Failure of Brittle Materials Under Static Loading Problem Determine the safety factors for the bracket rod shown in Figure 5-9 (repeated on next page) based on the modified-Mohr theory. Given The material is class 50 gray cast iron with Sut = 52 500 psi and Suc = –164 000 psi. The rod length l = 6

Question

Failure of Brittle Materials Under Static Loading

Problem Determine the safety factors for the bracket rod shown in Figure 5-9 (repeated on next page) based on the modified-Mohr theory.

Given The material is class 50 gray cast iron with [latex]S_{ut}[/latex] = 52 500 psi and [latex]S_{uc}[/latex] = –164 000 psi. The rod length [latex]l[/latex] = 6 in and arm [latex]a[/latex] = 8 in. The rod outside diameter [latex]d[/latex] = 1.5 in. Load [latex]F[/latex] = 1 000 lb.

Assumptions The load is static and the assembly is at room temperature. Consider shear due to transverse loading as well as other stresses.

Accepted Answer

See Figures 5-9 and 4-33 (opp. page) and Examples 4-9 (p. 180) and 5-1.

1 The rod of Figure 5-9 is loaded both in bending (as a cantilever beam) and in torsion. The largest tensile bending stress will be in the top outer fiber at point [latex]A[/latex]. The largest torsional shear stress will be all around the outer circumference of the rod. First take a differential element at point [latex]A[/latex] where both of these stresses combine. Find the normal bending stress and torsional shear stress on point [latex]A[/latex] using equations 4.11b (p. 156) and 4.23b (p. 178), respectively.

[latex] \sigma_{\max }=\frac{M c}{I} [/latex] (4.11b)

[latex] au_{\max }=\frac{T r}{J} [/latex] (4.23b)

[latex] \sigma_x=\frac{M c}{I}=\frac{(F l) c}{I}=\frac{1000(6)(0.75)}{0.249}=18108 psi [/latex] (a)

[latex] au_{x z}=\frac{T r}{J}=\frac{(F a) r}{J}=\frac{1000(8)(0.75)}{0.497}=12072 psi [/latex] (b)

2 Find the maximum shear stress and principal stresses that result from this combination of applied stresses using equations 4.6.

[latex] \begin{aligned} \sigma_a, \sigma_b &=\frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left\lgroup\frac{\sigma_x-\sigma_y}{2} ight group ^2+ au_{x y}^2} \ \sigma_c &=0 \end{aligned} [/latex] (4.6a)

[latex] au_{\max }= au_{13}=\frac{\left|\sigma_1-\sigma_3 ight|}{2} [/latex] (4.6b)

[latex] \begin{aligned} au_{\max } &=\sqrt{\left\lgroup\frac{\sigma_{ x }-\sigma_z}{2} ight group ^2+ au_{x z}^2}=\sqrt{\left\lgroup\frac{18108-0}{2} ight group ^2+12072^2}=15090 psi \ \sigma_1 &=\frac{\sigma_{ x }+\sigma_z}{2}+ au_{\max }=\frac{18108}{2}+15090=24144 psi \ \sigma_2 &=0 \ \sigma_3 &=\frac{\sigma_{ x }+\sigma_z}{2}- au_{\max }=\frac{18108}{2}-15090=-6036 psi \end{aligned} [/latex] (c)

Note that these stresses are identical to those of Example 5-1.

3 The principal stresses for point [latex]A[/latex] can now be plotted on a modified-Mohr diagram as shown in Figure 5-14a (p. 264). This shows that the load line crosses the failure envelope above the [latex]S_{ut},–S_{ut}[/latex] point, making equation 5.12a (p. 260) appropriate for the safety-factor calculation.

[latex] N=\frac{S_{u t}}{\sigma_1}=\frac{52400}{24144}=2.2 [/latex] (d)

4 An alternative approach that does not require drawing the modified Mohr diagram is to find the Dowling factors [latex]C_{1}, C_{2}, C_{3}[/latex] using equation 5.12c (p. 262).

[latex] \begin{array}{l} C_1=\frac{1}{2}\left[\left|\sigma_1-\sigma_2 ight|+\frac{2 S_{u t}-\left|S_{u c} ight|}{-\left|S_{u c} ight|}\left(\sigma_1+\sigma_2 ight) ight] \ C_2=\frac{1}{2}\left[\left|\sigma_2-\sigma_3 ight|+\frac{2 S_{u t}-\left|S_{u c} ight|}{-\left|S_{u c} ight|}\left(\sigma_2+\sigma_3 ight) ight] \ C_3=\frac{1}{2}\left[\left|\sigma_3-\sigma_1 ight|+\frac{2 S_{u t}-\left|S_{u c} ight|}{-\left|S_{u c} ight|}\left(\sigma_3+\sigma_1 ight) ight] \end{array} [/latex] (5.12c)

[latex] \begin{aligned} C_1 &=\frac{1}{2}\left[\left|\sigma_1-\sigma_2 ight|+\frac{2 S_{u t}-\left|S_{u c} ight|}{-\left|S_{u c} ight|}\left(\sigma_1+\sigma_2 ight) ight] \ &=\frac{1}{2}\left[|24144-0|+\frac{2(52500)-164000}{-164000}(24144+0) ight]=16415 psi \end{aligned} [/latex] (e)

[latex] \begin{aligned} C_2 &=\frac{1}{2}\left[\left|\sigma_2-\sigma_3 ight|+\frac{2 S_{u t}-\left|S_{u c} ight|}{-\left|S_{u c} ight|}\left(\sigma_2+\sigma_3 ight) ight] \ &=\frac{1}{2}\left[|0-(-6036)|+\frac{2(52500)-164000}{-164000}(0-6036) ight]=1932 psi \end{aligned} [/latex] (f)

[latex] \begin{aligned} C_3 &=\frac{1}{2}\left[\left|\sigma_3-\sigma_1 ight|+\frac{2 S_{u t}-\left|S_{u c} ight|}{-\left|S_{u c} ight|}\left(\sigma_3+\sigma_1 ight) ight] \ &=\frac{1}{2}\left[|24144-(-6036)|+\frac{2(52500)-164000}{-164000}(24144-6036) ight]=18348 \end{aligned} [/latex] (g)

5 Then find the largest of the six stresses [latex]C_{1}, C_{2}, C_{3}, \sigma _{1}, \sigma _{2}, \sigma _{3}[/latex]:

[latex] \begin{array}{l} ilde{\sigma}=\operatorname{MAX}\left(C_1, C_2, C_3, \sigma_1, \sigma_2, \sigma_3 ight) \ ilde{\sigma}=\operatorname{MAX}(16415,1932,18348,24144,0,-6036)=24144 \end{array} [/latex] (h)

which is the modified-Mohr effective stress.

6 The safety factor for point [latex]A[/latex] can now be found using equation 5.12e (p. 262):

[latex] N=\frac{S_{u t}}{ ilde{\sigma}}=\frac{52.500}{24144}=2.2 [/latex] (i)

which is the same as was found in step 3.

7 Since the rod is a short beam, we need to check the shear stress due to transverse loading at point [latex]B[/latex] on the neutral axis. The maximum transverse shear stress at the neutral axis of a solid round rod was given as equation 4.15c (p. 161).

[latex] au_{\max }=\frac{4}{3} \frac{V}{A} [/latex] (4.15c)

[latex] au_{ ext {bending }}=\frac{4 V}{3 A}=\frac{4(1000)}{1.767}=755 psi [/latex] (j)

Point [latex]B[/latex] is in pure shear. The total shear stress at point [latex]B[/latex] is the algebraic sum of the transverse shear stress and the torsional shear stress, which both act on the same planes of the differential element and, in this case, act in the same direction as shown in Figure 4-33b (p. 185).

[latex] au_{ ext {max }}= au_{ ext {torsion }}+ au_{ ext {bending }}=12072+755=12827 psi [/latex] (k)

8 Find the principal stresses for this pure shear loading:

[latex] \begin{array}{l} \sigma_1= au_{\max }=12827 psi \ \sigma_2=0 \ \sigma_3=- au_{\max }=-12827 psi \end{array} [/latex] (l)

9 These principal stresses for point [latex]B[/latex] can now be plotted on a modified-Mohr diagram as shown in Figure 5-14b. Because this is a pure shear loading, the load line crosses the failure envelope at the [latex] S_{u t}{ }^{\prime}-S_{u t} [/latex] point, making equation 5.12a (p. 260) appropriate for the safety-factor calculation.

[latex] N=\frac{S_{u t}}{\sigma_1}=\frac{52400}{12827}=4.1 [/latex] (m)

10 To avoid drawing the modified-Mohr diagram, find the Dowling factors [latex]C_{1}, C_{2}, C_{3}[/latex] using equations 5.12c:

[latex] \begin{aligned} C_1 &=8721 psi \ C_2 &=4106 psi \ C_3 &=12827 psi \end{aligned} [/latex] (n)

11 And find the largest of the six stresses [latex]C_{1}, C_{2}, C_{3}, \sigma _{1}, \sigma _{2}, \sigma _{3}[/latex]:

[latex] ilde{\sigma}=12827 psi [/latex] (o)

which is the modified-Mohr effective stress.

12 The safety factor for point [latex]B[/latex] can now be found using equation 5.12e (p. 262):

[latex] N=\frac{S_{u t}}{ ilde{\sigma}}=\frac{52500}{12827}=4.1 [/latex] (p)

and it is the same as was found in step 9.

13 The files EX05-02 are on the CD-ROM.

Question 5.2: Failure of Brittle Materials Under Static Loading Problem De...

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