#### Question

A bar having a square cross section of 30 mm by 30 mm is 2 m long and is held upward. If it has a mass of 5 kg/m, determine the largest angle θ, measured from the vertical, at which it can be supported before it is subjected to a tensile stress along its axis near the grip.

#### Solutions

Verified#### Step 1

1 of 7We have the following data:

An inclined bar with angle $\theta$ with respect to the y-axis

$\bullet$ Cross-section of the bar Dimensions: 30 mm $\times$ 30 mm $\bullet$ Length of the bar $L$ = 2 m $\bullet$ Mass per unit length $m$ = 5 $\frac{\text{kg}}{\text{m}}$

Required: $\Rightarrow$ Largest angle $\theta$ without developing any tensile stress

#### Step 1

1 of 3First, it is convenient to draw a free body diagram of the bar in order to obtain the reactions at the grip: $M$, $N$ and $V$, in terms of the angle $\theta$. Since the system is static, we can use equilibrium equations.

$\begin{align*} \left(\sum F \right)_x &= N-mg\cos\left(\theta\right) =0\\ \implies N &= mg\cos\left(\theta\right)\\ \left( \sum F \right)_y &= V- mg\sin\left(\theta\right) =0\\ \implies V &= mg\sin\left(\theta\right)\\ \left( \sum M \right)_O &= M - mg\left(\dfrac{l}{2} \right) \sin\left(\theta\right) =0\\ \implies M &= \dfrac{mgl\sin\left(\theta\right)}{2} \end{align*}$

#### Step 1

1 of 6**Given Data:**

- $A=30\;\text{mm}\cdot30\;\text{mm}$ - square cross section
- $l=2\;\text m$ - length of the bar
- $\dot m=5\;\frac{\text{kg}}{\text m}$ - mass per unit length

**To find:**

- $\phi$ - angle