2. The "C" frame shown below has a square (1 inch by 1 inch) cross section. It is supporting a test rig that applies the force F at the ends, as shown.
\frac{1}{12} = 0.083 \frac{in^4}{ft}
Material is hot rolled mild steel with E = 30(10$^6$) psi, G = 11(10$^6$) psi.
S$_y$ = 50,000 psi, S$_u$ = 80,000 psi
$\frac{6M}{I} = \frac{1}{1000 - (5)^2} = \frac{1}{1000 - 25} = \frac{1}{975}$
$\frac{6M}{I} = \frac{1}{1000 - (5)^2} = \frac{1}{1000 - 25} = \frac{1}{975}$
-50(0.083) ft$^4$
$\tau = 0$ Motorque
$\sigma = \frac{Mc}{I} = \frac{31120476}{975}$
$\sigma = \frac{31120476}{975} = 31870.24$
$\sigma_{cr} = \frac{\pi^2 E I}{L^2} = \frac{-15560.24}{1000} = -15.560.24$
$R = \sqrt{\sigma_{cr}^2 + \tau^2} = \sqrt{-15560.24^2 + 0^2} = 15560.24$
$\sigma = \sigma_c + R = 31870.4716$
$\sigma = 6C + R_0$
$\sigma = 6C$
6>0
1. Determine the safety factor if a steady force F = 1000 lbs is to be applied at the ends, as shown.
2. On a separate application, the force cycles, with F = 1000 sin($\omega$t). Determine the fatigue safety factor. Take:
Gradient (size) factor: C$_g$ = 0.8, Bending: k$_b$ = 2.1; Torsion: k$_t$ = 1.6
Axial: k$_{yp}$ = 1.9
