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Lecture Mechanics of materials (Third edition) - Chapter 8: Principle stresses under a given loading. In chapter 8, you will learn how to determine the stress in a structural member or machine element due to a combination of loads and how to find the corresponding principal stresses and maximum shearing stress.
Third Edition CHAPTER MECHANICS OF MATERIALS Ferdinand P Beer E Russell Johnston, Jr John T DeWolf Lecture Notes: J Walt Oler Texas Tech University Principle Stresses Under a Given Loading © 2002 The McGraw-Hill Companies, Inc All rights reserved Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Principle Stresses Under a Given Loading Introduction Principle Stresses in a Beam Sample Problem 8.1 Sample Problem 8.2 Design of a Transmission Shaft Sample Problem 8.3 Stresses Under Combined Loadings Sample Problem 8.5 © 2002 The McGraw-Hill Companies, Inc All rights reserved 8-2 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Introduction • In Chaps and 2, you learned how to determine the normal stress due to centric loads In Chap 3, you analyzed the distribution of shearing stresses in a circular member due to a twisting couple In Chap 4, you determined the normal stresses caused by bending couples In Chaps and 6, you evaluated the shearing stresses due to transverse loads In Chap 7, you learned how the components of stress are transformed by a rotation of the coordinate axes and how to determine the principal planes, principal stresses, and maximum shearing stress at a point • In Chapter 8, you will learn how to determine the stress in a structural member or machine element due to a combination of loads and how to find the corresponding principal stresses and maximum shearing stress © 2002 The McGraw-Hill Companies, Inc All rights reserved 8-3 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Principle Stresses in a Beam • Prismatic beam subjected to transverse loading My Mc σm = I I VQ VQ τ xy = − τm = It It σx = − • Principal stresses determined from methods of Chapter • Can the maximum normal stress within the cross-section be larger than σm = © 2002 The McGraw-Hill Companies, Inc All rights reserved Mc I 8-4 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Principle Stresses in a Beam © 2002 The McGraw-Hill Companies, Inc All rights reserved 8-5 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Principle Stresses in a Beam • Cross-section shape results in large values of τxy near the surface where σx is also large max may be greater than m â 2002 The McGraw-Hill Companies, Inc All rights reserved 8-6 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.1 SOLUTION: • Determine shear and bending moment in Section A-A’ • Calculate the normal stress at top surface and at flange-web junction A 160-kN force is applied at the end of a W200x52 rolled-steel beam • Evaluate the shear stress at flangeweb junction Neglecting the effects of fillets and of stress concentrations, determine whether the normal stresses satisfy a design specification that they be equal to or less than 150 MPa at section A-A’ • Calculate the principal stress at flange-web junction © 2002 The McGraw-Hill Companies, Inc All rights reserved 8-7 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.1 SOLUTION: • Determine shear and bending moment in Section A-A’ M A = (160 kN )(0.375 m ) = 60 kN - m V A = 160 kN • Calculate the normal stress at top surface and at flange-web junction MA 60 kN ⋅ m = S 512 × 10− m3 = 117.2 MPa y 90.4 mm σb = σ a b = (117.2 MPa ) c 103 mm = 102.9 MPa σa = © 2002 The McGraw-Hill Companies, Inc All rights reserved 8-8 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.1 • Evaluate shear stress at flange-web junction Q = (204 × 12.6 )96.7 = 248.6 × 103 mm3 = 248.6 × 10− m3 ( ) V AQ (160 kN ) 248.6 × 10− m3 τb = = It 52.7 × 10− m (0.0079 m ) ( ) = 95.5 MPa • Calculate the principal stress at flange-web junction σ max = 12 σ b + (12 σ b )2 + τ b2 102.9 ⎛ 102.9 ⎞ = + ⎜ ⎟ + (95.5) ⎝ ⎠ = 169.9 MPa (> 150 MPa ) Design specification is not satisfied © 2002 The McGraw-Hill Companies, Inc All rights reserved 8-9 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.2 SOLUTION: • Determine reactions at A and D • Determine maximum shear and bending moment from shear and bending moment diagrams The overhanging beam supports a uniformly distributed load and a concentrated load Knowing that for the grade of steel to used σall = 24 ksi and τall = 14.5 ksi, select the wideflange beam which should be used • Calculate required section modulus and select appropriate beam section • Find maximum normal stress • Find maximum shearing stress © 2002 The McGraw-Hill Companies, Inc All rights reserved - 10 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.2 SOLUTION: • Determine reactions at A and D ∑ M A = ⇒ RD = 59 kips ∑ M D = ⇒ R A = 41kips • Determine maximum shear and bending moment from shear and bending moment diagrams M max = 239.4 kip ⋅ in V max = 43 kips with V = 12.2 kips • Calculate required section modulus and select appropriate beam section M max 24 kip ⋅ in = = 119.7 in S = 24 ksi σ all select W21ì 62 beam section â 2002 The McGraw-Hill Companies, Inc All rights reserved - 11 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.2 • Find maximum shearing stress Assuming uniform shearing stress in web, τ max = Vmax 43 kips = = 5.12 ksi < 14.5 ksi Aweb 8.40 in • Find maximum normal stress σa = 60 kip ⋅ in M max = 2873 = 22.6 ksi S 127in 9.88 y = 21.3 ksi σb = σ a b = (22.6 ksi ) 10.5 c τb = 12.2 kips V = = 1.45 ksii Aweb 8.40 in 2 21.3 ksi ⎛ 21.3 ksi ⎞ + ⎜ σ max = ⎟ + (1.45 ksi ) ⎝ ⎠ = 21.4 ksi < 24 ksi © 2002 The McGraw-Hill Companies, Inc All rights reserved - 12 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Design of a Transmission Shaft • If power is transferred to and from the shaft by gears or sprocket wheels, the shaft is subjected to transverse loading as well as shear loading • Normal stresses due to transverse loads may be large and should be included in determination of maximum shearing stress • Shearing stresses due to transverse loads are usually small and contribution to maximum shear stress may be neglected © 2002 The McGraw-Hill Companies, Inc All rights reserved - 13 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Design of a Transmission Shaft • At any section, Mc I Tc τm = J σm = where M = M y2 + M z2 • Maximum shearing stress, 2 σ Mc ⎞ ⎛ Tc ⎞ τ max = ⎛⎜ m ⎞⎟ + (τ m )2 = ⎛⎜ ⎟ +⎜ ⎟ ⎝ 2I ⎠ ⎝ J ⎠ ⎝ ⎠ for a circular or annular cross - section, I = J τ max = c M2 +T2 J • Shaft section requirement, ⎛J⎞ = ⎜ ⎟ c ⎝ ⎠ © 2002 The McGraw-Hill Companies, Inc All rights reserved ⎛⎜ M + T ⎞⎟ ⎝ ⎠ max τ all - 14 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.3 SOLUTION: • Determine the gear torques and corresponding tangential forces • Find reactions at A and B • Identify critical shaft section from torque and bending moment diagrams Solid shaft rotates at 480 rpm and transmits 30 kW from the motor to gears G and H; 20 kW is taken off at gear G and 10 kW at gear H Knowing that σall = 50 MPa, determine the smallest permissible diameter for the shaft © 2002 The McGraw-Hill Companies, Inc All rights reserved • Calculate minimum allowable shaft diameter - 15 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.3 SOLUTION: • Determine the gear torques and corresponding tangential forces TE = 30 kW P = = 597 N ⋅ m 2πf 2π (80 Hz ) T 597 N ⋅ m = 3.73 kN FE = E = rE 0.16 m TC = 20 kW = 398 N ⋅ m 2π (80 Hz ) FC = 6.63 kN TD = 10 kW = 199 N ⋅ m 2π (80 Hz ) FD = 2.49 kN • Find reactions at A and B Ay = 0.932 kN Az = 6.22 kN B y = 2.80 kN Bz = 2.90 kN © 2002 The McGraw-Hill Companies, Inc All rights reserved - 16 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.3 • Identify critical shaft section from torque and bending moment diagrams ⎛⎜ M + T ⎞⎟ = ⎝ ⎠ max (11602 + 3732 )+ 5972 = 1357 N ⋅ m © 2002 The McGraw-Hill Companies, Inc All rights reserved - 17 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.3 • Calculate minimum allowable shaft diameter J M + T 1357 N ⋅ m = = = 27.14 × 10−6 m3 50 MPa τ all c For a solid circular shaft, J π = c = 27.14 × 10− m3 c c = 0.02585 m = 25.85 m d = 2c = 51.7 mm © 2002 The McGraw-Hill Companies, Inc All rights reserved - 18 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stresses Under Combined Loadings • Wish to determine stresses in slender structural members subjected to arbitrary loadings • Pass section through points of interest Determine force-couple system at centroid of section required to maintain equilibrium • System of internal forces consist of three force components and three couple vectors • Determine stress distribution by applying the superposition principle © 2002 The McGraw-Hill Companies, Inc All rights reserved - 19 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stresses Under Combined Loadings • Axial force and in-plane couple vectors contribute to normal stress distribution in the section • Shear force components and twisting couple contribute to shearing stress distribution in the section © 2002 The McGraw-Hill Companies, Inc All rights reserved - 20 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Stresses Under Combined Loadings • Normal and shearing stresses are used to determine principal stresses, maximum shearing stress and orientation of principal planes • Analysis is valid only to extent that conditions of applicability of superposition principle and Saint-Venant’s principle are met © 2002 The McGraw-Hill Companies, Inc All rights reserved - 21 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.5 SOLUTION: • Determine internal forces in Section EFG • Evaluate normal stress at H • Evaluate shearing stress at H Three forces are applied to a short steel post as shown Determine the principle stresses, principal planes and maximum shearing stress at point H • Calculate principal stresses and maximum shearing stress Determine principal planes © 2002 The McGraw-Hill Companies, Inc All rights reserved - 22 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.5 SOLUTION: • Determine internal forces in Section EFG Vx = −30 kN P = 50 kN Vz = −75 kN M x = (50 kN )(0.130 m ) − (75 kN )(0.200 m ) = −8.5 kN ⋅ m M y = M z = (30 kN )(0.100 m ) = kN ⋅ m Note: Section properties, A = (0.040 m )(0.140 m ) = 5.6 × 10−3 m (0.040 m )(0.140 m )3 = 9.15 × 10 − m I x = 12 (0.140 m )(0.040 m )3 = 0.747 × 10− m I z = 12 © 2002 The McGraw-Hill Companies, Inc All rights reserved - 23 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.5 • Evaluate normal stress at H σy =+ = P Mz a Mx b + − A Iz Ix (3 kN ⋅ m )(0.020 m ) + 5.6 × 10-3 m 0.747 × 10−6 m 50 kN − (8.5 kN ⋅ m )(0.025 m ) 9.15 × 10−6 m = (8.93 + 80.3 − 23.2 ) MPa = 66.0 MPa • Evaluate shearing stress at H Q = A1 y1 = [(0.040 m )(0.045 m )](0.0475 m ) = 85.5 × 10−6 m3 ( ) ( 75 kN ) 85.5 × 10−6 m3 Vz Q = τ yz = I xt 9.15 × 10−6 m (0.040 m ) ( ) = 17.52 MPa © 2002 The McGraw-Hill Companies, Inc All rights reserved - 24 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 8.5 • Calculate principal stresses and maximum shearing stress Determine principal planes τ max = R = 33.02 + 17.522 = 37.4 MPa σ max = OC + R = 33.0 + 37.4 = 70.4 MPa σ = OC − R = 33.0 − 37.4 = −7.4 MPa tan 2θ p = CY 17.52 2θ p = 27.96° = CD 33.0 θ p = 13.98° τ max = 37.4 MPa σ max = 70.4 MPa σ = −7.4 MPa θ p = 13.98° © 2002 The McGraw-Hill Companies, Inc All rights reserved - 25 ... principal stresses, maximum shearing stress and orientation of principal planes • Analysis is valid only to extent that conditions of applicability of superposition principle and Saint-Venant’s principle. .. Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Principle Stresses Under a Given Loading Introduction Principle Stresses in a Beam Sample Problem 8.1 Sample Problem 8.2 Design of a Transmission... DeWolf Principle Stresses in a Beam • Cross-section shape results in large values of τxy near the surface where σx is also large max may be greater than m â 2002 The McGraw-Hill Companies, Inc All