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.
Centroid of the Cross Section. We locate the centroid D of the cross section
| \bar{r}Σ A_{i} = Σ \bar{r}_{i}A_{i} | A_{i}, mm² | \bar{r}_{i}, mm | \bar{r}_{i}A_{i}, mm³ | |
| \bar{r} (2400) = 120 × 10³ | (20)(80) = 1600 | 40 | 64 × 10³ | 1 |
| \bar{r} = 50 mm = 0.050 m | (40)(20) = 800 | 70 | 56 × 10³ | 2 |
| Σ A_{i} = 2400 | Σ \bar{r}_{i}A_{i} = 120 × 10³ |
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
M = P(50 mm + 60 mm) = (0.110 m)P
Superposition. The centric force P causes a uniform compressive stress on section a-a. The bending couple M causes a varying stress distribution [Eq. (4.71)]. We note that the couple M tends to increase the curvature of the member and is therefore positive (cf. Fig. 4.70). The total stress at a point of section a-a located at distance r from the center of curvature C is
σ_{x} = \frac{M(r – R)}{Aer} (4.71)
σ = – \frac{P}{A} + \frac{M(r – R)}{Aer} (1)
Radius of Neutral Surface. We now determine the radius R of the neutral surface by using Eq. (4.66).
R = \frac{A}{\int{\frac{dA}{r} } } = \frac{2400 mm^{2}}{\int_{r_{1}}^{r_{2}}{\frac{(80 mm)dr}{r} }+{\int_{r_{2}}^{r_{3}}{\frac{(20 mm)dr}{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
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 σ_{all} = 50 MPa and using Eq. (1), we write
-50 × 10^{6} Pa = – \frac{P}{2.4 × 10^{-3} m²} + \frac{(0.110 P) (0.030 m – 0.04561 m)}{(2.4 × 10^{-3} m²) (0.00439 m) (0.030 m)} – 50 × 10^{6} = – 417p – 5432 p
p = 8.55 kN
