ch01-03 stress & strain & properties

01 Solutions 46060 5/6/10 2:43 PM Page 71
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1–106. The bearing pad consists of a 150 mm by 150 mm
block of aluminum that supports a compressive load of
6 kN. Determine the average normal and shear stress acting
on the plane through section a–a. Show the results on a
differential volume element located on the plane.
Equation of Equilibrium:
6 kN
a
+Q©F x = 0; V a - a - 6 cos 60° = 0 V a - a = 3.00 kN
a+©F y = 0; N a - a - 6 sin 60° = 0 N a - a = 5.196 kN
30
Averge Normal Stress And Shear Stress: The cross sectional Area at section a–a is
A = a 0.15 b(0.15) = 0.02598 m2.
sin 60°
a
150 mm
s a - a = N a - a
A = 5.196(103 )
= 200 kPa
0.02598
Ans.
t a - a = V a - a
A = 3.00(103 )
= 115 kPa
0.02598
Ans.
1–107. The yoke-and-rod connection is subjected to a
tensile force of 5 kN. Determine the average normal stress
in each rod and the average shear stress in the pin A
between the members.
5 kN
40 mm
For the 40 – mm – dia rod:
s 40 = P A = 5 (103 )
p
4
(0.04)2
= 3.98 MPa
Ans.
A
25 mm
30 mm
For the 30 – mm – dia rod:
5 kN
s 30 = V A = 5 (103 )
p
4
(0.03)2
= 7.07 MPa
Ans.
Average shear stress for pin A:
t avg = P A = 2.5 (103 )
= 5.09 MPa
(0.025)2
p
4
Ans.
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