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.

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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

[latex]V_{x}=-30 kN[/latex] P = 50 kN [latex]V_{z}=-75 kN[/latex]

[latex]M_{x}=(50 kN )(0.130 m )-(75 kN )(0.200 m )=-8.5 kN \cdot m[/latex]

[latex]M_{y}=0[/latex] [latex]M_{z}=(30 kN )(0.100 m )=3 kN \cdot m[/latex]

We note that there is no twisting couple about the y axis. The geometric
properties of the rectangular section are

[latex]A=(0.040 m )(0.140 m )=5.6 imes 10^{-3} m ^{2}[/latex]

[latex]I_{x}=\frac{1}{12}(0.040 m )(0.140 m )^{3}=9.15 imes 10^{-6} m ^{4}[/latex]

[latex]I_{z}=\frac{1}{12}(0.140 m )(0.040 m )^{3}=0.747 imes 10^{-6} m ^{4}[/latex]

Normal Stress at H. We note that normal stresses [latex]s_{y}[/latex] are produced by the centric force P and by the bending couples [latex]M_{x}[/latex] and [latex]M_{z}[/latex]. We determine the sign of each stress by carefully examining the sketch of the force-couple system at C.

[latex]s _{y}=+\frac{P}{A}+\frac{\left|M_{z} ight| a}{I_{z}}-\frac{\left|M_{x} ight| b}{I_{x}}[/latex]

[latex]=\frac{50 kN }{5.6 imes 10^{-3} m ^{2}}+\frac{(3 kN \cdot m )(0.020 m )}{0.747 imes 10^{-6} m ^{4}}-\frac{(8.5 kN \cdot m )(0.025 m )}{9.15 imes 10^{-6} m ^{4}}[/latex]

[latex]s _{y}=8.93 MPa +80.3 MPa -23.2 MPa[/latex] [latex]s _{y}=66.0 MPa[/latex]

Shearing Stress at H. Considering first the shearing force [latex]V_{x}[/latex], we note that Q = 0 with respect to the z axis, since H is on the edge of the cross section. Thus [latex]V_{x}[/latex] produces no shearing stress at H. The shearing force [latex]V_{z}[/latex] does produce a shearing stress at H and we write

[latex]Q=A_{1} \bar{y}_{1}=[(0.040 m )(0.045 m )](0.0475 m )=85.5 imes 10^{-6} m ^{3}[/latex]

[latex]t _{y z}=\frac{V_{z} Q}{I_{x} t}=\frac{(75 kN )\left(85.5 imes 10^{-6} m ^{3} ight)}{\left(9.15 imes 10^{-6} m ^{4} ight)(0.040 m )}[/latex] [latex]t _{y z}=17.52 MPa[/latex]

Principal Stresses, Principal Planes, and Maximum Shearing Stress at H. We draw Mohr’s circle for the stresses at point H

[latex] an 2 u _{p}=\frac{17.52}{33.0}[/latex] [latex]2 u_{p}=27.96^{\circ}[/latex] [latex]u _{p}=13.98^{\circ}[/latex]

[latex]R=\sqrt{(33.0)^{2}+(17.52)^{2}}=37.4 MPa[/latex] [latex]t _{\max }=37.4 MPa[/latex]

[latex]s _{\max }=O A=O C+R=33.0+37.4[/latex] [latex]s_{\max }=70.4 MPa[/latex]

[latex]s _{\min }=O B=O C-R=33.0-37.4[/latex] [latex]s_{\min }=-7.4 MPa[/latex]

Question 8.S-P.5: Three forces are applied as shown at points A, B, and D of a...

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