Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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- Two cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that the average normal stress must not exceed 175 MPa in rod AB and 150 MPa in rod BC, determine the smallest allowable values of d1 and d2. 300 mm 250 mm A B C -d₁ 40 kN -d₂. 30 kNarrow_forwardQ2: The pin-coected structure shown in figure below consists of a rigid member BED and three steel rods (E- 200 GPa). Each of the rods AB and CD has a 200 mm cross-sectional area and rod EF has a 625 mm2 cross-sectional area. If a load of P- 36 KN is applied at E, determine: (a) the change in length of rod EF. (b) the stress in each rod. 300 mm P-36 kN 300 mm 400 mm * 500 mmarrow_forward1/2 Problem 1: Two solid cylindrical rods AB and BC are welded together at B and loaded as shown in. If the stress must not exceed 14OMPA in either rod, determine the smallest allowable value of the diameter of rod AB and ВС. 300mm B 40 kN 250mm - d2 30 kNarrow_forward
- The C83400-red-brass rod AB and 2014-T6- aluminum rod BC are joined at the collar B and fixed connected at their ends. If there is no load in the members when T = 10 °C, determine the average normal stress in each member when T2 = 45 °C. Also, how far will the collar be displaced? The cross-sectional area of each member is 1130 mm?. B. -1 m - -0.6 m-arrow_forward2. Axial loads are applied to the compound rod that is composed of an aluminum segment rigidly connected between steel and bronze segments. What is the stress in each material given that P = 10 kN? Specify whether tension or compression. Bronze Aluminum Steel A =400 mm2 A= 600 mm2 A = 300 mm2 2P ЗР 4P P -3 m 5 m 4 m Stress in Bronze (MPa) %D Stress in Aluminum (MPa) %3D Stress in Steel (MPa)arrow_forward7.A cast-iron machine part is acted upon by the 3kN.m couple shown. Knowing that E=165GPA and neglecting the effect of fillets. i) Determine the maximum tensile and compressive stress in the casting ii) Determine the radius of curvature of the casting. 90 mm 20 mm 40 mm 30 mm M = 3 kN marrow_forward
- The rigid bar AB, attached to two vertical rods as shown in Fig. is horizontal before the load P is applied. Determine the maximum value for P If the stress must not exceed: 160 MPa for steel and 80 MPa for Bronze. Then determine the vertical movement of P. BRONZE L-5 m A 500 mm E-80 GPa STEEL L-4 m A-400 mm E-200 GPa 3 m. 5 m. VParrow_forwardthe assembly consists of three disk a b and c that are used to support the load of 140kn if d1=25mm and d2=70mm and d3=30mm determine the compressive stress in disk A, DISK B AND DISK Carrow_forwardTwo steel rods are welded together (see figure);the seam is oriented at angle θ = 50°. The stresses onthe rotated element are σ x1 = 10 ksi, σy1 = -12 ksi,and τx1y1 = -5 ksi. Find the state of plane stress onthe element if it is rotated clockwise to align the x1axis with the longitudinal axis of the rods.arrow_forward
- The light rigid bar ABCD shown below is pinned at B and connected to two vertical rods. Assuming that the bar was initially horizontal and the rods stress-free, determine the stress in each rod after the load P = 20 kN is applied. steel L=1.5 m A=300 mm2 2 m- E= 200 GPa B D Aluminun L=2 m P = 20 kN - 1m A= 500 mm2 E= 70 GPaarrow_forwardThe compound shaft consists of a solid aluminum segment (1) and a hollow brass segment (2) that are connected at flange B and securely attached to rigid supports at A and C. Aluminum segment (1) has a diameter of 0.55 in., a length of L₁ = 40 in., a shear modulus of 3800 ksi, and an allowable shear stress of 5.8 ksi. Brass segment (2) has an outside diameter of 0.45 in., a wall thickness of 0.15 in., a length of L₂ = 30 in., a shear modulus of 6400 ksi, and an allowable shear stress of 8.2 ksi. Determine: (a) the allowable torque Te that can be applied to the compound shaft at flange B. (b) the magnitudes of the internal torques in segments (1) and (2). (c) the rotation angle of flange B (relative to support A) that is produced by the allowable torque TB. L₁ L2 TB (1) Answer: (a) TB= (b) T₁ = T2= (c) OB= B lb-ft lb-ft lb-ft rad (2) Carrow_forwarda=2 b=5 80*a=160 90*b=450 100*b=500arrow_forward
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