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## Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a 1200-N vertical shear. Knowing that the allowable

Question

Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a 1200-N vertical shear. Knowing that the allowable shearing force in each nail is 700 N, determine the largest permissible spacing s between the nails.

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Physics
5 months
2021-08-03T00:55:37+00:00
2021-08-03T00:55:37+00:00 2 Answers
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## Answers ( )

Answer:Using equation

qs=F

q=VQ/I V=1200

q=998.45 N

qs=F

s=F/q

s=700/998.45

spacing=s=0.701 mAnswer:The question is incomplete. The mass of the board is needed in order to calculate the moment of inertia of the beam. 91.67 * 10⁶mm⁴ is used as the moment of inertia in the calculation below. You can always substitute to get your answer.

The spacing is

95.1mmExplanation:Given’vertical shear V = 1200NShearing force = 700NThickness of each board = 50mmThe spacing can be calculated using the formula;F = q * s—————————1

but, q = VQ/I———————————2

Therefore,

F = VQ/I * s——————————3where,

s = spacing

v = vertical shear

f = shearing force

I = moment of inertia

Making s subject formula in equation 3, we have

s = FI/VQ——————————-4but Q = YA

A = area = 50 *(150)

Y = 75

therefore, Q = 75 *50*150

Q= 5.625* 10⁶mm⁴

Substituting into equation 4, we have

s = 700 * 91.67 *10⁶) / (1200*5.625* 10⁶)

= 6.42*10¹⁰ /6.75*10⁸

= 95.1mm