Question

Three 4 cm thick and 10 cm wide boards are connected together by two parallel rows of uniformly distributed nails separated by longitude distance δto form a beam that is subject to constant vertical shear of 1000 N. Knowing the maximum allowed shearing force for each nail is 200 N, determine the largest longitude separation δbetween nails?

Answer 1

Answer:

The largest longitude separation is 3.6 cm.

Explanation:

Given that,

Thickness$\bar{y}= 4\ cm$

Wide = 10 cm

Vertical shear = 1000 N

Shearing force = 200 N

We need to calculate the shear flow

Using formula of shear flow

$q=\dfrac{VQ}{I}$

Where, q = shear flow

V = shear force

$Q = \bar{y}\times A$

Where, A = area of cross section

$\bar{y}$ =distance from natural axis to centroid of A

I = moment of inertia

We need to calculate the Area

Using formula of area

$A=4\times10=40 cm^2$

We need to calculate the moment of inertia

$I=\dfrac{bd^3}{12}$

$I=\dfrac{10\times12^3}{12}$

$I=1440\ cm^4$

Put the value into the formula of shear flow

$q=\dfrac{1000\times40\times4}{1440}$

$q=111.11\ N/cm$

We need to calculate the largest longitude separation

Using formula of separation

$q=\dfrac{F}{\delta}$

Put the value into the formula

$111.11=\dfrac{2\times200}{\delta}$

$\delta=\dfrac{2\times200}{111.11}$

$\delta=3.6\ cm$

Hence, The largest longitude separation is 3.6 cm.