Question 4.SP.11: A machine component has a T-shaped cross section and is load......
A machine component has a T-shaped cross section and is loaded as shown. Knowing that the allowable compressive stress is 50 MPa, determine the largest force P that can be applied to the component.
STRATEGY: The properties are first determined for the singly symmetric cross section. The force and couple at the critical section are used to calculate the maximum compressive stress, which is obtained by superposing the axial stress and the bending stress determined from Eqs. (4.66) and (4.71). This stress is then equated to the allowable compressive stress to determine the force P.
R=\frac{A}{\int \frac{d A}{r}} (4.66)
\sigma_x=\frac{M(r-R)}{A e r} (4.71)
MODELING and ANALYSIS:
Centroid of the Cross Section. Locate the centroid D of the cross section (Fig. 1).
Force and Couple at D. The internal forces in section a–a are equivalent to a force P acting at D and a couple M of moment (Fig. 2)
M = P(50 mm + 60 mm) = (0.110 m)P
Superposition. The centric force P causes a uniform compressive stress on section a–a, shown in Fig. 3a. The bending couple M causes a varying stress distribution [Eq. (4.71)], shown in Fig. 3b. We note that the couple M tends to increase the curvature of the member and is therefore positive (see Fig. 4.58). The total stress at a point of section a–a located at distance r from the center of curvature C is
\sigma=-\frac{P}{A}+\frac{M(r-R)}{A e r} (1)
Radius of Neutral Surface. Using Fig. 4, we now determine the radius R of the neutral surface by using Eq. (4.66).
\begin{aligned} R & =\frac{A}{\int \frac{d A}{r}}=\frac{2400 mm ^2}{\int_{r_1}^{r_2} \frac{(80 mm ) d r}{r}+\int_{r_2}^{r_3} \frac{(20 mm ) d r}{r}} \\ & =\frac{2400}{80 \ln \frac{50}{30}+20 \ln \frac{90}{50}}=\frac{2400}{40.866+11.756}=45.61 mm \\ & =0.04561 m \end{aligned}
We also compute: e = \bar{r} − R = 0.05000 m − 0.04561 m = 0.00439 m
Allowable Load. We observe that the largest compressive stress will occur at point A where r = 0.030 m. Recalling that \sigma_{\text {all }} = 50 MPa and using Eq. (1), write
\begin{aligned} & -50 \times 10^6 Pa =-\frac{P}{2.4 \times 10^{-3} m ^2}+\frac{(0.110 P )(0.030 m -0.04561 m )}{\left(2.4 \times 10^{-3} m ^2\right)(0.00439 m )(0.030 m )} \\ & -50 \times 10^6=-417 P -5432 P \end{aligned}
P = 8.55 kN
| \bar{r} \sum A_i=\sum \bar{r}_i A_i | \bar{r}_l A , mm ^3 | \bar{r}_l, mm | A _l, mm ^2 | |
| \bar{r}(2400)=120 \times 10^3 | 64 × 10³ | 40 | (20)(80) = 1600 | 1 |
| \bar{r}=50 mm =0.050 m | 56 × 10³ | 70 | (40)(20) = 800 | 2 |
| \Sigma \bar{r}_i A_i=120 \times 10^3 | Σ Ai = 2400 |
