ch01-03 stress & strain & properties
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01 Solutions 46060 5/6/10 2:43 PM Page 42<br />
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1–59. The open square butt joint is used to transmit a<br />
force of 50 kip from one plate to the other. Determine the<br />
average normal and average shear <strong>stress</strong> components that<br />
this loading creates on the face of the weld, section AB.<br />
50 kip<br />
30<br />
30<br />
Equations of Equilibrium:<br />
a + © F y = 0; N - 50 cos 30° = 0 N = 43.30 kip<br />
+ Q© F x = 0; -V + 50 sin 30° = 0 V = 25.0 kip<br />
A<br />
B<br />
6 in.<br />
2 in.<br />
50 kip<br />
Average Normal and Shear Stress:<br />
2<br />
A¿ =a b(6) = 13.86 in2<br />
sin 60°<br />
s = N = 43.30 = 3.125 ksi<br />
A¿ 13.86<br />
t avg = V = 25.0 = 1.80 ksi<br />
A¿ 13.86<br />
Ans.<br />
Ans.<br />
*1–60. If P = 20 kN, determine the average shear <strong>stress</strong><br />
developed in the pins at A and C. The pins are subjected to<br />
double shear as shown, and each has a diameter of 18 mm.<br />
C<br />
Referring to the FBD of member AB,Fig.a<br />
a+©M A = 0; F BC sin 30° (6) - 20(2) - 20(4) = 0 F BC = 40 kN<br />
: + ©F x = 0; A x - 40 cos 30° = 0 A x = 34.64 kN<br />
A<br />
2 m<br />
2 m<br />
30<br />
2 m<br />
B<br />
+ c ©F y = 0; A y - 20 - 20 + 40 sin 30° A y = 20 kN<br />
Thus, the force acting on pin A is<br />
F A = 2 A x 2 + A y 2 = 2 34.64 2 + 20 2 = 40 kN<br />
Pins A and C are subjected to double shear. Referring to their FBDs in Figs. b and c,<br />
V A = F A<br />
2 = 40<br />
2 = 20 kN V C = F BC<br />
= 40 = 20 kN<br />
2 2<br />
= 81(10 - 6 )p m 2 . Thus<br />
The cross-sectional area of Pins A and C are A A = A C = p 4 (0.0182 )<br />
t A = V A<br />
A A<br />
= 20(1<strong>03</strong> )<br />
81(10 - 6 )p = 78.59(106 ) Pa = 78.6 MPa<br />
t C = V C<br />
A C<br />
= 20(1<strong>03</strong> )<br />
81(10 - 6 )p = 78.59(106 ) Pa = 78.6 MPa<br />
Ans.<br />
Ans.<br />
P<br />
P<br />
42