Third Edition CHAPTER MECHANICS OF MATERIALS 4Ferdinand P. Beer E. Russell Johnston, Jr. Pure Bending John T. DeWolf Lecture Notes:J. Walt OlerTexas Tech University © 2002 The McGraw-Hill Companies, Inc. All rights reserved.

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Pure Bending Pure Bending Example 4.03 Other Loading Types Reinforced Concrete Beams Symmetric Member in Pure Bending Sample Problem 4.4 Bending Deformations Stress Concentrations Strain Due to Bending Plastic Deformations Beam Section Properties Members Made of an Elastoplastic Material Properties of American Standard Shapes Plastic Deformations of Members With a Single Plane of S… Deformations in a Transverse Cross Section Residual Stresses Sample Problem 4.2 Example 4.05, 4.06 Bending of Members Made of Several Materials Eccentric Axial Loading in a Plane of Symmetry Example 4.03 Example 4.07 Reinforced Concrete Beams Sample Problem 4.8 Sample Problem 4.4 Unsymmetric Bending Stress Concentrations Example 4.08 Plastic Deformations General Case of Eccentric Axial Loading Members Made of an Elastoplastic Material © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 2

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Pure Bending Pure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 3

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Other Loading Types • Eccentric Loading: Axial 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 may be combined with the normal stress due to axial loading and shear stress due to shear loading to find the complete state of stress. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 4

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Symmetric Member in Pure Bending • Internal forces in any cross section are equivalent to a couple. The moment of the couple is the section bending moment. • From statics, a couple M consists of two equal and opposite forces. • The sum of the components of the forces in any direction is zero. • The moment is the same about any axis perpendicular to the plane of the couple and zero about any axis contained in the plane. • These requirements may be applied to the sums of the components and moments of the statically indeterminate elementary internal forc F es.x = ∫σ dA x = 0 M y = ∫ zσ dA x = 0 M z = ∫ − yσ dA x = M © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 5

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Bending Deformations Beam with a plane of symmetry in pure bending: • member remains symmetric • bends uniformly to form a circular arc • cross-sectional plane passes through arc center and remains planar • length of top decreases and length of bottom increases • a neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not change • stresses and strains are negative (compressive) above the neutral plane and positive (tension) below it © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 6

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Strain Due to Bending Consider a beam segment of length L. After deformation, the length of the neutral surface remains L. At other sections, L′ = ( ρ − y)θ δ = L − L′ = ( ρ − y)θ − ρθ = −yθ δ yθ y εx = = − = − ( strain varies l inearly) L ρθ ρ c c εm = o r ρ = ρ εm y εx = − εm c © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 7

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Stress Due to Bending • For a linearly elastic material, y σ x = Eεx = − Eεm c y = − σ (stres s varies li nearly) m c • For static equilibrium, • For static equilibrium, Fx = 0 = ∫σ x = ∫− y dA σ dA y c m M = ∫ − yσ dA =∫ − y − σ dA x c m 0 = −σ m ∫ y dA c σ 2 σ I M m = y dA m ∫ = c c First moment with respect to neutral Mc M plane is zero. Therefore, the neutral σm = = I S surface must pass through the y section centroid. Substitutin g σ x = − σ c m My σ x = − I © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 8

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Beam Section Properties • The maximum normal stress due to bending, σ = Mc m = M I S I = s ection moment of i nertia = I S = section modulus c A beam section with a larger section modulus will have a lower maximum stress • Consider a rectangular beam cross section, 1 bh3 I S 12 1 = = = bh3 1 = Ah c h 6 6 2 Between two beams with the same cross sectional area, the beam with the greater depth will be more effective in resisting bending. • Structural steel beams are designed to have a large section modulus. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 9

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Properties of American Standard Shapes © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 10

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Deformations in a Transverse Cross Section • Deformation due to bending moment M is quantified by the curvature of the neutral surface 1 ε σ 1 Mc m m = = = ρ c Ec Ec I M = EI • Although cross sectional planes remain planar when subjected to bending moments, in-plane deformations are nonzero, νy ν ε y = ν − εx = εz = ν − ε y = ρ x ρ • Expansion above the neutral surface and contraction below it cause an in-plane curva 1 ture ν , = = a nticlasti c c urvature ρ′ ρ © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 11

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Sample Problem 4.2 SOLUTION: • Based on the cross section geometry, calculate the location of the section centroid and moment of inertia. ∑ Y = yA 2 Ix′ = ∑(I + Ad ) ∑ A • Apply the elastic flexural formula to find the maximum tensile and compressive stresses. Mc σm = I A cast-iron machine part is acted upon by a 3 kN-m couple. Knowing E = • Calculate the curvature 165 GPa and neglecting the effects of 1 M = fillets, determine (a) the maximum ρ EI tensile and compressive stresses, (b) the radius of curvature. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 12

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Sample Problem 4.2 SOLUTION: Based on the cross section geometry, calculate the location of the section centroid and moment of inertia. 2 3 Area, m m y m , m yA m , m 1 20×90 = 1800 50 90× 3 10 2 40×30 = 1200 20 4 2 × 3 10 ∑ A = 3000 ∑ yA = 114× 3 10 ∑ yA 114×103 Y = = = 38 m m ∑ A 3000 Ix′ = ∑( 2 I + Ad ) = ∑( 1 3 2 bh + Ad 12 ) = ( 1 3 2 90× 20 +1800 ×12 + 30× 40 +1200 ×18 12 ) (1 3 2 12 ) 3 -9 4 I = 868×10 m m = 8 68×10 m © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 13

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Sample Problem 4.2 • Apply the elastic flexural formula to find the maximum tensile and compressive stresses. σm = Mc I M c k 3 N A ⋅m× . 0 022 m σ σ A = 7 + 6 M 0 . Pa A = = −9 4 I 868×10 mm M c k 3 N B ⋅m× . 0 038 m σ B = − = − σ B = 1 − 31 M 3 . Pa −9 4 I 868×10 mm • Calculate the curvature 1 = M ρ EI k 3 N ⋅ m 1 = = 20 9 . 5×10−3m 1 – (165 G Pa )( -9 4 868×10 m ) ρ ρ = 47 m 7 . © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 14

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Bending of Members Made of Several Materials • Consider a composite beam formed from two materials with E and E . 1 2 • Normal strain varies linearly. ε y x = − ρ • Piecewise linear normal stress variation. σ = E 1 ε E y 1 2 1 x = − σ = E 2 ε E y = − ρ 2 x ρ Neutral axis does not pass through section centroid of composite section. • Elemental forces on the section are E y E y dF1 = σ dA 1 = − dA dF 1 2 = σ dA 2 = − dA ρ 2 ρ My σ x = − • Define a transformed section such that I (n 1 E ) y 1 E y E σ1 = σ x σ2 = nσ x d 2 F = − dA = − (ndA) 2 n = ρ ρ 1 E © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 15

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Example 4.03 SOLUTION: • Transform the bar to an equivalent cross section made entirely of brass • Evaluate the cross sectional properties of the transformed section • Calculate the maximum stress in the transformed section. This is the correct maximum stress for the brass pieces of the bar. Bar is made from bonded pieces of steel (E = 29×106 psi) and brass • Determine the maximum stress in the s (E = 15×106 psi). Determine the steel portion of the bar by multiplying b maximum stress in the steel and the maximum stress for the transformed brass when a moment of 40 kip*in section by the ratio of the moduli of is applied. elasticity. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 16

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Example 4.03 SOLUTION:• Transform the bar to an equivalent cross section made entirely of brass. E 29 s ×106psi n = = = . 1 933 b E 15×106psi T b = . 0 i 4 n + 9 . 1 33× 7 . 0 5 i n + . 0 i 4 n = 2 . 2 5 i n • Evaluate the transformed cross sectional properties 1 3 1 I = b h T = ( 2 . 2 5 i n.)( i 3 n)3 12 12 4 = 0 . 5 63 i n • Calculate the maximum stresses Mc (40 k ip⋅in)( .1 i 5 n) σm = = = 11.85 k si I 5.063 i n4 (σb) σ (σb) max = m = 11 8 . 5 k si max (σs) = nσ m = 9 . 1 33×11 8 . 5 k si (σs) max = 22.9 k si max © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 17

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Reinforced Concrete Beams • Concrete beams subjected to bending moments are reinforced by steel rods. • The steel rods carry the entire tensile load below the neutral surface. The upper part of the concrete beam carries the compressive load. • In the transformed section, the cross sectional area of the steel, A , is replaced by the equivalent area s nA where n = E /E . s s c • To determine the location of the neutral axis, ( ) x bx − n As(d − x) = 0 2 1 2 b x + n A x s − n A d s = 0 2 • The normal stress in the concrete and steel My σ x = − Iσc = σ x σ s = nσ x © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 18

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Sample Problem 4.4 SOLUTION: • Transform to a section made entirely of concrete. • Evaluate geometric properties of transformed section. • Calculate the maximum stresses in the concrete and steel. A concrete floor slab is reinforced with 5/8-in-diameter steel rods. The modulus of elasticity is 29x106psi for steel and 3.6x106psi for concrete. With an applied bending moment of 40 kip*in for 1-ft width of the slab, determine the maximum stress in the concrete and steel. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 19

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Sample Problem 4.4 SOLUTION:• Transform to a section made entirely of concrete. 6 E 29 s ×10 p si n = = = 0 . 8 6 6 c E . 3 6×10 p si nA = 0 . 8 6 2 × π s (5i n = .495in 4 8 )2 2 • Evaluate the geometric properties of the transformed section. x 12 x − 4 9 . ( 5 4 − x) = 0 x = 1 4 . 50in 2 1 I = (12in)(1 4 . 5in)3 + 4 9 . 5in 2.55in = 44 4 . in 3 ( 2 )( )2 4 • Calculate the maximum stresses. M 1 c 40 kip ⋅in ×1.45in σc = = σc = . 1 306 ksi 4 I 44.4in Mc2 40 kip ⋅in × 5 . 2 5in σ σ s = s = n = 0 . 8 6 18.52 ksi 4 I 44.4in © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 20

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Stress Concentrations Stress concentrations may occur: Mc σ K m = I • in the vicinity of points where the loads are applied • in the vicinity of abrupt changes in cross section © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 21

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Plastic Deformations • For any member subjected to pure bending y εx = − εm strain varies linearly across the c section • If the member is made of a linearly elastic material, the neutral axis passes through the section centroid My and σ x = − I • For a material with a nonlinear stress-strain curve, the neutral axis location is found by satisfying Fx = ∫σ dA x = 0 M = ∫ − yσ dA x • For a member with vertical and horizontal planes of symmetry and a material with the same tensile and compressive stress-strain relationship, the neutral axis is located at the section centroid and the stress-strain relationship may be used to map the strain distribution from the stress distribution. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 22

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Plastic Deformations • When the maximum stress is equal to the ultimate strength of the material, failure occurs and the corresponding moment M is referred to as the U ultimate bending moment. • The modulus of rupture in bending, R , is found B from an experimentally determined value of M U and a fictitious linear stress distribution. M c R U B = I • R may be used to determine M of any member B U made of the same material and with the same cross sectional shape but different dimensions. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 23

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Members Made of an Elastoplastic Material • Rectangular beam made of an elastoplastic material Mc σ x ≤ σY σm = I I σm = σY MY = σ Y = m aximum e lastic m oment c • If the moment is increased beyond the maximum elastic moment, plastic zones develop around an elastic core. 2 y 3 M = M 1 1 Y − Y Y y = e lastic c ore h alf – thickness 2 3 2 c • In the limit as the moment is increased further, the elastic core thickness goes to zero, corresponding to a fully plastic deformation. 3 M p = MY = p lastic m oment 2 M k = p = s hape f actor ( depends o nly o n c ross s ection shape) MY © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 24

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Plastic Deformations of Members With a Single Plane of Symmetry • Fully plastic deformation of a beam with only a vertical plane of symmetry. • The neutral axis cannot be assumed to pass through the section centroid. • Resultants R and R of the elementary 1 2 compressive and tensile forces form a couple. R = R 1 2 A σ 1 Y = A σ 2 Y The neutral axis divides the section into equal areas. • The plastic moment for the member, M 1 p = ( AσY 2 )d © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 25

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Residual Stresses • Plastic zones develop in a member made of an elastoplastic material if the bending moment is large enough. • Since the linear relation between normal stress and strain applies at all points during the unloading phase, it may be handled by assuming the member to be fully elastic. • Residual stresses are obtained by applying the principle of superposition to combine the stresses due to loading with a moment M (elastoplastic deformation) and unloading with a moment -M (elastic deformation). • The final value of stress at a point will not, in general, be zero. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 26

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Example 4.05, 4.06 A member of uniform rectangular cross section is subjected to a bending moment M = 36.8 kN-m. The member is made of an elastoplastic material with a yield strength of 240 MPa and a modulus of elasticity of 200 GPa. Determine (a) the thickness of the elastic core, (b) the radius of curvature of the neutral surface. After the loading has been reduced back to zero, determine (c) the distribution of residual stresses, (d) radius of curvature. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 27

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Example 4.05, 4.06 • Thickness of elastic core: 2 y 3 M = M 1 1 Y − Y 2 3 2 c 2 y 36.8kN ⋅ m 3 = (28.8kN ⋅m)1 1 − Y 2 3 2 c Y y = Yy = 6. 0 66 2 Y y = 80 mm c 60 mm • Radius of curvature: • Maximum elastic moment: 6 σ 240 Y ×10 Pa εY = = 9 I E 200 ×10 Pa 2 2 2 = bc = × − m × − m 3 ( 2 50 10 3 3 )(60 10 3 ) c −3 = . 1 2×10 = 120×10−6m3 ε = Y y Y I ρ MY = σY = (120×10−6m3)(240MPa) c −3 y 40 Y ×10 m = 28 k 8 . N ⋅ m ρ = = −3 ε ρ = 33.3m Y . 1 2×10 © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 28

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Example 4.05, 4.06 • M = 36.8 kN-m • M = -36.8 kN-m • M = 0 Y y = 40 mm Mc 36 8 . kN ⋅ m σ ′ At the e dge o f the e lastic c ore, m = = σ 240 MPa 6 3 6 Y = I 120 ×10 m σ x −35.5×10 Pa = 306.7MPa < 2σ εx = = Y 9 E 200 ×10 Pa −6 = 1 − 77.5×10 −3 y 40 Y ×10 m ρ = − = −6 εx 177 5 . ×10 ρ = 225m © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 29

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Eccentric Axial Loading in a Plane of Symmetry • Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due a pure bending moment σ x = (σ x ) + σ centric ( x)bending P My = − A I • Eccentric loading • Validity requires stresses below proportional F = P limit, deformations have negligible effect on M = Pd geometry, and stresses not evaluated near points of load application. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 30

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Example 4.07 SOLUTION: • Find the equivalent centric load and bending moment • Superpose the uniform stress due to the centric load and the linear stress due to the bending moment. • Evaluate the maximum tensile and compressive stresses at the inner and outer edges, respectively, of the An open-link chain is obtained by superposed stress distribution. bending low-carbon steel rods into the shape shown. For 160 lb load, • Find the neutral axis by determine (a) maximum tensile and determining the location where the compressive stresses, (b) distance normal stress is zero. between section centroid and neutral axis © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 31

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Example 4.07 • Normal stress due to a centric load 2 A = c π = π ( . 0 25in)2 = 1 . 0 963in2 P 160 lb σ0 = = A . 0 1963in2 = 815psi • Equivalent centric load • Normal stress due to and bending moment bending moment P = 160lb 1 4 1 I = c π = π ( 2 . 0 5)4 4 4 M = Pd = (160lb)( . 0 6in) = . 3 068×10−3 in4 = 104lb⋅in Mc (104lb⋅in)( .025in) σm = = I 0 . 68×10−3 in4 = 8475psi © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 32

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Example 4.07 • Maximum tensile and compressive • Neutral axis location stresses P My0 σ 0 = − t = σ 0 + σ m A I = 815 + 8475 σt = 9260psi P I 0 . 3 68×10−3in4 σ 0 y = = (815psi) c = σ 0 − σ m A M 105lb ⋅in = 815 −8475 σc = 7 − 660 psi y 0 . 0 240 in 0 = © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 33

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Sample Problem 4.8 The largest allowable stresses for the cast iron link are 30 MPa in tension and 120 MPa in compression. Determine the largest force P which can be applied to the link. SOLUTION: • Determine an equivalent centric load and bending moment. • Superpose the stress due to a centric load and the stress due to bending. • Evaluate the critical loads for the allowable From Sample Problem 2.4, tensile and compressive stresses. −3 2 A = 3×10 m • The largest allowable load is the smallest Y = . 0 038 m of the two critical loads. −9 4 I = 868×10 m © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 34

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Sample Problem 4.8 • Determine an equivalent centric and bending loads. d = 0 . 0 38 − . 0 010 = . 0 028 m P =centric l oadM = Pd = . 0 028 P = b ending m oment • Superpose stresses due to centric and bending loads P Mc P A ( .0028P)( 0. 0 22) σ = − + = − + = 3 + 77 P A A I 3×10−3 868×10−9 P Mc P A ( 0. 0 28 P)( . 0 022) σ = − − = − − = 1 − 559 P B A I 3×10−3 868×10−9 • Evaluate critical loads for allowable stresses. σ A = 3 + 77 P = 30MPa P = 79.6kN σ B = 1 − 559 P = 1 − 20 MPa P = 79.6kN • The largest allowable load P = 77.0 k N © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 35

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Unsymmetric Bending • Analysis of pure bending has been limited to members subjected to bending couples acting in a plane of symmetry. • Members remain symmetric and bend in the plane of symmetry. • The neutral axis of the cross section coincides with the axis of the couple • Will now consider situations in which the bending couples do not act in a plane of symmetry. • Cannot assume that the member will bend in the plane of the couples. • In general, the neutral axis of the section will not coincide with the axis of the couple. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 36

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Unsymmetric Bending • y 0 = Fx = ∫σ dA x = ∫− σm dA c or 0 = ∫ y dA neutral axis passes through centroid y Wish to determine the conditions under • M = M = −∫ y− σ dA z c m which the neutral axis of a cross section σ I m of arbitrary shape coincides with the or M = I = I = moment of inertia c z axis of the couple as shown. defines stress distribution • The resultant force and moment from the distribution of • y 0 = M = ∫ zσ dA = ∫ z− σ dA y x elementary forces in the section c m or 0 = must satisfy ∫ yz dA = I = product of inertia yz F = 0 = M M = M = applied couple x y z couple vector must be directed along a principal centroidal axis © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 37

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Unsymmetric Bending Superposition is applied to determine stresses in the most general case of unsymmetric bending. • Resolve the couple vector into components along the principle centroidal axes. M = M cos z θ M = M sin y θ • Superpose the component stress distributions M y M y z y σ x = − + Iz I y • Along the neutral axis, M y M y z y (M cosθ ) y (M sinθ ) σ y = 0 x = − + = − + Iz I y Iz I y y I tanφ z = = tanθ z I y © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 38

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Example 4.08 SOLUTION: • Resolve the couple vector into components along the principle centroidal axes and calculate the corresponding maximum stresses. M = M cos z θ M = M sin y θ • Combine the stresses from the component stress distributions. M y M y z y σ = − + A 1600 lb-in couple is applied to a x Iz I y rectangular wooden beam in a plane • Determine the angle of the neutral forming an angle of 30 deg. with the axis. vertical. Determine (a) the maximum y I stress in the beam, (b) the angle that the tanφ z = = tanθ z I y neutral axis forms with the horizontal plane. © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 39

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Example 4.08 • Resolve the couple vector into components and calculate the corresponding maximum stresses. M z = (1600lb⋅in) cos30 = 1386lb⋅inM y = (1600lb⋅in) sin 30 = 800lb⋅in 1 Iz = (1 5 . in)( . 3 5in)3 = . 5 359in4 12 1 I y = ( . 3 5in)( . 1 5in)3 = 9 . 0 844in4 12 The l argest tensile s tress d u e t o M o ccurs a lon g z AB M z y (1386lb⋅in)( . 1 75in) σ = = = 452.6psi 1 Iz . 5 359in4 The l argest tensile s tress d u e t o M o ccurs a lon g z AD M yz (800lb⋅in)( . 0 75in) σ = = = 609.5psi 2 I y . 0 9844 in4 • The largest tensile stress due to the combined loading occurs at A. σ = σ +σ = 452 6 . + 609 5 . max 1 2 σ 1062 psi max = © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 40

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n Example 4.08 • Determine the angle of the neutral axis. I 3 . 5 59in4 tanφ = z tanθ = tan 30 I y 9 . 0 844 in4 = 1 . 3 43 o φ = 72 4 . © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 41

E T d h i i t r i d o MECHANICS OF MATERIALS Beer • Johnston • DeWolf n General Case of Eccentric Axial Loading • Consider a straight member subject to equal and opposite eccentric forces. • The eccentric force is equivalent to the system of a centric force and two couples. P = c entric f orce M = Pa M = Pb y z • By the principle of superposition, the combined stress distribution is P M y M z z y σ x = − + A Iz I y • If the neutral axis lies on the section, it may be found from M M z y P y − z = I I A z y © 2002 The McGraw-Hill Companies, Inc. All rights reserved. 4 – 42