## Related questions with answers

Rods $A B$ and $B C$have diameters of$4$ mm and $6$ mm, respectively. If the $3$ kN force is applied to the ring at $B$ , determine the angle $\theta$ so that the average normal stress in each rod is equivalent. What is this stress?

Solutions

VerifiedWe have to determine the angle of the 3 kN force and the normal stress of each rod, if it has to be equal in both rods. The arrangement is given in Prob. 1-56.

Given data:

$\begin{aligned} &\text{Diameter of rod $AB$: }&&d_{AB} = 4\text{ mm}\\ &\text{Diameter of rod $BC$: }&&d_{BC} = 6\text{ mm}\\ \end{aligned}$

We will begin by calculating the angle of the rod $BC$ with the tangent definition.

$\begin{aligned} \tan{\alpha}&=\dfrac{3}{4}\\ \alpha&=\arctan{\left( \dfrac{3}{4} \right)}\\ &=36.87^{\circ} \end{aligned}$

Rods $AB$ and $BC$ have a diameter of $4$ mm and $6$ mm. If the work force has $3$ kN and works at an angle of $\Theta$. Determine the normal stress and the aunt angle so that the normal stress in the $AB$ stick is the same as in the $BC$ stick.

We apply the conditions of equilibrium:

$\xrightarrow{+} \sum F_{x}=0$

${+}\uparrow\sum F_{y}=0$

From picture $b)$:

$\sum F_{x}=0$

$\begin{align} \Rightarrow -F_{AB}+F_{BC}\cdot\cos36.87^{\circ}-3\cdot\sin \Theta=0 \end{align}$

$\sum F_{x}=0$

$\begin{align} \Rightarrow +F_{BC}\cdot\sin36.87^{\circ}-3\cdot \cos\Theta=0 \end{align}$

$\begin{align} \textcolor{#4257b2} {F_{BC}=5\cdot\cos\Theta} \end{align}$

Now we insert the expression $(3)$ into expression $(1)$:

$\begin{align} \textcolor{#4257b2} {F_{AB}=4\cdot\cos\Theta-3\cdot\sin\Theta} \end{align}$

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