERIA... 13-2 13.4.1 imit States... 13-2 13.4.2 Forces... 13-2 13.5 AROXIMAE

More information

ETABS. Integrated Building Design Software. Concrete Shear Wall Design Manual. Computers and Structures, Inc. Berkeley, California, USA

ETABS. Integrated Building Design Software. Concrete Shear Wall Design Manual. Computers and Structures, Inc. Berkeley, California, USA ETABS Integrated Building Design Software Concrete Shear Wall Design Manual Computers and Structures, Inc. Berkeley, California, USA Version 8 January 2002 Copyright The computer program ETABS and all

More information

CHAPTER 15 FORCE, MASS AND ACCELERATION

CHAPTER 15 FORCE, MASS AND ACCELERATION CHAPTER 5 FORCE, MASS AND ACCELERATION EXERCISE 83, Page 9. A car initially at rest accelerates uniformly to a speed of 55 km/h in 4 s. Determine the accelerating force required if the mass of the car

More information

MCB4UW Optimization Problems Handout 4.6

MCB4UW Optimization Problems Handout 4.6 MCB4UW Optimization Problems Handout 4.6 1. A rectangular field along a straight river is to be divided into smaller fields by one fence parallel to the river and 4 fences perpendicular to the river. Find

More information

MCE380: Measurements and Instrumentation Lab. Chapter 9: Force, Torque and Strain Measurements

MCE380: Measurements and Instrumentation Lab. Chapter 9: Force, Torque and Strain Measurements MCE380: Measurements and Instrumentation Lab Chapter 9: Force, Torque and Strain Measurements Topics: Elastic Elements for Force Measurement Dynamometers and Brakes Resistance Strain Gages Holman, Ch.

More information

CE 201 (STATICS) DR. SHAMSHAD AHMAD CIVIL ENGINEERING ENGINEERING MECHANICS-STATICS

CE 201 (STATICS) DR. SHAMSHAD AHMAD CIVIL ENGINEERING ENGINEERING MECHANICS-STATICS COURSE: CE 201 (STATICS) LECTURE NO.: 28 to 30 FACULTY: DR. SHAMSHAD AHMAD DEPARTMENT: CIVIL ENGINEERING UNIVERSITY: KING FAHD UNIVERSITY OF PETROLEUM & MINERALS, DHAHRAN, SAUDI ARABIA TEXT BOOK: ENGINEERING

More information

AISI CHEMICAL COMPOSITION LIMITS: Nonresulphurized Carbon Steels

AISI CHEMICAL COMPOSITION LIMITS: Nonresulphurized Carbon Steels AISI CHEMICAL COMPOSITION LIMITS: Nonresulphurized Carbon Steels AISI No. 1008 1010 1012 1015 1016 1017 1018 1019 1020 1021 1022 1023 1024 10 1026 1027 1029 10 1035 1036 1037 1038 1039 10 1041 1042 1043

More information

Laterally Loaded Piles

Laterally Loaded Piles Laterally Loaded Piles 1 Soil Response Modelled by p-y Curves In order to properly analyze a laterally loaded pile foundation in soil/rock, a nonlinear relationship needs to be applied that provides soil

More information

FOUNDATION DESIGN. Instructional Materials Complementing FEMA 451, Design Examples

FOUNDATION DESIGN. Instructional Materials Complementing FEMA 451, Design Examples FOUNDATION DESIGN Proportioning elements for: Transfer of seismic forces Strength and stiffness Shallow and deep foundations Elastic and plastic analysis Foundation Design 14-1 Load Path and Transfer to

More information

Solutions Manual. Failure, Fracture, Fatigue. An Introduction

Solutions Manual. Failure, Fracture, Fatigue. An Introduction Solutions Manual to problems in Failure, Fracture, Fatigue An Introduction by Tore Dahlberg Anders Ekberg Studentlitteratur, Lund 2002, ISBN 91-44-02096-1. This manual contains solutions to problems in

More information