A steel plane truss is loaded at B and C by forces P =200 kN. The cross-sectional area of each member $\text { is } A = 3970 \mathrm {~mm} ^ { 2 }$ . Truss dimensions are H=3 m and L = 4 m. The maximum shear stress in bar AB is approx- imately:

Question

Quizplus · Accepted Answer

First, we need to find the reactions at A and D. By taking moments about point A, we have:

\begin{align*}
\sum M_A &= 0 \
-200	ext{ kN}\cdot 4	ext{ m} - 200	ext{ kN}\cdot 8	ext{ m} + D_y\cdot 3	ext{ m} &= 0 \
D_y &= 133.33	ext{ kN}
\end{align*}

By taking vertical forces, we have:

\begin{align*}
\sum F_y &= 0 \
A_y + D_y - 200	ext{ kN} - 200	ext{ kN} &= 0 \
A_y &= 266.67	ext{ kN}
\end{align*}

Now, we can analyze the forces in the members. We can see that AB and BC are zero-force members, so they do not carry any load. The forces in the other members can be found by analyzing the joints.

At joint B, we have:

\begin{align*}
\sum F_x &= 0 \
-200	ext{ kN} + BC\cdot\cos 30^\circ &= 0 \
BC &= 231.13	ext{ kN} \
\sum F_y &= 0 \
A_y + AB\cdot\cos 60^\circ &= BC\cdot\sin 30^\circ \
AB &= 239.61	ext{ kN} \
\end{align*}

At joint C, we have:

\begin{align*}
\sum F_x &= 0 \
-200	ext{ kN} + CD\cdot\cos 30^\circ &= 0 \
CD &= 231.13	ext{ kN} \
\sum F_y &= 0 \
A_y + AB\cdot\sin 60^\circ &= CD\cdot\sin 30^\circ \
AB &= 239.61	ext{ kN} \
\end{align*}

Now, we can find the maximum shear stress in member AB. The shear force in member AB can be found by taking vertical forces at joint B:

\begin{align*}
\sum F_y &= 0 \
A_y + AB\cdot\cos 60^\circ &= BC\cdot\sin 30^\circ \
AB &= 174.08	ext{ kN} \
\end{align*}

The maximum shear stress in member AB is at point B where the force is the highest. The cross-sectional area of member AB is 3970 mm$^2$. Therefore, the maximum shear stress is:

\begin{align*}
	au_{max} &= \frac{V}{A} \
&= \frac{174080	ext{ N}}{3970	ext{ mm}^2} \
&= 43.79	ext{ MPa} \
\end{align*}

However, this is the shear stress at point B. We need to find the shear stress at a point where it is the highest. By using the maximum shear stress theory, we have:

\begin{align*}
	au_{max} &= \frac{\sqrt{	au_x^2 + 	au_y^2}}{2} \
&= \frac{\sqrt{\left(\frac{AB}{A}}ight)^2 + \left(\frac{BC}{A}}ight)^2}{2} \
&= \frac{\sqrt{\left(\frac{239610	ext{ N}}{3970	ext{ mm}^2}}ight)^2 + \left(\frac{231130	ext{ N}}{3970	ext{ mm}^2}}ight)^2}{2} \
&= 50.03	ext{ MPa} \
\end{align*}

Therefore, the best choice is C with a maximum shear stress of approximately 50 MPa.

A Steel Plane Truss Is Loaded at B and C is A=3970 mm2\text { is } A = 3970 \mathrm {~mm} ^ { 2 } is A=3970 mm2

A Steel Plane Truss Is Loaded at B and C $\text { is } A = 3970 \mathrm {~mm} ^ { 2 }$