Question

Two 6-in wide wooden boards are to be joined by splice plates that will be fully glued on the contact surfaces.The glue to be used can safely provide a shear strength of 120 psi. Determine the smallest allowable length L that can be used for the splice plates for an applied load of P = 15,000 lb. Note that a gap of 0.5 in is required between boards 1 and 2.which one of these is the right answer and show work to prove why.a) 19.7 inb) 15.7 inc) 24.3 ind) 21.3 ine) 11.6 in

Answer 1

Answer:

D

Explanation:

The smallest possible allowable length that can be used:

Calculate area of contact between plates A

A = 2 * 6 * a

A = 12a in^2

Where a is an arbitrary constant length along the direction L.

Shear stress = Shear force / Area

Area = 15000 lb / 120 psi = 12*a

Evaluate a

a = 10.411 in

Total Length = 2*a + gap = 2*10.411 + 0.5 = 21.33 in

Answer: L = 21.33