Q1: A solid aluminum [E = 70 GPa] rod (1) is connected to a solid bronze [E = 100 GPa] rod at flange B, as shown in the Figure. Aluminum rod (1) has an outside diameter of 35 mm and bronze rod (2) has an outside diameter of 20 mm. The normal stress in the aluminum rod must be limited to 160 MPa, and the normal stress in the bronze rod must be limited to 110 MPa. Determine: (a) the maximum downward load P that may be applied at flange B. (b) the deflection of flange B at the load determined in part (a). (2) 340 mm B 175 mm (1)

Q1: A solid aluminum [E = 70 GPa] rod (1) is connected to a solid bronze [E = 100 GPa] rod at flange B, as shown in the Figure. Aluminum rod (1) has an outside diameter of 35 mm and bronze rod (2) has an outside diameter of 20 mm. The normal stress in the aluminum rod must be limited to 160 MPa, and the normal stress in the bronze rod must be limited to 110 MPa. Determine: (a) the maximum downward load P that may be applied at flange B. (b) the deflection of flange B at the load determined in part (a). (2) 340 mm B 175 mm (1)

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

6
A hollow titanium [G = 43 GPa] shaft has an outside diameter of D = 46 mm and a wall thickness of t = 1.23 mm. The maximum shear stress in the shaft must be limited to 118 MPa. Determine: (a) the maximum power P that can be transmitted by the shaft if the rotation speed must be limited to 22 Hz. (b) the magnitude of the angle of twist in a 830-mm length of the shaft when 17 kW is being transmitted at 9 Hz. Answers: (a) P = (b) = i kW. O
Normal stress is calculated dividing the reaction by the cross-sectional area. Is this approach valid for any point of a structural member? Why or why not? To answer this question, consider the loading point, cross- sectional shape, etc.
Normal stress is calculated dividing the reaction by the cross-sectional area. Is this approach valid for any point of a structural member? Why or why not? To answer this question, consider the loading point, cross- sectional shape, etc.
please just solve the second part
1. At a temperature of 5°C, a 3-mm gap exists between the end of polymer bar (1) and support A as shown. Bar (1) is 50 mm wide and 20 mm thick (E = 800 MPa, a = 140 x 10-6/°C). Bar (2) is 75 mm wide and 25 mm thick (E = 2.7 GPa, a = 67 x 106/°C). The supports at A and C are rigid (i.e., fixed). Determine: a) the temperature at which the normal stress in bar (2) will be equal to -3.0 MPa, and b) the lengths of the two polymer bars at this temperature. (1) 3-mm gap 50 mm 500 mm B 75 mm 400 mm (2)
Two triangular wedges are glued together as shown in the figure. The stress acting normal to the interface, o, is MPa. 100 MPa 100 MPa on 45% 100 MPa -100 MPa
The solid shaft, made of three different materials, carries the two torques shown. (a) Calculate the maximum shear stress in each material. (b) Find the angle of rotation of the free end of the shaft. The shear moduli are 28 GPa for the aluminum, 83 GPa for steel and 35 GPa for bronze. Aluminum 100 mm -3 m- 4 kN.m Steel 75 mm -2 m 1.5 kN - m Bronze |--15m-l
Determine induced stress
In a laboratory test of a beam loaded by end couples, the fiber at layer AB as shown are found to increase 20 x 103 mm while those at CD decrease 60 x 103 mm in the 230- mm-gage length. Using E = 180 GPa, determine the flexural stress at the top fiber. Answer must be in MPa. Given: a = 40 mm, b = 130 mm, and c= 70 mm. gage length a mm A B. b mm c mm
Narrow bars of aluminum are bonded to the two sides of a thick steel plate as shown. Initially, at T₁ = 70°F, all stresses are zero. Knowing that the temperature will be slowly raised to T₂ and then reduced to T₁, determine (a) the highest temperature T₂ that does not result in residual stresses, (b) the temperature T₂ that will result in a residual stress in the aluminum equal to 58 ksi. Assume aa = 12.8 x 10-6/°F for the aluminum and a = 6.5 × 10-6/°F for the steel. Further assume that the aluminum is elastoplastic with E = 10.9 × 106 psi and ay = 58 ksi. (Hint: Neglect the small stresses in the plate.) Fig. P2.121