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ch01-03 stress & strain & properties

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01 Solutions 46060 5/6/10 2:43 PM Page 34<br />

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1–50. The block is subjected to a compressive force of<br />

2 kN. Determine the average normal and average shear<br />

<strong>stress</strong> developed in the wood fibers that are oriented along<br />

section a–a at 30° with the axis of the block.<br />

a<br />

50 mm<br />

150 mm<br />

2 kN 2 kN<br />

a<br />

30<br />

Force equilibrium equations written perpendicular and parallel to section a–a gives<br />

+Q©F x¿ = 0; V a - a - 2 cos 30° = 0 V a - a = 1.732 kN<br />

+a©F y¿ = 0; 2 sin 30° - N a - a = 0 N a - a = 1.00 kN<br />

The cross sectional area of section a–a is A = a 0.15 b(0.05) = 0.015 m2. Thus<br />

sin 30°<br />

(s a - a ) avg = N a - a<br />

Ans.<br />

A = 1.00(1<strong>03</strong> )<br />

= 66.67(10 3 )Pa = 66.7 kPa<br />

0.015<br />

(t a - a ) avg = V a - a<br />

A = 1.732(1<strong>03</strong> )<br />

0.015<br />

= 115.47(10 3 )Pa = 115 kPa<br />

Ans.<br />

34

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