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3 years ago

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Three full-size 50 × 100-mm boards are nailed together to form a beam that is subjected to a vertical shear of 1500 N. Knowing t

hat the allowable shearing force in each nail is 400 N, determine the largest longitudinal spacing s that can be used between each pair of nails. (Round the final answer to one decimal place.)

2 answers:

EleoNora [17]3 years ago

8 0

Answer:

<u><em>note:</em></u>

<u><em>solution is attached due to error in mathematical equation. please find the attachment</em></u>

Nadya [2.5K]3 years ago

5 0

Answer:

Largest longitudinal spacing = 60mm

Explanation:

If we can find the shear stress and allowable load in one nail, then we can compute the needed spacing of the nails.

Now, moment of Inertial of rectangular section is given as;

I = bh³/12

Since nailed together, the effective total height will be 150mm

Thus, I = (100 x 150³)/12 = 28.125 x 10^6 mm⁴ = 28.125 x 10^(-6) m⁴

Also, the area of the section = 100 x 50 = 5000 mm²

Now, from the centre of the 150mm length to the centre of the 50mm joined part, the centroid is calculated as 50mm.

Thus, centroid ȳ1 = 50mm

So moment of area = 5000mm² x 50mm = 250,000 mm³ or 250 x 10^(-6) m³

Formula of shear stress(q) = VQ/I

q = (1500 x 250 x 10^(-6))/(28.125 x 10^(-6)) = 13.333 x 10³ N/m

For the required spacing, we'll use the formula;

qs = (Shearing force in each nail) x 2

Where q = shear stress and s = spacing

Thus;s = (400 x 2)/13.333 x 10³ = 60 x 10^(-3)m = 60mm

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3 years ago

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Answer:

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Let $U_B$ be the initial velocity of rod B

Let $V_A$ be the final velocity of rod A

Let $V_B$ be the final velocity of rod B

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Substitute equation (2) into equation (1)

$M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B$ ..............(3)

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$V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}$ ....................(4)

From equation (2):

$V_B = V_A + e(U_A -U_B)$ ......(5)

Substitute equation (5) into (1)

$M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))$ ..........(6)

Solving for $V_A$ in equation (6) above:

$V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}$ .........(7)

b)

$M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?$

Rod A is said to be at rest after the impact, $V_A = 0 m/s$

Substitute these parameters into equation (7)

$0 = \frac{(2 - 0.65*1)U_A - (1*3)(1+0.65)}{2+1}\\U_A = 3.66 m/s$

To calculate the final velocity, $V_B$ , substitute the given parameters into (4):

$V_B = \frac{(2*3.66)(1+0.65) - (1 - (0.65*2))*3}{2+1}\\V_B = 4.32 m/s$

c) Impulse, $I = M_AV_A + M_BV_B - (M_AU_A + M_BU_B)$

$I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))$

I = 0 $kg m/s^2$

d) % $\triangle KE = \frac{(0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2)}{0.5 M_A U_A^2 + 0.5 M_B U_B^2} * 100\%$

% $\triangle KE = \frac{((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2))}{ (0.5 *2*3.66^2) + 0.5*1*(-3)^2)} * 100\%$

% $\triangle KE = -47.85 \%$

7 0

4 years ago