Question 4.SP.2: A cast-iron machine part is acted upon by the 3 kN⋅m couple ......
A cast-iron machine part is acted upon by the 3 kN⋅m couple shown. Knowing that E = 165 GPa and neglecting the effect of fillets, determine (a) the maximum tensile and compressive stresses in the casting and (b) the radius of curvature of the casting.
STRATEGY: The moment of inertia is determined, recognizing that it is first necessary to determine the location of the neutral axis. Then Eqs. (4.15) and (4.21) are used to determine the stresses and radius of curvature.
\frac{1}{\rho}=\frac{M}{E I} (4.21)
\sigma_m=\frac{M c}{I} (4.15)
MODELING and ANALYSIS:
Centroid. Divide the T-shaped cross section into two rectangles as shown in Fig. 1 and write
\bar{Y}(3000)=114 \times 10^6
\bar{Y}=38 mm
Centroidal Moment of Inertia. The parallel-axis theorem is used to determine the moment of inertia of each rectangle (Fig. 2) with respect to the axis x′ that passes through the centroid of the composite section. Adding the moments of inertia of the rectangles, write
\begin{aligned} I_{x^{\prime}} & =\Sigma\left(\bar{I}+A d^2\right)=\Sigma\left(\frac{1}{12} b h^3+A d^2\right) \\ & =\frac{1}{12}(90)(20)^3+(90 \times 20)(12)^2+\frac{1}{12}(30)(40)^3+(30 \times 40)(18)^2 \\ & =868 \times 10^3 mm ^4 \\ I & =868 \times 10^{-9} m ^4 \end{aligned}
a. Maximum Tensile Stress. Since the applied couple bends the casting downward, the center of curvature is located below the cross section. The maximum tensile stress occurs at point A (Fig. 3), which is farthest from the center of curvature.
\sigma_A=\frac{M c_A}{I}=\frac{(3 kN \cdot m )(0.022 m )}{868 \times 10^{-9} m ^4} \quad \sigma_A=+76.0 MPa
Maximum Compressive Stress. This occurs at point B (Fig. 3):
\sigma_B=-\frac{M c_B}{I}=-\frac{(3 kN \cdot m )(0.038 m )}{868 \times 10^{-9} m ^4} \quad \sigma_B=-131.3 MPa
b. Radius of Curvature. From Eq. (4.21), using Fig. 3, we have
\begin{aligned} \frac{1}{\rho}=\frac{M}{E I} & =\frac{3 kN \cdot m }{(165 GPa )\left(868 \times 10^{-9} m ^4\right)} \\ & =20.95 \times 10^{-3} m ^{-1} \end{aligned}
ρ = 47.7 m
REFLECT and THINK: Note the T section has a vertical plane of symmetry, with the applied moment in that plane. Thus the couple of this applied moment lies in the plane of symmetry, resulting in symmetrical bending. Had the couple been in another plane, we would have unsymmetric bending and thus would need to apply the principles of Sec. 4.8.
| \bar{y} A , mm ^3 | \bar{y}, mm | Area, mm² | |
| 90 × 10³ | 50 | (20)(90) = 1800 | 1 |
| 24 × 10³ | 20 | (40)(30) = 1200 | 2 |
| \Sigma \bar{y} A=114 \times 10^3 | Σ A = 3000 |
