Problem 918 | Stress at Each Corner of Eccentrically Loaded Rectangular Section | Strength of Materials Review at MATHalino

Problem 918
A compressive load P = 12 kips is applied, as in Fig. 9-8a, at a point 1 in. to the right and 2 in. above the centroid of a rectangular section for which h = 10 in. and b = 6 in. Compute the stress at each corner and the location of the neutral axis. Illustrate the answers with a sketch.

Solution 918

Click here to expand or collapse this section

Section Properties

$\text{Area} = bh = 6(10)$

$\text{Area} = 60 ~ \text{in.}^2$

$I_x = \dfrac{hb^3}{12} = \dfrac{10(6^3)}{12}$

$I_x = 180 ~ \text{in.}^4$

$I_y = \dfrac{bh^3}{12} = \dfrac{6(10^3)}{12}$

$I_x = 500 ~ \text{in.}^4$

Moments

$M_x = Pe_y = 12(1000) \times 2$

$M_x = 24,000 ~ \text{lb}\cdot\text{in.}$

$M_y = Pe_x = 12(1000) \times 1$

$M_y = 12,000 ~ \text{lb}\cdot\text{in.}$

Stresses

$\sigma_1 = \dfrac{P}{\text{Area}} = \dfrac{12(1000)}{60}$

$\sigma_1 = 200 ~ \text{psi}$

$f_{bx} = \dfrac{M_x (b/2)}{I_x} = \dfrac{24,000(3)}{180}$

$f_{bx} = 400 ~ \text{psi}$

$f_{by} = \dfrac{M_y (h/2)}{I_y} = \dfrac{12,000(5)}{500}$

$f_{by} = 120 ~ \text{psi}$

Stresses at Corners

$\sigma_A = -\sigma_1 - f_{bx} + f_{by} = -200 - 400 + 120$

$\sigma_A = -480 ~ \text{psi}$

$\sigma_B = -\sigma_1 + f_{bx} + f_{by} = -200 + 400 + 120$

$\sigma_B = 320 ~ \text{psi}$

$\sigma_C = -\sigma_1 + f_{bx} - f_{by} = -200 + 400 - 120$

$\sigma_C = 80 ~ \text{psi}$

$\sigma_D = -\sigma_1 - f_{bx} - f_{by} = -200 - 400 - 120$

$\sigma_D = -720 ~ \text{psi}$

Line of Zero Stress (NA)

$\dfrac{d_A}{480} = \dfrac{6}{480 + 320}$

$d_A = 3.6 ~ \text{in.}$

$\dfrac{d_D}{720} = \dfrac{6}{720 + 80}$

$d_D = 5.4 ~ \text{in.}$

Problem 918 | Stress at Each Corner of Eccentrically Loaded Rectangular Section

Navigation

Recent comments