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Unformatted text preview: Sec_2.8-2.12.qxd 244 9/25/08 11:44 AM Page 244 CHAPTER 2 Axially Loaded Members STRESSES From Eq. (3): FB
Wire C has the larger stress. Therefore, it will yield first. 2( sB (a) YIELD LOAD
C FC YA sY
2 (From Eq. 7) 1
2Y FB 1
2 a sYA b + 3(sYA)
PY YA YA) 5
From Eq. (2):
dB dD From Eq. (3): P PP 3( From Eq. (4):
2 sB Y P YA) dP 2dB 2sYL
E ; (c) LOAD-DISPLACEMENT DIAGRAM
PP ; From Eq. (4): P 5
2 Y FB L
From Eq. (2):
dD dY sY L
E 2dB ; (b) PLASTIC LOAD
At the plastic load, both wires yield.
B Y C FB FC YA Problem 2.12-10 Two cables, each having a length L of approximately 40 m, support a l
oaded container of weight W (see figure). The cables, which have effective cross-sectional area
A 48.0 mm2 and effective modulus of elasticity E 160 GPa, are identical except that one
cable is longer than the other when they are hanging separately and unloaded. The difference
in lengths is d 100 mm. The cables are made of steel having an elastoplastic stress-strain
diagram with Y 500 MPa. Assume that the weight W is initially zero and is slowly increased
by the addition of material to the container.
L (a) Determine the weight WY that first produces yielding of the shorter cable. Also, determine
the corresponding elongation Y of the shorter cable.
(b) Determine the weight WP that produces yielding of both cables. Also, determine the
elongation P of the shorter cable when the weight W just reaches the value WP.
(c) Construct a load-displacement diagram showing the weight W as ordinate and the
elongation of the shorter cable as abscissa. (Hint: The load displacement diagram is
not a single straight line in the region 0 W WY.)
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.
- Fall '11