CHPTER 4 Pure MECHNCS OF MTERLS Bending
Pure Bending Pure Bending Other Loading Tpes Smmetric Member in Pure Bending Bending Deformations Strain Due to Bending Stress Due to Bending Beam Section Properties Bending of Members Made of Several Materials Reinforced Concrete Beams Unsmmetric Bending 4 -
MECHNCS OF MTERLS Pure Bending Pure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane 4-3
Pure/Non-Pure Bending 4-4
Other Loading Tpes Eccentric Loading: xial loading which does not pass through section centroid produces internal forces equivalent to an axial force and a couple Transverse Loading: Concentrated or distributed transverse load produces internal forces equivalent to a shear force and a couple Principle of Superposition: The normal stress due to pure bending ma be combined with the normal stress due to axial loading and shear stress due to shear loading to find the complete state of stress. 4-- 5
Smmetric Member in Pure Bending nternal forces in a cross section are equivalent to a couple moment, i.e., the section bending moment. Couple moment is same about an axis perpendicular to the plane of the couple. Due to smmetr of loading and smmetr of cross-section, beam bends uniforml into arc of circle and plane sections remain plane. 4--6
Terminologies/ssumptions 4-7
Phsical meaning of assumptions. ssumption that plane sections remain plane and perpendicular to deflection curve after deformation implies shear strain γ = 0. Thus Hooke s law gives shear stress τ = 0. This is exactl true in pure bending (since no Shear force). f plane sections remain plane but not perpendicular to deflection curve after deformation, γ (hence τ ) is nonero but constant through depth (-direction). f plane sections don t remain plane or perpendicular to deflection curve after deformation, γ (hence τ ) is nonero and varies through depth (-direction). This is what we will obtain later when we find shear stresses. 4-8
Longitudinal Strain 4-9
Normal Stress 4-10
Resultant of Normal Stress Distribution 4-11
Resultant Force 4-1
Resultant Moment 4-13
Relationship between Bending Stress and Moment 4-14
rea Moments of nertia 4-15 d d d d d d d tan axes, principal i.e., 0, For sin sin cos sin cos sin cos sin d ; cos sin ; sin cos axes : Rotation of )d )( ( Similarl,. 0 d use expand,, d ) ( d axes : of Translation 0 d d so axes, centroidal are,, Here ; d ; d ; d C=centroid
rea centroids and rea Moments of nertia 4-16
rea centroids and rea Moments of nertia 4-17
Deformations in a Transverse Cross Section Recall: deformation due to bending moment M is quantified b curvature of neutral surface 1 x x E M E lthough cross sectional planes remain planar when subjected to bending moments, in-plane deformations are nonero, x x So expansion occurs above neutral surface and contraction below it, causing in-plane curvature, 1 anticlastic curvature 4-18
Section Modulii 4-19
Doubl Smmetric Shape Larger the section modulus lower the max normal stress Review Calculations of Centroid & for various cross-sections ppendix, Beer and Johnston c h c d For two rectangular beams with same area, the one with larger depth is better S bh 6 h 6 4-0
Square Vs. Circular Cross-Section 4-1
Design of Beams for Bending The largest normal stress is found at the surface where the maximum bending moment occurs. m M max c M max S Safe design requires that maximum normal stress be less than allowable stress for material used. This leads to determination of minimum acceptable section modulus. S m min all M max all mong beam section choices which have an acceptable section modulus, the one with smallest weight per unit length or cross sectional area will be least expensive and hence best choice. See ppendix C of BJ book. 5 -
Design of Beams in Bending 4-3
Example 5.1 Cast-iron machine part, acted upon b 3 kn-m couple. E = 165 Gpa, neglect effects of fillets, determine (a) max tensile and compressive stresses, (b) radius of curvature. 4-4
Example 5.1 SOLUTON: Based on cross section geometr, calculate location of section centroid and moment of inertia. 1 rea, mm 090 1800 4030 100 3000, mm 50 0, mm 9010 410 11410 3 3 3 3 Y x 11410 3000 3 38 mm 3 d 1 bh d 1 3 3 1 90 0 18001 1 30 40 10018 1 86810 3 mm 86810-9 m 1 4 4-5
Example 5.1 ppl the elastic flexural formula to find the maximum tensile and compressive stresses. m B Mc M c M c 3 kn m0.0m 9 4 86810 mm 3 kn m0.038m 9 4 86810 mm B B 76.0 MPa 131.3 MPa Calculate the curvature 1 M E 3 kn m 165 GPa 9 4 868 10 - m 1 3-1 0.9510 47.7 m m 4-6
Example 5. 4-7
MECHNCS OF MTERLS Example 5. 4-8
MECHNCS OF MTERLS Example 5. 4-9
MECHNCS OF MTERLS Example 5.3 4-30
Example 5.4 Simpl supported steel beam is to carr loads as shown. llowable normal stress for steel used is 160 Mpa. Select the wide-flange shape that should be used (section properties of wide flange sections are given). 5-31
Example 5.4 Determine reactions at and D. M F D 0 D 58.0 kn 0 5.0 kn 5m 60kN1.5m 50kN4m 58.0 kn 60kN 50kN Develop shear force diagram and determine maximum bending moment. V V V B B V 8kN 5.0 kn area under load curve 60kN Maximum bending moment occurs at V = 0 or x =.6 m. M max area under shear curve, to E 67.6 kn 5-3
Example 5.4 Determine minimum acceptable section modulus. S min M max all 4.510 67.6 kn m 160MPa 6 m 3 4.510 3 mm 3 Shape W41038.8 W3603.9 W310 38.7 W50 44.8 W00 46.1 S 10 3, 69, 4950 475, 4190 547, 4940 531, 5700 451, 5880 Choose best standard section which meets this criteria. W 3603.9 5-33
Bending of Members Made of Several Materials Consider a composite beam formed from two materials with E 1 and E. Normal strain (still) varies linearl (cannot violate kinematics). x Piecewise linear normal stress variation. E E 1 1 1 x E x E Neutral axis does not pass through section centroid of composite section. x 1 M x n x Elemental forces on the section are df E E d 1 d df d 1 d 1 Define a transformed section such that ne1 E1 E df d n d n E 1 4-34
lternate derivation - neutral axis for composite beam F 0 1 1 n d 0 1 E 1 E 1 E 1 x1 d d E 1 E E x 1 d E 1 d 1 1 1 d n E E 1 x1 d d E x d So neutral axis is centroid of 1 n Note: widening ( n 1 ) or narrowing ( n 1 ) done parallel to neutral axis, so that and hence remain unaltered. x 4-35
Example 5.5 SOLUTON: Transform to an equivalent cross section made entirel of (sa) brass Evaluate cross sectional properties of transformed section Calculate maximum stress in transformed section. This is the correct maximum stress for brass portion of the bar. Bar made from bonded pieces of steel (E s = 00 GPa) and brass (E b = 100 GPa). Determine maximum stress in steel and in brass when a moment of 4.5 KNm is applied. Determine the maximum stress in steel portion of b multipling maximum stress for transformed section b the modular ratio. Don t forget this important step!! 4--36
Example 5.5 SOLUTON: Transform to equivalent brass section. E n E b Evaluate transformed cross sectional properties T 10 mm 18 mm 10 mm 56 mm 1 1 s b b T h 00GPa 100GPa 3 1 1 1.9687510 56 mm75 mm 6.0 mm Calculate maximum stresses m Mc 4 4500 Nm0.0375 m 1.9687510 m 3-6 4 85.7 MPa b max m b 85.7 MPa max n 85.7 MPa 171.4 MPa s max m s max 4--37
Reinforced Concrete Beams Concrete beams subjected to bending moments are reinforced b steel rods. The steel rods assumed to carr the entire tensile load below neutral axis (since concrete is weak in tension). Upper part of the concrete beam carries the compressive load. n the transformed section, cross sectional area of steel, s, replaced b equivalent area n s where n = E s /E c. To determine the location of the neutral axis, x bx n d x 0 s 1 b x n x n d s s The normal stress in the concrete and steel x c M x s n x 0 4-38
Example 5.6 Concrete floor slab reinforced with 16- mm-diameter steel rods. Modulus of elasticit is 00 GPa for steel and 5 GPa for concrete. pplied bending moment is 4.5 knm per 0.3 m width of slab. Find maximum stress in concrete and in steel. 4- -39 39
Example 5.6 SOLUTON: Transform to section made entirel of concrete. n n s E E s c 8.0 00 GPa 5 GPa 8.0 16mm 316 4 mm Evaluate geometric properties of transformed section. x 300x 3 316 100 x 0 x 36.8mm 3 6 4 300mm 36.8mm 316mm 63. mm 17.8310 1 mm Check: F s = F c Calculate maximum stresses. Mc1 4500 Nm0.0368m c -6 4 17.8310 m Mc 4500 Nm 0.063 m s n 8.0-6 4 17.8310 m c s 9.9 MPa 17.61MPa 4--40
Example 5.6 4-41
Unsmmetric Bending Thus far analsis of pure bending limited to members subjected to bending couples acting in a plane of smmetr. Members remain smmetric and bend in the plane of smmetr. Neutral axis of cross section coincides/parallel with axis of couple Now consider situations in which bending couple does not act in a plane of smmetr. Cannot assume that member will bend in the plane of the couples. n general, neutral axis of section will not coincide with axis of couple. 4-4
Unsmmetric Bending 0 or 0 F xd x d E d neutral axis passes through centroid Wish to determine conditions under which neutral axis of section of arbitrar shape coincides with axis of couple, as shown above. M M or M E d E x, defines stress distribution moment of inertia Resultant force and moment from the stress distribution in the section must satisf: F x 0 M M M appliedcouple 0 M or 0 d x d product of So couple vector must be directed along a principal centroidal axis. M d inertia 4-43
Unsmmetric Bending Superposition applied to determine stresses in the most general case of unsmmetric bending. Resolve couple vector into components along principle centroidal axes. M M cos M M sin Superpose stresses due to M and M M x M long neutral axis we have, M cos sin tan tan M M M x 0 4-44
Example 5.7 180 Nm couple applied to rectangular wooden beam in a plane at 30 deg. to vertical. Find (a) maximum stress in beam, (b) angle that neutral axis makes with horiontal plane. 4-45 4-45
Example 5.7 Resolve couple vector along principal axes and calculate corresponding maximum stresses. M M 180 Nmcos30 180 Nmsin 30 0.04 m0.09m 0.09 m0.04 m The largest tensile stress due to M 1 155.9 Nm 0.045 m.4310 The largest tensile stress due to M 1 1 1 1 M M 90 Nm0.0 m 0.4810 3 3-6 155.9 Nm 90 Nm.4310 0.4810-6 m m 4 4-6 -6 m m 4 4 occurs along.89 MPa occurs along 3.75 MPa Largest tensile stress due to combined M and M occurs at. B D max 1.89 3.75 6.64 max MPa 4--46
Example 5.7 Determine angle of neutral axis. -6 4.4310 m tan tan -6 4 0.4810 m.9 tan 30 o 71 4--47
MECHNCS OF MTERLS Example 5.8 4-48
Example 5.8 4-49
Example 5.8 4-50
General Case of Eccentric xial Loading Consider straight member subject to equal and opposite eccentric forces. Eccentric force equivalent to sstem comprising centric force and two couples. P M centric force Pa M Pb B principle of superposition, combined stress distribution is x P M M f neutral axis lies on section, it is found from P M M 0 4-51
Example 5.9 Open-link chain obtained b bending low-carbon steel rods into shape shown. For 700 N load, find (a) maximum tensile and compressive stresses, (b) distance between section centroid and neutral axis 4-5
Example 5.9 Normal stress due to a centric load c 0 113.1mm P 6. MPa 6 mm 700 N 113.110 6 m Equivalent centric load and bending moment P 700 N M Pd 11. Nm 700 N 0.016 m Normal stress due to bending moment m 1017.9 mm 1 4 c Mc 4 66 MPa 1 4 6 mm 4 11. Nm 0.006 m 4 1017.910-1 m 4 4-53
Example 5.9 Maximum tensile and compressive stresses t c 0 6. 66 0 m m 6. 66 t c 7. MPa 59.8MPa Neutral axis location P M0 0 P 0 M 0 0.56 mm 6.10 6 Pa 1017.910 11. Nm 1 m 4 4-54
Sample Problem 5.10 Largest allowable stresses for cast iron link are 30 MPa in tension and 10 MPa in compression. Determine largest force P which can be applied. 310 3 Y 0.038m 86810 m 9 m 4 4-55
Sample Problem 5.10 Determine equivalent centric load & bending moment. d 0.038 0.010 0.08m P centric load M Pd 0.08P bending moment Superpose stresses due to P and M, Mc P 0.08 P0.0 Evaluate critical loads for allowable stresses. B B P 377 P 1559P P Mc 310 P 310 30MPa 10MPa 3 3 86810 0.08 P0.0 86810 P 79.6 kn P 77.0 kn 9 9 377 P 1559P Largest allowable load is P 77.0 kn 4-56
MECHNCS OF MTERLS Example 5.11 lso locate N.. 4-57
Example 5.11 4-58