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Unformatted text preview: 11Ch11.qxd 9/27/08 3:39 PM Page 887 887 SECTION 11.6 The Secant Formula Solution 11.6-11 Column with two loads Pinned-end column. (a) MAXIMUM COMPRESSIVE STRESS W 12 * 87 L
E 18 ft 216 in.
P2 75 k
sY 36 ksi
29,000 ksi P P1 + P2 255 k A 25.6 in.2 I r2 I
2 c P
c 1 + 2 sec a
2r A EA
r Secant formula (Eq. 11-59): DATA e
I1 5.0 in. smax Substitute into Eq. (1): P2s
740 in.4 ; 13.4 ksi (b) FACTOR OF SAFETY WITH RESPECT TO YIELDING 1.471 in.
d (1) smax 12.53 in. pY
[1 + 0.3188sec(0.02332 1PY)]
sY 36 ksi P PY Substitute into Eq. (1):
2r A EA 9.961 ksi smax r 2 5.376 in.
0.3188 Solve numerically:
P 255 k PY
P n 0.3723 664.7 k 664.7 k
255 k 2.61 ; Problem 11.6-12 The wide-flange pinned-end column shown in
the figure carries two loads, a force P1 450 kN acting at the
centroid and a force P2 270 kN acting at distance s 100 mm,
from the centroid. The column is a W 250 * 67 shape with
L 4.2 m, E 200 GPa, and sY 290 MPa.
(a) What is the maximum compressive stress in the column?
(b) If the load P1 remain at 450 kN, what is the largest permissible
value of the load P2 in order to maintain a factor of safety of 2.0 with respect to yielding? Solution 11.6-12
W 250 * 67 L P1 450 kN P2 270 kN s 100 mm E 200 GPa sy 290 MPa P 720 kN A 8580 mm2 d 257 mm c d
2 P c I
smax P1 + P2
P e I
AA I1 128.5 mm P
a 1 + 2 sec a
2r A EA
r (a) MAXIMUM COMPRESSION STRESS 4.2 m e 37.5 mm
103 * 106 mm4 I
r 109.6 mm 120.4 MPa ; (b) LARGEST VALUE OF LOAD P2 WHEN
P1 450 kN n 2.0
sy n(P1 + P2)
Solve for P2 Py
A P22 a1
* c1 P2 Py
2r A EA L n1P1 P22
ec r 2 sec a ec 2 sec c ...
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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