Abronze pipe (1) is to be connected to an aluminum alloy pipe (2) at flange B. When put in place, however, a gap of A = 0.35 in. exists etween the two pipes. Bronze pipe (1) has an elastic modulus of E1 = 16,000 ksi, a cross-sectional area of A, = 2.66 in.2, and a length of L1 = 6.7 ft. Aluminum alloy pipe (2) has an elastic modulus of E2 = 10,000 ksi, a cross-sectional area of A2 = 1.37 in.2, and a length of 2 = 10.2 ft. If bolts are inserted in the flanges and tightened so that the gap at B is closed, determine: a) the normal stresses produced in each of the members. b) the final position of flange B with respect to support A. L2 B (2) (1)
Abronze pipe (1) is to be connected to an aluminum alloy pipe (2) at flange B. When put in place, however, a gap of A = 0.35 in. exists etween the two pipes. Bronze pipe (1) has an elastic modulus of E1 = 16,000 ksi, a cross-sectional area of A, = 2.66 in.2, and a length of L1 = 6.7 ft. Aluminum alloy pipe (2) has an elastic modulus of E2 = 10,000 ksi, a cross-sectional area of A2 = 1.37 in.2, and a length of 2 = 10.2 ft. If bolts are inserted in the flanges and tightened so that the gap at B is closed, determine: a) the normal stresses produced in each of the members. b) the final position of flange B with respect to support A. L2 B (2) (1)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:A bronze pipe (1) is to be connected to an aluminum alloy pipe (2) at flange B. When put in place, however, a gap of A = 0.35 in. exists
between the two pipes. Bronze pipe (1) has an elastic modulus of E1 = 16,000 ksi, a cross-sectional area of A1 = 2.66 in.?, and a length
of L1 = 6.7 ft. Aluminum alloy pipe (2) has an elastic modulus of E2 = 10,000 ksi, a cross-sectional area of A2 = 1.37 in.2, and a length of
L2 = 10.2 ft. If bolts are inserted in the flanges and tightened so that the gap at B is closed, determine:
(a) the normal stresses produced in each of the members.
(b) the final position of flange B with respect to support A.
L2
В
(1)
Answer:
(a) 01 =
ksi
02 =
ksi
(b) ÕB/A =
in.
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