Combined Loading Solutions Part 3

Stresses at points A and B (A) The axial force P1 produces uniform compressive stresses throughout the post. These stresses are σP1 = P1 / A where A is the cross section area of the post A = b2 – (b – 2t)2 = 4t(b-t) = 4 (0.5in)(6in – 0.5in) = 11.0in2 σP1 = P1 / A = 3240lb / 11.00in2 = 295psi (B) The bending moment M1 produces compressive stresses σM1 at points A and B. These stresses are obtained from the flexure formula σM1 = M1 (b / 2) / Ι where Ι is the moment of inertia of the cross section. The moment of inertia is Ι = [b4 - (b -2t)4] / 12 = [(6in)4 – (5in)4] / 12 = 55.92in4 Thus, σM1 = M1 b / 2Ι = (29160lb.in)(6in) / (2)(55.92in4) = 1564psi (C) The shear force P2 produces a shear stress at point B but not at point A. We know that an approximate value of the shear stress can be obtained by dividing the shear force by the web area. τ P2 = P2 / Aweb =P2 /(2t(b – 2t)) =800lb / (2)(0.5in)(6in–1in)= 160psi The stress τp2 acts at point B in the direction shown in the above figure. We can calculate the shear stress τP2 from the more accurate formula. The result of this calculation is τ P2 = 163psi, which shows that the shear stress obtained from the approximate formula is satisfactory. D) The bending moment M2 produces a compressive stress at point A but no stress at point B. The stress at A is σM2 = M2 b / 2Ι = (41600lb.in)(6in) / (2)(55.92in4) = 2232psi. This stress is also shown in the above figure. Stress Elements Each element is oriented so that the y-axis is vertical (i.e. parallel to the longitudinal axis of the post) and the x-axis is horizontal axis Point A : The only stress in point A is a compressive stress σa in the y direction σA = σP1 + σM1 + σM2 σA = 295psi + 1564psi + 2232psi = 4090psi (compression) Thus, this element is in uniaxial stress. Principal Stresses and Maximum Shear Stress σx = 0 σy = - σa = - 4090psi τ xy = 0 Since the element is in uniaxial stress, σ1 = σ x and σ2 = σy = - 4090psi And the maximum in-plane shear stress is τ max = (σ1 - σ2) / 2 = ½ (4090psi) = 2050psi The maximum out-of-plane shear stress has the same magnitude. Point B: Here the compressive stress in the y direction is σB = σP1 + σ M1 σB = 295psi + 1564psi = 1860psi (compression) And the shear stress is τB = τP2 = 160psi The shear stress acts leftward on the top face and downward on the x face of the element. Principal Stresses and Maximum Shear Stress σx =0 2 ⎛σx +σy ⎞ ⎛σx −σy ⎞ σy = - σB = - 1860psi σ 1, 2 = ⎜⎜ ⎟⎟ ± ⎜⎜ ⎟⎟ + (τ xy )2 2 2 ⎝ ⎠ ⎝ ⎠ τxy = - τP2 = - 160psi 944psii Substituting σ1,2 = - 930psi +/- 944ps σ1 = 14psi σ2 = - 1870psi and The maximum in-plane shear stresses can be obtained from the equation Because the principal stresses have opposite signs, the maximum in-plane shear stresses are larger than the maximum out-of-plane shear stresses. Then, τmax = 944psi. 2 τ MAX = ⎛σ x ⎞ ⎜ ⎟ + (τ ⎝ 2 ⎠ ) 2 xy = 944 psi Three forces are applied to a short steel post as shown. Determine the principle stresses, principal planes and maximum shearing stress at point H. 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 ) M x = −8.5 kN ⋅ m M y =0 M z = (30 kN )(0.100 m ) = 3 kN ⋅ m A = (0.040 m )(0.140 m ) = 5.6 ×10 −3m 2 Note: Section properties, 1 (0.040 m )(0.140 m )3 = 9.15 ×10− 6 m 4 I x = 12 Evaluate the stresses at H. 1 (0.140 m )(0.040 m )3 = 0.747 × 10−6 m4 I z = 12 Normal stress at H. σy =+ = Q = A1y1 = [(0.040 m )(0.045 m )](0.0475 m ) ( ) ( Vz Q 75 kN ) 85.5 ×10 −6 m 3 = τ yz = I xt 9.15 × 10−6 m4 (0.040 m ) ( = 17.52 MPa 50 kN 5.6 × 10-3 m2 − Shear stress at H. = 85.5 × 10− 6 m3 P Mz a Mx b + − Ix A Iz ) + (3 kN ⋅ m )(0.020 m ) 0.747 × 10 −6 m 4 (8.5 kN ⋅ m )(0.025 m ) 9.15 × 10− 6 m4 = (8.93 + 80.3 − 23.2 )MPa = 66.0 MPa Calculate principal stresses and maximum shearing stress. τ max = R = 33.02 + 17.522 = 37.4 MPa σ max = OC + R = 33.0 + 37.4 = 70.4 MPa σ min = 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 min = −7.4 MPa θ p = 13.98° The cantilever tube shown is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276MPa. We wish to select a stock size tube (according to the table below). Using a design factor of n=4. The bending load is F=1.75kN, the axial tension is P=9.0kN and the torsion is T=72N.m. What is the realized factor of safety? Consider the critical area ( top surface). σ VM ≤ Sy n = σx = 0.276 GPa = 0.0690GPa 4 P Mc + A I Maximum bending moment = 120F d 120mm × 1.75kNx⎛⎜ ⎞⎟ 9 kN ⎝ 2⎠ + σx = A I ⎛d ⎞ 72 × ⎜ ⎟ Tr ⎝ 2 ⎠ = 36d τ zx = = J J J σ VM = (σ x2 + 3τ zx2 ) 1 2 For the dimensions of that tube n= Sy σ VM = 0.276 = 4.57 0.06043 A certain force F is applied at D near the end of the 15-in lever, which is similar to a socket wrench. The bar OABC is made of AISI 1035 steel, forged and heat treated so that it has a minimum (ASTM) yield strength of 81kpsi. Find the force (F) required to initiate yielding. Assume that the lever DC will not yield and that there is no stress concentration at A. Solution: 1) Find the critical section The critical sections will be either point A or Point O. As the moment of inertia varies with r4 then point A in the 1in diameter is the weakest section.