Compound axial member ABC has a uniform diameter of d= 1.50 in. • Segment (1) is an aluminum [E1 = 10000 ksi] alloy. The length of segment (1) is L1 - 90 in. • Segment (2) is a copper [E2 - 17000 ksi] alloy. The length of segment (2) is L2 = 126 in. When axial force Pis applied, a strain gage attached to copper segment (2) measures a normal strain of ɛ2- 2100 µin./in. in the longitudinal direction. What is the total elongation of member ABC?

Compound axial member ABC has a uniform diameter of d= 1.50 in. • Segment (1) is an aluminum [E1 = 10000 ksi] alloy. The length of segment (1) is L1 - 90 in. • Segment (2) is a copper [E2 - 17000 ksi] alloy. The length of segment (2) is L2 = 126 in. When axial force Pis applied, a strain gage attached to copper segment (2) measures a normal strain of ɛ2- 2100 µin./in. in the longitudinal direction. What is the total elongation of member ABC?

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

H.W 3 Two aluminum alloy plates (2) are attached to the sides of a wooden beam (1) as shown in Figure 3. mensions of the composite cross section are b1 = 80 mm, d¡ = 240 mm, b2 = 10 mm, d2 = b2 The di- |di 120 mm, and a = 60 mm. Deter- (2) (2) %3D mine the maximum bending (1) a stresses produced in both the wooden beam and the aluminum b1 plates if a bending moment FIGURE 3 +6,000 N m is applied M2 %3D about the z axis. Assume that Ej = 12.5 GPa and E2 = 70 GPa.
Problem # 2. At room temperature (20°C) a 0.5-mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 140 °C, determine (a) the normal stress in the aluminum rod Ans: A -300 mm Aluminum A = 2000 mm² E = 75 GPa α = 23 x 16-6/°C 0.5 mm -250 mm Stainless steel A = 800 mm² B E = 190 GPa a = 17.3 x 10-6/°C
Two polymer bars are connected to a rigid plate at B, as shown. Bar (1) has a cross-sectional area of 1.58 in. and an elastic modulus of 2430 ksi. Bar (2) has a cross-sectional area of 0.714 in and an elastic modulus of 4110 ksi. Assume Ly-18 in., Lz-47in, Q=6,4 kips, P-3.2 kips, and R-128 kips. Determine the horizontal deflection of end Crelative to end A. (1) Answer: uCA in.
A 5.2-m-long aluminum tube (1) is to be connected to a 2.2-m-long bronze pipe (2) at B. When put in place, however, a gap of 9 mm exists between the two members, as shown. Aluminum tube (1) has an elastic modulus of 69 GPa and a cross-sectional area of 1980 mm2. Bronze pipe (2) has an elastic modulus of 93 GPa and a cross-sectional area of 3970 mm2. If bolts are inserted in the flanges and tightened so that the gap at B is closed, determine: (a) the normal stresses ?1, ?2 produced in each of the members. (b) the displacement (change in length of material (2), up is positive) uB of flange B with respect to support A. 5.2 m 2.2 m Answers: (a) 0₁ = (b) Ug = C (1) B (2) -95.40 1.126 9 mm MPa, 0₂ = mm. -47.58 MPa
The pin-connected structure shown in the figure consists of a rigid beam ABCD and two supporting bars. Bar (1) is a bronze alloy [E = 102 GPa] with a cross-sectional area of A₁ = 200 mm². Bar (2) is an aluminum alloy [E = 72 GPa] with a cross-sectional area of A₂ = 640 mm². All bars are unstressed before the load P is applied; however, there is a 2-mm clearance in the pin connection at A. Assume L₁ = 1880 mm, a = 1060 mm, b = 620 mm, c = 480 mm, and L₂ = 1270 mm. If a load of P = 71 kN is applied at B, determine (a) the normal stresses in both bars (1) and (2). (b) the downward deflection of point A on the rigid bar. (1) L₁ b tc B (2) D L2
Problem 1 An axial force of P is applied to the center of the rigid end plate shown below. If the portion of the load carried by the aluminum plates is half the portion of the load carried by the brass core, (a) determine the value of h, (b) the total load if the stress in the brass is 50 MPa. Brass core (E = 105 GPa) Aluminum plates (E = 70 GPa) Rigid end plate 300 mm 60 mim 40 mm
Compound axial member ABC has a uniform diameter of d = 1.25 in. • Segment (1) is an aluminum [E1 = 10000 ksi] alloy. The length of segment (1) is L1 = 100 in. • Segment (2) is a copper [E2 = 17000 ksi] alloy. The length of segment (2) is L2 = 140 in. When axial force Pis applied, a strain gage attached to copper segment (2) measures a normal strain of ɛ2 = 2300 uin./in. in the longitudinal direction. What is the total elongation of member ABC? L2 (1) Aluminum B (2) Copper Answer: in, AL = i
The following information about the o-e curve is given for a steel alloy. E = 0.001527 at o = 300 MPa and ɛ = 0.003054 for o = 600 MPa. (a) Draw the stress-strain diagram and calculate the E for this alloy.
At room temperature (20°C) a 0.7-mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 150°C, determine: (a.)The normal stress in the aluminum rod, (b.) The change in length of the aluminum rod. 0.7 mm 300 mm A Aluminum A =2,000 mm² E= 75 GPa α = 23 x 106/°C 250 mm B Stainless steel A = 800 mm² E = 190 GPa a = 17.3 x 10-6/°C
TWO SOLID CYLINDRICAL RODS ARE JOINED AT B AND LOADED AS SHOWN. ROD AB IS MADE OF STEEL (E = 210 GPa) AND ROD BC OF BRASS (E = 108 GPa). DETERMINE (A) THE TOTAL DEFORMATION OF THE COMPOSITE ROD ABC, (B) THE DEFLECTION OF POINT B. 40 KN 400 mm 600 mm A B C 30 mm 50 KN 50 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
Question 2) Knowing that a 0.02-in. gap exists when the temperature is 75 °F, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to -11 ksi, (b) the corresponding exact length of the aluminum bar. 0.02 in. -14 in. Bronze A = 2.4 in² E = 15 x 10 psi a = 12 x 10-6/°F -18 in.- Aluminum A = 2.8 in² E = 10.6 x 105 psi a = 12.9 x 10-6/°F