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3 years ago

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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

1 answer:

Zigmanuir [339]3 years ago

8 0

Answer:

3.03 INCHES

Explanation:

According to ASTM D198 ;

Modulus of rupture = ( M / I ) * y ----- ( 1 )

M ( bending moment ) = R * length of span / 2

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

I ( moment of inertia ) = bd^3 / 12

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

b = 2 in , d = ?

length of span = 4 * 12 = 48 inches

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

y ( centroid distance ) = d / 2 inches

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 ( divide both sides with d )

4d^2 = 34560000 / 940300

d^2 = 9.188 ∴ Value of d ≈ 3.03 in

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3 years ago