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 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.
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