A 1.25-in.-diameter solid shaft is subjected to an axial force of P = 360 lb, a vertical force of V = 215 lb, and a concentrated torque of T = 430 lb-in., acting in the directions shown in Fig. P15.43. Assume L = 4.5 in. Determine the normal and shear stresses at (a) point H and (b) point K.
Question 15.43: A 1.25-in.-diameter solid shaft is subjected to an axial for...
Section properties:
\begin{array}{ll}A=\frac{\pi}{4}(1.25 in .)^{2}=1.227185 in .^{2} & J=\frac{\pi}{32}(1.25 in .)^{4}=0.239684 in. ^{4} \\Q=\frac{(1.25 in .)^{3}}{12}=0.162760 in .^{3} & I_{y}=I_{z}=\frac{\pi}{64}(1.25 in .)^{4}=0.119842 in .^{4}\end{array}
Equivalent forces at H and K:
\begin{aligned}&F_{x}=360 lb \\&F_{y}=-215 lb \\&F_{z}=0 lb\end{aligned}
Equivalent moments at H and K:
\begin{aligned}M_{x} &=430 lb – in. \\M_{y} &=0 lb – in . \\M_{z} &=-(215 lb )(4.5 in .)=-967.5 lb – in.\end{aligned}
Each of the non-zero forces and moments will be evaluated to determine whether stresses are created at the point of interest.
(a) Consider point H.
Force F_{x} creates an axial stress at H. The magnitude of this normal stress is:
Force F_{y} does not cause either a normal stress or a shear stress at H.
Moment M _{x}, which is a torque, creates a torsion shear stress in the xz plane at H. The magnitude of this shear stress is:
\tau_{x z}=\frac{M_{x} c}{J}=\frac{(430 lb – in .)(1.25 in . / 2)}{0.239684 in .^{4}}=1,121.267 psiMoment M _{z}, which is simply a bending moment, creates bending stress at H. The magnitude of this stress is:
\sigma_{x}=\frac{M_{z} y}{I_{z}}=\frac{(967.5 lb – in. )(1.25 in ./ 2)}{0.119842 in .^{4}}=5,045.694 psi
Summary of stresses at H:
\begin{aligned}\sigma_{x} &=293.354 psi +5,045.694 psi \\&=5,339.048 psi =5,340 psi ( T ) \\\sigma_{z} &=0 psi \\\tau_{x z} &=1,121.267 psi =1,121 psi\end{aligned}
(b) Consider point K.
Force F _{x} creates an axial stress at K. The magnitude of this normal stress is:
Force F _{y} creates a transverse shear stress in the xy plane at K. The magnitude of this shear stress is:
\tau_{x y}=\frac{(215 lb )\left(0.162760 in .^{3}\right)}{\left(0.119842 in .^{4}\right)(1.25 in. )}=233.597 psiMoment M _{x}, which is a torque, creates a torsion shear stress in the xy plane at K. The magnitude of this shear stress is:
\tau_{x y}=\frac{M_{x} c}{J}=\frac{(430 lb – in .)(1.25 in. / 2)}{0.239684 in .^{4}}=1,121.267 psiMoment M _{z} does not create bending stress at K because K is located on the neutral axis for bending about the z axis.
Summary of stresses at K:
\begin{aligned}\sigma_{x} &=293.354 psi =293 psi ( T ) \\\sigma_{y} &=0 psi \\\tau_{x y} &=-233.597 psi -1,121.267 psi \\&=-1,354.864 psi =-1,355 psi\end{aligned}
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