Question 8.S-P.5: Three forces are applied as shown at points A, B, and D of a...
Three forces are applied as shown at points A, B, and D of a short steel post. Knowing that the horizontal cross section of the post is a 40 × 140-mm rectangle, determine the principal stresses, principal planes and maximum shearing stress at point H.
Internal Forces in Section EFG. We replace the three applied forces by an equivalent force-couple system at the center C of the rectangular section EFG. We have
V_{x}=-30 kN P = 50 kN V_{z}=-75 kN
M_{x}=(50 kN )(0.130 m )-(75 kN )(0.200 m )=-8.5 kN \cdot m
M_{y}=0 M_{z}=(30 kN )(0.100 m )=3 kN \cdot m
We note that there is no twisting couple about the y axis. The geometric
properties of the rectangular section are
I_{x}=\frac{1}{12}(0.040 m )(0.140 m )^{3}=9.15 \times 10^{-6} m ^{4}
I_{z}=\frac{1}{12}(0.140 m )(0.040 m )^{3}=0.747 \times 10^{-6} m ^{4}
Normal Stress at H. We note that normal stresses s_{y} are produced by the centric force P and by the bending couples M_{x} and M_{z}. We determine the sign of each stress by carefully examining the sketch of the force-couple system at C.
s _{y}=+\frac{P}{A}+\frac{\left|M_{z}\right| a}{I_{z}}-\frac{\left|M_{x}\right| b}{I_{x}}=\frac{50 kN }{5.6 \times 10^{-3} m ^{2}}+\frac{(3 kN \cdot m )(0.020 m )}{0.747 \times 10^{-6} m ^{4}}-\frac{(8.5 kN \cdot m )(0.025 m )}{9.15 \times 10^{-6} m ^{4}}
s _{y}=8.93 MPa +80.3 MPa -23.2 MPa s _{y}=66.0 MPa
Shearing Stress at H. Considering first the shearing force V_{x}, we note that Q = 0 with respect to the z axis, since H is on the edge of the cross section. Thus V_{x} produces no shearing stress at H. The shearing force V_{z} does produce a shearing stress at H and we write
Q=A_{1} \bar{y}_{1}=[(0.040 m )(0.045 m )](0.0475 m )=85.5 \times 10^{-6} m ^{3}
t _{y z}=\frac{V_{z} Q}{I_{x} t}=\frac{(75 kN )\left(85.5 \times 10^{-6} m ^{3}\right)}{\left(9.15 \times 10^{-6} m ^{4}\right)(0.040 m )} t _{y z}=17.52 MPa
Principal Stresses, Principal Planes, and Maximum Shearing Stress at H. We draw Mohr’s circle for the stresses at point H
\tan 2 u _{p}=\frac{17.52}{33.0} 2 u_{p}=27.96^{\circ} u _{p}=13.98^{\circ}
R=\sqrt{(33.0)^{2}+(17.52)^{2}}=37.4 MPa t _{\max }=37.4 MPa
s _{\max }=O A=O C+R=33.0+37.4 s_{\max }=70.4 MPa
s _{\min }=O B=O C-R=33.0-37.4 s_{\min }=-7.4 MPa
