Question 7.S-P.1: A single horizontal force P of magnitude 150 lb is applied t...

Force-Couple System.     We replace the force P by an equivalent force-couple system at the center C of the transverse section containing point H:

P = 150 lb                      T = (150 lb) (18 in.) = 2.7 kip · in.

a. Stresses S_{x}, S _{y}, T _{x y} at Point H. Using the sign convention shown in Fig. 7.2, we determine the sense and the sign of each stress component by carefully examining the sketch of the force-couple system at point C:

s _{x}=0                      s _{y}=+\frac{M c}{I}=+\frac{(1.5 kip \cdot in .)(0.6 in .)}{\frac{1}{4} p (0.6 in .)^{4}}                          s _{y}=+8.84 ksi

t _{x y}=+\frac{T c}{J}=+\frac{(2.7 kip \cdot in .)(0.6 in .)}{\frac{1}{2} p (0.6 in .)^{4}}                          t _{x y}=+7.96 ksi

We note that the shearing force P does not cause any shearing stress at point H.

b. Principal Planes and Principal Stresses.     Substituting the values of the stress components into Eq. (7.12), we determine the orientation of the principal planes:

\tan 2 u _{p}=\frac{2 t _{x y}}{ s _{x}- s _{y}}                    (7.12)

2 u _{p}=-61.0^{\circ}                 and              180° – 61.0° = +119°

u _{p}= -30.5°                  and                  +59.5°

Substituting into Eq. (7.14), we determine the magnitudes of the principal stresses:

s _{\max , \min }=\frac{ s _{x}+ s _{y}}{2} \pm \sqrt{\left(\frac{ s _{ x }- s _{ y }}{2}\right)^{2}+ t _{x y}^{2}}                            (7.14)

Considering face ab of the element shown, we make u_{p}=-30.5^{\circ} in Eq. (7.5) and find s_{x^{\prime}}=-4.68 ksi. We conclude that the principal stresses are as shown.

s _{x^{\prime}}=\frac{ s _{x}+ s _{y}}{2}+\frac{ s _{x}- s _{y}}{2} \cos 2 u + t _{x y} \sin 2 u                                  (7.5)