A single horizontal force P of magnitude 150 lb is applied to end D of lever ABD. Knowing that portion AB of the lever has a diameter of 1.2 in., determine (a) the normal and shearing stresses on an element located at point H and having sides parallel to the x and y axes, (b) the principal planes

Question

A single horizontal force P of magnitude 150 lb is applied to end D of lever ABD. Knowing that portion AB of the lever has a diameter of 1.2 in., determine (a) the normal and shearing stresses on an element located at point H and having sides parallel to the x and y axes, (b) the principal planes and the principal stresses at point H.

Accepted Answer

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.

[latex]M_{x}=(150 lb )(10 in .)=1.5 kip \cdot ext { in. }[/latex]

a. Stresses [latex]S_{x}, S _{y}, T _{x y}[/latex] 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:

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

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

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:

[latex] an 2 u _{p}=\frac{2 t _{x y}}{ s _{x}- s _{y}}[/latex] (7.12)

[latex] an 2 u _{p}=\frac{2 t _{x y}}{ s _{x}- s _{y}}=\frac{2(7.96)}{0-8.84}=-1.80[/latex]

[latex]2 u _{p}=-61.0^{\circ}[/latex] and 180° - 61.0° = +119°

[latex]u _{p}[/latex]= -30.5° and +59.5°

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

[latex]s _{\max , \min }=\frac{ s _{x}+ s _{y}}{2} \pm \sqrt{\left(\frac{ s _{ x }- s _{ y }}{2} ight)^{2}+ t _{x y}^{2}}[/latex] (7.14)

[latex]=\frac{0+8.84}{2} \pm \sqrt{\left(\frac{0-8.84}{2} ight)^{2}+(7.96)^{2}}=+4.42 \pm 9.10[/latex]

[latex]s _{\max }=+13.52 ksi[/latex]

[latex]s _{\min }=-4.68 ksi[/latex]

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

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

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

Related Answered Questions

Verified Answer:

Verified Answer:

For the state of plane stress already considered in Example 7.01, (a) construct Mohr’s circle, (b) determine the principal stresses, (c) determine the maximum shearing stress and the corresponding normal stress. ...

Verified Answer:

For the state of plane stress shown in Fig. 7.13, determine (a) the principal planes, (b) the principal stresses, (c) the maximum shearing stress and the corresponding normal stress. ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

A state of plane stress consists of a tensile stress s0 = 8 ksi exerted on vertical surfaces and of unknown shearing stresses. Determine (a) the magnitude of the shearing stress t0 for which the largest normal stress is 10 ksi, (b) the corresponding maximum shearing stress. ...

Verified Answer:

For the state of plane stress shown, determine (a) the principal planes and the principal stresses, (b) the stress components exerted on the element obtained by rotating the given element counterclockwise through 30°. ...

Verified Answer:

Verified Answer: