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wel
2 months ago
12

A center-point bending test was performed on a 2 in. x d in. wood lumber according to ASTM D198 procedure with a span of 4 ft an

d the 4 in. side is positioned vertically. If the maximum load was 240 kips and the modulus of rupture was 940.3 ksi, what is the value of d
Engineering
1 answer:
pantera1 [306]2 months ago
8 0

Response:

3.03 INCHES

Clarification:

Following ASTM D198;

The formula for the modulus of rupture is: ( M / I ) * y ----- ( 1 )

Where M (bending moment) is calculated as R * span length / 2

= (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in

I (moment of inertia) is derived from the formula bd^3 / 12

= ( 2 )*( d )^3 / 12 = 2d^3 / 12

Given b = 2 in, and d is unknown

The span length = 4 * 12 = 48 inches

R = P / 2 = 240 * 10^3 / 2 = 120 * 10^3 Ib

y (distance to centroid) equals d / 2 inches

Refer back to equation ( 1 )

( M / I ) * y

940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2

= ( 288 * 10^4 * 12 ) / 2d^3 ) * d / 2

940300 = 34560000* d / 4d^3

4d^3 ( 940300 ) = 34560000 d (by dividing both sides by d )

4d^2 = 34560000 / 940300

d^2 = 9.188 ∴ Thus, d is approximately 3.03 in

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