Problem

Sketch the intensity of the shear-stress distribution acting over the beam’s cross-secti...

Sketch the intensity of the shear-stress distribution acting over the beam’s cross-sectional area, and determine the resultant shear force acting on the segment AB . The shear acting at the section is V = 35 kip . Show that INA = 872.49 in4.

Step-by-Step Solution

Solution 1

Sketch the cross section of the beam.

10.jpg

Consider \(\bar{y}\) as the distance of the neutral axis \(x-x\) of the section from the bottom. Calculate the centroid for the cross section at neutral axis using the following equation:

$$ \bar{y}=\frac{A_{1} y_{1}+A_{2} y_{2}}{A_{1}+A_{2}} $$

Here, \(A_{1}\) and \(A_{2}\) are the areas of the sections \((1)\) and \((2)\) and \(y_{2}, y_{1}\) is the distance of the centroids for the sections (1) and (2).

Calculate the value \(y_{1}\) as follows:

$$ \begin{aligned} y_{1} &=\left(6+\frac{8}{2}\right) \\ &=10 \mathrm{in} \end{aligned} $$

Calculate the value \(y_{2}\) as follows:

$$ \begin{aligned} y_{2} &=\frac{6}{2} \\ &=3 \mathrm{in} . \end{aligned} $$

Substitute \((8 \times 8)\) in. \(^{2}\) for \(A_{1}, 10\) in for \(y_{1},(6 \times 2)\) in. \(^{2}\) for \(A_{2}\), and 3 in. for \(y_{2}\).

$$ \begin{aligned} \bar{y} &=\frac{(8 \times 8) \times 10+(6 \times 2) \times 3}{(8 \times 8)+(6 \times 2)} \\ &=8.895 \mathrm{in} . \end{aligned} $$

Determine moment of inertia \(I_{N 4}\) of the section about the neutral axis by apply the parallel axis theorem.

$$ \begin{aligned} I_{\mathrm{MA}} &=\left(I_{1}+A_{1} d_{\mathrm{x}}^{2}\right)+\left(I_{2}+A_{2} d_{2}^{2}\right) \\ &=\left[\frac{\left(b_{1}\right) \times\left(h_{\mathrm{t}}\right)^{3}}{12}+\left(A_{1}\right) \times\left(y_{1}-\bar{y}\right)^{2}+\frac{\left(b_{2}\right) \times\left(h_{2}\right)^{3}}{12}+\left(A_{2}\right) \times\left(\bar{y}-y_{2}\right)^{2}\right] \end{aligned} $$

Here, \(b_{1}\) is the width of the section (1), \(h_{1}\) is the height of the section (1), \(b_{2}\) is the width of the section (1), \(h_{2}\) is the height of the section \((2), d_{1}\) and \(d_{2}\) are the distance of the centroid from the to the neutral axis to that particular section, \(\bar{y}\) is the distance of the neutral axis \(x-x\) of the section from the bottom, and \(y_{2}, y_{1}\) are the distance of the centroids for the sections (1) and (2).

Substitute 8 in. for \(b_{1}, 2\) in. for \(b_{2}, 4\) in. for \(h_{1}\), and 2 in. for \(h_{2},(10-8.895)\) in. for \(d_{1}\), and \((2.667-2)\) in. for \(d_{2}\)

$$ \begin{aligned} I_{\text {MA }} &=\frac{(8) \times(8)^{3}}{12}+(8 \times 8) \times(10-8.895)^{2}+\frac{(2) \times(6)^{3}}{12}+(2 \times 6) \times(8.895-3)^{2} \\ &=872.49 \text { in. }^{4} \end{aligned} $$

Compute the shear stress in portion \(B C\) :

Calculate the moment of area about neutral axis.

$$ Q_{1}=A_{1} \times y_{1} $$

Here, \(A_{1}\) is the area of cross section above \(y_{1}\) attached to the beam and \(y_{1}\) is the distance of the centroid of section (1) from the neutral axis.

Substitute \(\left[\left(8+6-8.895-y_{1}\right) \times(8)\right]\) in. \(^{2}\) for \(A_{1}\) and \(\left(y_{1}+\frac{8+6-8.895-y_{1}}{2}\right)\) for \(y_{1}\)

$$ \begin{aligned} Q_{1} &=\left(8+6-8.895-y_{1}\right) \times(8) \times\left(y_{1}+\frac{8+6-8.895-y_{1}}{2}\right) \\ &=4 \times\left(26.061-y_{1}^{2}\right) \mathrm{in}^{3} \end{aligned} $$

Calculate the shear stress at a distance \(y_{1}\) from neutral axis using the following relation:

$$ \tau_{1}=\frac{V Q_{1}}{I_{N A} t_{1}} $$

Here, \(V\) is the shear stress, \(Q_{1}\) is the moment of area \(A_{1}\) about neutral axis, \(I_{N A}\) is the moment of inertial about neutral axis, and \(t_{1}\) is the thickness of the section at \(y_{1}\).

Substitute \(35 \times 10^{3} \mathrm{lb}\) for \(V, 4 \times\left(26.061-y_{1}^{2}\right)\) in. \({ }^{3}\) for \(Q_{1}, 872.49 \mathrm{in} .{ }^{4}\) for \(I_{N A}\) and 8 in. for \(t_{1}\).

$$ \begin{aligned} \tau_{1} &=\frac{\left(35 \times 10^{3}\right) \times 4 \times\left(26.061-y_{1}^{2}\right)}{(872.49) \times(8)} \ldots \ldots .(1) \\ &=20.0575 \times\left(26.061-y_{1}^{2}\right) \mathrm{psi} \end{aligned} $$

Consider the portion \(A B\).

Determine moment of area about neutral axis.

$$ Q_{2}=A_{2} \times y_{2} $$

Here, \(A_{2}\) is the area of cross section below \(y_{2}\) attached to the beam, and \(y_{2}\) is the distance of the centroid of the section from the neutral axis.

Substitute \(\left[\left(8.895\right.\right.\) in \(\left.-y_{2}\right) \times(2\) in \(\left.)\right]\) for \(A_{2}\) and \(\left(y_{2}+\frac{8.895 \text { in }-y_{2}}{2}\right)\) for \(y_{2}\).

$$ \begin{aligned} Q_{2} &=\left(8.895 \text { in }-y_{2}\right) \times(2 \text { in }) \times\left(y_{2}+\frac{8.895 \text { in }-y_{2}}{2}\right) \\ &=\left(79.121-y_{2}^{2}\right) \text { in }^{3} \end{aligned} $$

Calculate the shear stress \(\tau_{2}\) at a distance \(y_{2}\) from neutral axis using the following relation:

$$ \tau_{2}=\frac{V Q_{2}}{I_{M A} t_{2}} $$

Here, \(V\) is the shear stress, \(Q_{2}\) is the moment of area \(A_{1}\) about neutral axis, \(I_{N A}\) is the moment of inertial about neutral axis, and \(t_{2}\) is the thickness of the section at \(y_{1}\).

Substitute \(35 \times 10^{3} \mathrm{lb}\) for \(V,\left(79.121-y_{2}^{2}\right)\) in. \(^{3}\) for \(Q_{2}, 872.49\) in \(^{4}\) for \(I_{N A}\) and 2 in. for \(t_{2}\) in the equation for \(\tau_{2}\).

$$ \begin{aligned} \tau_{2} &=\frac{\left(35 \times 10^{3}\right) \times\left(79.121-y_{2}^{2}\right)}{(872.49) \times(2)} \\ &=20.0575 \times\left(79.121-y_{2}^{2}\right) \mathrm{psi} \end{aligned} $$

Draw the sketch of the section and show the shear stress distribution acting over the beam’s cross sectional area.

2.jpg

Consider an element of thickness \(d y_{2}\) at a depth \(y_{2}\) from neutral axis.

Calculate the shear force over the section \(A B\).

$$ V_{A B}=\int_{y-6}^{y} \tau_{2} \times 2 d y_{2} $$

Here, thickness of the element is \(d y_{2}\) and \(\tau_{2}\) is the shear stress at a distance \(y_{2}\) from neutral axis.

Substitute \(20.0575 \times\left(79.121-y_{2}^{2}\right)\) psi for \(\tau_{2}\) and \(8.895\) in. for \(\bar{y}\).

$$ \begin{aligned} V_{A B} &=\int_{8.895-6}^{8.899} 20.0575 \times\left(79.121-y_{2}^{2}\right) \times 2 \times d y_{2} \\ &=20.0575 \times 2\left[79.121 y_{2}-\frac{y_{2}^{3}}{3}\right]_{8.895-6}^{8.895} \\ &=40.115 \times\left[79.121(6)-\left(\frac{679.51845}{3}\right)\right] \\ &=9957.33 \mathrm{ps} \\ &=9.96 \mathrm{kips} \end{aligned} $$

Therefore, the resultant shear force acting on the segment \(A B\) is \(9.96 \mathrm{kips}\).

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