Question 8.75: The 20-kg drum is suspended from the hook mounted on the woo...

The 20-kg drum is suspended from the hook mounted on the wooden frame. Determine the state of stress at point E
on the cross section of the frame at section a–a. Indicate the results on an element.

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Support Reactions: Referring to the free-body diagram of member B C shown in Fig. a,

 

\curvearrowleft +\Sigma M_{B}=0 ; \quad F \sin 45^{\circ}(1)-20(9.81)(2)=0 \quad F=554.94 \mathrm{N}

 

\overset{+}{\rightarrow }\Sigma F_{x}=0 ; \quad 554.94 \cos 45^{\circ}-B_{x}=0  \quad B_{x}=392.4 \mathrm{N}

 

+\uparrow \Sigma F_{y}=0 ; \quad 554.94 \sin 45^{\circ}-20(9.81)-B_{y}=0 \quad B_{y}=196.2 \mathrm{N}

 

Internal Loadings: Consider the equilibrium of the free-body diagram of the right segment shown in Fig. b.

 

\overset{+}{\rightarrow }\Sigma F_{x}=0 \quad N-392.4=0 \quad N=392.4 \mathrm{N}

 

+\uparrow \Sigma F_{y}=0 ;\quad V-196.2=0 \quad V=196.2 \mathrm{N}

 

\curvearrowleft +\Sigma M_{C}=0 ; \quad 196.2(0.5)-M=0\quad M=98.1 \mathrm{N} \cdot \mathrm{m}

 

Section Properties: The cross -sectional area and the moment of inertia of the cross section are

 

A=0.05(0.075)=3.75\left(10^{-3}\right) \mathrm{m}^{2}

 

I=\frac{1}{12}(0.05)\left(0.075^{3}\right)=1.7578\left(10^{-6}\right) \mathrm{m}^{4}

Referring to Fig. c, Q_{E} is

 

Q_{E}=\bar{y}^{\prime} A^{\prime}=0.025(0.025)(0.05)=31.25\left(10^{-6}\right) \mathrm{m}^{3}

Normal Stress: The normal stress is the combination of axial and bending stress. Thus,

 

\sigma=\frac{N}{A} \pm \frac{M y}{I}

For point E, y=0.0375-0.025=0.0125 m. Then

 

\sigma_{E}=\frac{392.4}{3.75\left(10^{-3}\right)}+\frac{98.1(0.0125)}{1.7578\left(10^{-6}\right)}=802 \mathrm{kPa}

Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,

 

\tau_{E}=\frac{V Q_{A}}{I t}=\frac{196.2\left[31.25\left(10^{-6}\right)\right]}{1.7578\left(10^{-6}\right)(0.05)}=69.8 \mathrm{kPa}

 

The state of stress at point $E$ is represented on the element shown in Fig. d.

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