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

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1 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 = = ()(1.5) + ()(5.5) + ()(1.5) = 8.85 in 1 y A = y B =.75 in in. MyA (5)(.75) σ A = = σ A =.8 ksi 8.85 MyB (5)(0.75) σ B = = σ B = 0.50 ksi 8.85 PROPRETARY MATERAL. 01 The McGrw-Hill Compnies, nc. All rights reserved. No prt of this Mnul my be displyed,

2 PROBLEM.7 Two W 1 rolled sections re welded together s shown. Knowing tht for the steel lloy used σ Y = ksi nd σ U = 58 ksi nd using fctor of sfety of.0, determine the lrgest couple tht cn be pplied when the ssembly is bent bout the z xis. SOLUTON Properties of W 1 rolled section. (See Appendix C.) Are =.8 in Depth =.1 in. x = 11. in For one rolled section, moment of inerti bout xis - is = + Ad = (.8)(.08) = 7.87 in x For both sections, z = = 55.7 in c = depth =.1 in. σ M ll ll σu 58 Mc = = = 19. ksi σ = F. S..0 σ ll (19.) (55.7) = = c.1 M = 59 kip in ll PROPRETARY MATERAL. 01 The McGrw-Hill Compnies, nc. All rights reserved. No prt of this Mnul my be displyed,

3 PROBLEM.9 Two verticl forces re pplied to bem of the cross section shown. Determine the mximum tensile nd compressive stresses in portion BC of the bem. SOLUTON r ()(5) A1 = π r (5) mm mm = π = y = π = π = h 5 A = bh = (50)(5) = 150 mm y = = = 1.5 mm Ay + Ay (981.7)(10.10) + (150)( 1.5) y = = =. mm A + A y y 1 π π A y r A y 8 8 d = y y = (.) = 1.9 mm 1 = x 1 = = (5) (981.7)(10.10) = mm 1 = 1+ = (981.7)(1.9) = mm Ad = bh = (50)(5) = mm d = y y = 1.5 (.) = 10.1 mm = + = (150)(10.1) = mm A d top bot 1 9 = + = mm = m = 5 +. = 7. mm = 0.07 m = 5 +. =. mm = 0.0 m M P = 0: M = P = ( 10)(00 10 ) = 100 N m σ σ My (100)(0.07) top top = = = bot bot P My (100)( 0.0) = = = P σ = 81.8 MP top σ = 7.8 MP bot PROPRETARY MATERAL. 01 The McGrw-Hill Compnies, nc. All rights reserved. No prt of this Mnul my be displyed,

4 PROBLEM.1 Knowing tht bem of the cross section shown is bent bout horizontl xis nd tht the bending moment is 50 kip in., determine the totl force cting () on the top flnge, (b) on the shded portion of the web. SOLUTON The stress distribution over the entire cross-section is given by the bending stress formul: where σ = where y is coordinte with its origin on the neutrl xis nd is the moment of inerti of the entire cross sectionl re. The force on the shded portion is clculted from this stress distribution. Over n re element da, the force is My df = σ xda = da The totl force on the shded re is then My M M F = df = da = yda = y A * y is the centroidl coordinte of the shded portion nd A * is its re. Clculte the moment of inerti. * * x My M = ( in.)(7 in.) ( in.)( in.) = in = 50 kip in () Top flnge: A* = ( in.)(1.5 in.) = 9 in y* = in in. =.75 in. F 50 kip in (9 in )(.75 in.) 8. kips = = F = 8. kips in (b) Hlf web: A* = ( in.)( in.) = in y* = 1 in. F 50 kip in ( in )(1 in.) 1. kips = = F = 1. kips in PROPRETARY MATERAL. 01 The McGrw-Hill Compnies, nc. All rights reserved. No prt of this Mnul my be displyed,

5 PROBLEM. A 0 N m couple is pplied to the steel br shown. () Assuming tht the couple is pplied bout the z xis s shown, determine the mximum stress nd the rdius of curvture of the br. (b) Solve prt, ssuming tht the couple is pplied bout the y xis. Use E = 00 GP. SOLUTON () Bending bout z-xis. = bh = (1)(0) = 8 10 mm = 8 10 m 0 c = = 10 mm = m 9 Mc (0)(0.010) σ = = = P M 0 = = = m 9 9 ρ E (00 10 )(8 10 ) 1 σ = 75.0 MP ρ =.7 m (b) Bending bout y-xis. = bh = (0)(1) = mm = m 1 c = = mm = 0.00 m Mc (0)(0.00) σ = = = P M 0 = = = m 9 9 ρ E (00 10 )( ) 1 σ = 15.0 MP ρ = 9.0 m PROPRETARY MATERAL. 01 The McGrw-Hill Compnies, nc. All rights reserved. No prt of this Mnul my be displyed,

6 PROBLEM. A br hving the cross section shown hs been formed by securely bonding brss nd luminum stock. Using the dt given below, determine the lrgest permissible bending moment when the composite br is bent bout horizontl xis. Aluminum Brss Modulus of elsticity 70 GP 105 GP Allowble stress 100 MP 10 MP SOLUTON Use luminum s the reference mteril. For luminum, n = 1.0 For brss, n = E / E = 105/70 = 1.5 b Vlues of n re shown on the sketch. For the trnsformed section, n = bh = (8)() = mm n 1.0 b ( H h ) ()( 1 ) mm = = = 1 1 = = mm 9 = + + = mm = m nmy σ σ = M = ny Aluminum: n= 1.0, y = 1 mm = 0.01 m, σ = P 9 ( )( ) M = = N m (1.0)(0.01) Brss: n= 1.5, y = 1 mm = 0.01 m, σ = P Choose the smller vlue. 9 (10 10 )( ) M = = 9. N m (1.5)(0.01) M = 887 N m PROPRETARY MATERAL. 01 The McGrw-Hill Compnies, nc. All rights reserved. No prt of this Mnul my be displyed,

7 PROBLEM.1 The 1-in. timber bem hs been strengthened by bolting to it the steel reinforcement shown. The modulus of elsticity for wood is psi nd for steel, 9 10 psi. Knowing tht the bem is bent bout horizontl xis by couple of moment M = 50 kip in., determine the mximum stress in () the wood, (b) the steel. SOLUTON Use wood s the reference mteril. For wood, n = 1 For steel, n= E / E = 9 /1.8 = Trnsformed section: = wood = steel s w 1.91 Y o = =.758 in. A, in na, in y o 7 7 nay o, in The neutrl xis lies.758 in. bove the wood-steel interfce. n1 1 1 = bh + n1a1d1 = ()(1) + (7)(.758) = in n b h n A d (5)(0.5) (0.78)( ) 7.87 in = + = + + = = + = in 1 nmy M = 50 kip in σ = () Wood: n= 1, y = = 8. in (1) (50) (8.) σ w = = ksi σ w = ksi (b) Steel: n= , y = =.58 in σ s (1.1111) (50) (.58) = = 1.8 ksi σ s = 1.8 ksi PROPRETARY MATERAL. 01 The McGrw-Hill Compnies, nc. All rights reserved. No prt of this Mnul my be displyed,

8 PROBLEM. The 1-in. timber bem hs been strengthened by bolting to it the steel reinforcement shown. The modulus of elsticity for wood is psi nd for steel, 9 10 psi. Knowing tht the bem is bent bout horizontl xis by couple of moment M = 50 kip in., determine the mximum stress in () the wood, (b) the steel. SOLUTON Use wood s the reference mteril. For wood, n = 1 Es 9 10 For steel, n = = = E For C chnnel section, w w y A=.8 in, t = 0.0 in., x = in., = 1. in For the composite section, the centroid of the chnnel (prt 1) lies in. bove the bottom of the section. The centroid of the wood (prt ) lies =. in. bove the bottom. Trnsformed section: Prt A, in na, in y, in in Y 0 = =.787 in. d = y 0 Y0 1.5 in nay, in d, in Σ The neutrl xis lies.787 in. bove the bottom of the section. 1 = + 1 = (1.1111)(1.) + (5.5)(.1) = 58.9 in n n Ad n 1 b h n A d ()(1) (7)(.) in = + = + = = + = in 1 M = 50 kip in nmy σ = () Wood: n= 1, y = = 8. in. (1)(50)(8.) σ w = =.0 ksi σ w =.0 ksi (b) Steel: n= , y =.787 in. σ s (1.1111) (50) (.787) = = 1.5 ksi σ s = 1.5 ksi PROPRETARY MATERAL. 01 The McGrw-Hill Compnies, nc. All rights reserved. No prt of this Mnul my be displyed,

9 PROBLEM.51 A concrete bem is reinforced by three steel rods plced s shown. The modulus of elsticity is 10 psi for the concrete nd 9 10 psi for the steel. Using n llowble stress of 150 psi for the concrete nd 0 ksi for the steel, determine the lrgest llowble positive bending moment in the bem. SOLUTON Locte the neutrl xis: Es 9 10 n = = = 9.7 E 10 c π 7 s π As = d = () in na 17.8 in = = 8 x 8 x (17.8)(1 x) = 0 x x.1= 0 Solve for x ()()(.1) x = = 5. in. ()() 1 x = 8.7 in. = 8 x + na (1 x) = (8)(5.) + (17.8)(8.7) = in s nmy σ σ = M = ny Concrete: n= 1.0, y = 5. in., σ = 150 psi (150)(197.5) M = = lb in = 07 kip in (1.0)(5.) Steel: n= 9.7, y = 8.7 in., σ = 0 10 psi Choose the smller vlue. M (0 10 )(197.5) = = 19.7 lb in = 0 kip in (9.7)(8.7) M = 07 kip in M =.9 kip ft PROPRETARY MATERAL. 01 The McGrw-Hill Compnies, nc. All rights reserved. No prt of this Mnul my be displyed,

10 PROBLEM.5 Knowing tht the bending moment in the reinforced concrete bem is +100 kip ft nd tht the modulus of elsticity is.5 10 psi for the concrete nd 9 10 psi for the steel, determine () the stress in the steel, (b) the mximum stress in the concrete. SOLUTON Es 9 10 n = = = 8.0 E.5 10 A s c π = () (1) =.11 in = 5.1 in nas Locte the neutrl xis. x ()()( x + ) + (1 x) (5.1)(17.5 x) = 0 9x x x = 0 or x x 17. = 0 Solve for x. x (11.1) + ()()(17.) = = in. ()() d = 17.5 x = 1.50 in. PROPRETARY MATERAL. 01 The McGrw-Hill Compnies, nc. All rights reserved. No prt of this Mnul my be displyed, bh Ad ()() ()()(.150) in 1 1 = bx = (1)(1.150) =.1 in 1 = + = + = = nad = (5.1)(1.50) = 8. in = + + = 90 in 1 nmy σ = where M = 100 kip ft = 100 kip in. () Steel: n = 8.0 y = 1.50 in. (b) Concrete: n = 1.0, y = = in. σ s (8.0)(100)( 1.50) = σ s =.1 ksi 90 (1.0)(100)(5.150) σ c = σ c = 1.5 ksi 90

11 PROBLEM.55 Five metl strips, ech 0 mm wide, re bonded together to form the composite bem shown. The modulus of elsticity is 10 GP for the steel, 105 GP for the brss, nd 70 GP for the luminum. Knowing tht the bem is bent bout horizontl xis by couple of moment 1800 N m, determine () the mximum stress in ech of the three metls, (b) the rdius of curvture of the composite bem. SOLUTON () Use luminum s the reference mteril. n = 1 in luminum. n= E / E = 10/70= in steel. s n= E / E = 105/70= 1.5in brss. b Due to symmetry of both the mteril rrngement nd the geometry, the neutrl xis psses through the center of the steel portion. For the trnsformed section, n1 1 1 = bh + n1ad = (0)(10) + (0)(10)(5) = mm n 1.5 b h n A d (0)(10) (1.5)(0)(10)(15) mm n.0 = bh = (0)(0) = mm = + = + = = = mm 5 = 1 = mm 9 nmy σ = where M = 1800 N m = = mm = m Aluminum: n = 1.0 y = 0 mm = 0.00 m σ (1.0)(1800)(0.00) = =. 10 P Brss: n= 1.5 y = 0 mm = 0.00 m σ =. MP (b) σb (1.5)(1800)(0.00) = =. 10 P Steel: n=.0 y = 10 mm = m σ s (.0)(1800)(0.010) = =. 10 P Rdius of curvture. 1 M 1800 = = = m 9 9 ρ E (70 10 )(8. 10 ) 1 σ =. MP b σ =. MP s ρ =.7 m PROPRETARY MATERAL. 01 The McGrw-Hill Compnies, nc. All rights reserved. No prt of this Mnul my be displyed,

12 PROBLEM.57 The composite bem shown is formed by bonding together brss rod nd n luminum rod of semicirculr cross sections. The modulus of elsticity is psi for the brss nd psi for the luminum. Knowing tht the composite bem is bent bout horizontl xis by couples of moment 8 kip in., determine the mximum stress () in the brss, (b) in the luminum. SOLUTON For ech semicircle, r = 0.8 in. π A = r = in, r ()(0.8) yo = π 0.95 in. bse r in π = π = = 8 = = bse Ay o = (1.0051)(0.95) = in Use luminum s the reference mteril. n = 1.0 in luminum Eb n = = = 1.5 in brss E Locte the neutrl xis. A, in na, in, in. o y nay o, in Σ Y o = = in..517 The neutrl xis lies in. bove the mteril interfce. d = = 0.71 in., d = = 0.07 in. 1 1 = 1 + = (1.5)(0.0957) + (1.5)(1.0051)(0.71) = in n n Ad = + = (1.0)(0.0957) + (1.0)(1.0051)(0.07) = in n n Ad = + 1 = in () Brss: n = 1.5, y = = in. σ = nmy (1.5)(8)(0.709) = σ = ksi (b) Aluminium: n = 1.0, y = = in. nmy (1.0)(8)( ) σ = = σ = ksi PROPRETARY MATERAL. 01 The McGrw-Hill Compnies, nc. All rights reserved. No prt of this Mnul my be displyed,

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