Question 8.41: The 500-kg engine is suspended from the jib crane at the pos...

The 500-kg engine is suspended from the jib crane at the position shown. Determine the state of stress at point B on the cross section of the boom at section a–a. Point B is just above the bottom flange.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

Support Reactions: Referring to the free-body diagram of the entire boom, Fig.a,

 

\curvearrowleft +\Sigma M_{C}=0 ; \quad F_{D E} \sin 30^{\circ}(6)+F_{D E} \cos 30^{\circ}(0.4)-500(9.81)(2)=0

 

F_{D E}=2931.50  \mathrm{N}

 

Internal Loadings: Considering the equilibrium of the free-body diagram of the boom’s right cut segment, Fig. b.

 

\stackrel{+}{\rightarrow} \Sigma F_{x}=0 ; \quad N-2931.50 \cos 30^{\circ}=0 \quad N=2538.75  \mathrm{N}

 

+\uparrow \Sigma F_{y}=0 \quad 2931.50 \sin 30^{\circ}-V=0 \quad V=1465.75  \mathrm{N}

 

\curvearrowleft +\Sigma M_{O}=0 ; \quad 2931.50 \sin 30^{\circ}(2)+2931.50 \cos 30^{\circ}(0.4)-M=0

 

M=3947.00  \mathrm{N} \cdot \mathrm{m}

 

Section Properties ‘The cross-sectional area and the moment of inertia about the centroidal axis of the boom’s cross section are

 

A=0.15(0.3)-0.13(0.26)=0.0112  \mathrm{m}^{2}

 

I=\frac{1}{12}(0.15)\left(0.3^{3}\right)-\frac{1}{12}(0.13)\left(0.26^{3}\right)=0.14709\left(10^{-3}\right)  \mathrm{m}^{4}

 

 

Referring to Fig. c, Q_{B} is

 

Q_{B}=\bar{y}_{2} A_{2}^{\prime}=0.14(0.02)(0.15)=0.42\left(10^{-3}\right)  \mathrm{m}^{3}

 

 

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

 

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

 

For point B, y=0.13 m . Then

 

\sigma_{B}=\frac{-2538.75}{0.0112}+\frac{3947.00(0.13)}{0.14709\left(10^{-3}\right)}=3.26  \mathrm{MPa}(T)

 

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

 

\tau_{B}=\frac{V Q_{B}}{I t}=\frac{1465.75\left[0.42\left(10^{-3}\right)\right]}{0.14709\left(10^{-3}\right)(0.02)}=0.209   \mathrm{MPa}

 

The state of stress at point B is represented on the element shown in Fig. d.8.41

2
3

Related Answered Questions