02-03ChapGere.0011 - -DISPLACEMENT RELATIONS (Eqs. 4, 5) S...

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116 CHAPTER 2 Axially Loaded Members Problem 2.5-15 Wires B and C are attached to a support at the left-hand end and to a pin-supported rigid bar at the right-hand end (see figure). Each wire has cross-sectional area A 5 0.03 in. 2 and modulus of elasticity E 5 30 3 10 6 psi. When the bar is in a vertical position, the length of each wire is L 5 80 in. However, before being attached to the bar, the length of wire B was 79.98 in. and of wire C was 79.95 in. Find the tensile forces T B and T C in the wires under the action of a force P 5 700 lb acting at the upper end of the bar. Solution 2.5-15 Wires B and C attached to a bar B C 80 in. 700 lb b b b B C L = 80 in. P = 700 lb b b b P 5 700 lb A 5 0.03 in. 2 E 5 30 3 10 6 psi L B 5 79.98 in. L C 5 79.95 in. E QUILIBRIUM E QUATION T C ( b ) 1 T B (2 b ) 5 P (3 b ) 2 T B 1 T C 5 3 P (Eq. 1) D ISPLACEMENT DIAGRAM S B 5 80 in. 2 L B 5 0.02 in. S C 5 80 in. 2 L C 5 0.05 in. © M pin 5 0 `~ Elongation of wires: d B 5 S B 1 2 d (Eq. 2) d C 5 S C 1 d (Eq. 3) F ORCE
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Unformatted text preview: -DISPLACEMENT RELATIONS (Eqs. 4, 5) S OLUTION OF EQUATIONS Combine Eqs. (2) and (4): (Eq. 6) Combine Eqs. (3) and (5): (Eq. 7) Eliminate d between Eqs. (6) and (7): (Eq. 8) Solve simultaneously Eqs. (1) and (8): S UBSTITUTE NUMERICAL VALUES : (Both forces are positive, which means tension, as required for wires.) T C 5 420 lb 2 90 lb 1 450 lb 5 780 lb T B 5 840 lb 1 45 lb 2 225 lb 5 660 lb EA 5 L 5 2250 lb / in. T C 5 3 P 5 2 2 EAS B 5 L 1 4 EAS C 5 L T B 5 6 P 5 1 EAS B 5 L 2 2 EAS C 5 L T B 2 2 T C 5 EAS B L 2 2 EAS C L T C L EA 5 S C 1 d T B L EA 5 S B 1 2 d d B 5 T B L EA d C 5 T C L EA P = 700 lb b b b Pin T B T C B C L = 80 in. S B S C 2 d d A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Institute of Technology.

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