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Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 4.2 SOLUTION: • Based on the cross section geometry, calculate the location of the section centroid and moment of inertia. Y = ∑ yA ∑A ( I x′ = ∑ I + A d 2 ) • Apply the elastic flexural formula to find the maximum tensile and compressive stresses. σm = A cast-iron machine part is acted upon by a 3 kN-m couple. Knowing E = 165 GPa and neglecting the effects of fillets, determine (a) the maximum tensile and compressive stresses, (b) the radius of curvature. Mc I • Calculate the curvature 1 ρ = M EI © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4-1 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 4.2 SOLUTION: Based on the cross section geometry, calculate the location of the section centroid and moment of inertia. Area, mm 2 y , mm yA, mm3 1 20 × 90 = 1800 50 90 × 103 2 40 × 30 = 1200 ∑ A = 3000 20 24 × 103 3 ∑ yA = 114 × 10 Y = 3 ∑ yA 114 × 10 = = 38 mm 3000 ∑A ( ) (121 bh3 + A d 2 ) 1 90 × 203 + 1800 × 12 2 ) + ( 1 30 × 403 + 1200 × 182 ) = (12 12 I x′ = ∑ I + A d 2 = ∑ I = 868 × 103 mm 4 = 868 × 10-9 m 4 © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4-2 1 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 4.2 • Apply the elastic flexural formula to find the maximum tensile and compressive stresses. Mc I M c A 3 kN ⋅ m × 0.022 m σ A = +76.0 MPa σA = = I 868 × 10−9 m 4 3 kN ⋅ m × 0.038 m σ = −131.3 MPa M cB B σB = − =− I 868 × 10−9 m 4 σm = • Calculate the curvature 1 ρ = M EI = (165 GPa )(868 ×10-9 m 4 ) 3 kN ⋅ m 1 = 20.95 × 10 −3 m -1 ρ ρ = 47.7 m © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4-3 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Bending of Members Made of Several Materials • Consider a composite beam formed from two materials with E1 and E2. • Normal strain varies linearly. εx = − y ρ • Piecewise linear normal stress variation. σ 1 = E1ε x = − E1 y ρ σ 2 = E2ε x = − E2 y ρ Neutral axis does not pass through section centroid of composite section. • Elemental forces on the section are Ey E y dF1 = σ 1dA = − 1 dA dF2 = σ 2 dA = − 2 dA ρ My σx = − I σ1 = σ x σ 2 = nσ x ρ • Define a transformed section such that © 2006 The McGraw-Hill Companies, Inc. All rights reserved. dF2 = − (nE1 ) y dA = − E1 y (n dA) ρ ρ E n= 2 E1 4-4 2 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf 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 (Es = 29x106 psi) and brass (Eb = 15x106 psi). Determine the maximum stress in the steel and brass when a moment of 40 kip*in is applied. • Determine the maximum stress in the steel portion of the bar by multiplying the maximum stress for the transformed section by the ratio of the moduli of elasticity. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. Fourth Edition MECHANICS OF MATERIALS 4-5 Beer • Johnston • DeWolf Example 4.03 SOLUTION: • Transform the bar to an equivalent cross section made entirely of brass. E 29 × 106 psi n= s = = 1.933 Eb 15 × 106 psi bT = 0.4 in + 1.933 × 0.75 in + 0.4 in = 2.25 in • Evaluate the transformed cross sectional properties 1 b h3 = 1 (2.25 in.)(3 in.)3 I = 12 T 12 = 5.063 in.4 • Calculate the maximum stresses σm = Mc (40 kip ⋅ in.)(1.5 in.) = = 11.85 ksi I 5.063 in.4 (σ b )max = σ m (σ s )max = nσ m = 1.933 ×11.85 ksi © 2006 The McGraw-Hill Companies, Inc. All rights reserved. (σ b )max = 11.85 ksi (σ s )max = 22.9 ksi 4-6 3 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf 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, As, is replaced by the equivalent area nAs where n = Es/Ec. • To determine the location of the neutral axis, (bx ) x − n As (d − x ) = 0 2 1 b x2 2 + n As x − n As d = 0 • The normal stress in the concrete and steel σx = − My I σc = σ x σ s = nσ x © 2006 The McGraw-Hill Companies, Inc. All rights reserved. Fourth Edition MECHANICS OF MATERIALS 4-7 Beer • Johnston • DeWolf 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. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4-8 4 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 4.4 SOLUTION: • Transform to a section made entirely of concrete. E 29 × 106 psi = 8.06 n= s = Ec 3.6 × 106 psi ( ) 2 nAs = 8.06 × 2⎡π4 85 in ⎤ = 4.95 in 2 ⎢⎣ ⎥⎦ • Evaluate the geometric properties of the transformed section. ⎛ x⎞ 12 x⎜ ⎟ − 4.95(4 − x ) = 0 ⎝ 2⎠ ( x = 1.450 in ) I = 13 (12 in )(1.45 in )3 + 4.95 in 2 (2.55 in )2 = 44.4 in 4 • Calculate the maximum stresses. Mc1 40 kip ⋅ in × 1.45 in = I 44.4 in 4 Mc 40 kip ⋅ in × 2.55 in σ s = n 2 = 8.06 I 44.4 in 4 σc = © 2006 The McGraw-Hill Companies, Inc. All rights reserved. σ c = 1.306 ksi σ s = 18.52 ksi 4-9 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stress Concentrations Stress concentrations may occur: • in the vicinity of points where the loads are applied σm = K Mc I • in the vicinity of abrupt changes in cross section © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 10 5 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Plastic Deformations • For any member subjected to pure bending y c ε x = − εm strain varies linearly across the section • If the member is made of a linearly elastic material, the neutral axis passes through the section centroid and σx = − My I • For a material with a nonlinear stress-strain curve, the neutral axis location is found by satisfying Fx = ∫ σ x dA = 0 M = ∫ − yσ x dA • 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 stressstrain relationship may be used to map the strain distribution from the stress distribution. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 11 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Plastic Deformations • When the maximum stress is equal to the ultimate strength of the material, failure occurs and the corresponding moment MU is referred to as the ultimate bending moment. • The modulus of rupture in bending, RB, is found from an experimentally determined value of MU and a fictitious linear stress distribution. RB = MU c I • RB may be used to determine MU of any member made of the same material and with the same cross sectional shape but different dimensions. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 12 6 Fourth Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Members Made of an Elastoplastic Material • Rectangular beam made of an elastoplastic material Mc I σ x ≤ σY σm = σ m = σY I M Y = σ Y = maximum elastic moment c • If the moment is increased beyond the maximum elastic moment, plastic zones develop around an elastic core. ⎛ y2 ⎞ M = 32 M Y ⎜1 − 13 Y2 ⎟ ⎜ c ⎟⎠ ⎝ yY = elastic core half - thickness • In the limit as the moment is increased further, the elastic core thickness goes to zero, corresponding to a fully plastic deformation. M p = 32 M Y = plastic moment Mp = shape factor (depends only on cross section shape) k= MY © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 13 7