Problem 2.17. Inclined load on a plate

On the edge of a plate with thickness t, an inclined concentrated force F/t = 600 kN/m acts (see the Figure). Based on the Boussinesq solution calculate the stresses in Points 1 and 2 and give their directions. Locations of the
points are given in the Figure, a = 1.5 m.

Solve Problem

SolveHide solve

Point 1

Radial stress, $σ_{r} [kN / m^{2}] =$

Point 2

Radial stress, $σ_{r} [kN / m^{2}] =$

Do you need help?

Steps

Step by stepHide steps

Apply Boussinesq solution. The solution is the same as a vertical force would act on inclined surface.

See Figure 2.34b.

Step 1. Calculate radial stress from the inclined load F/t at point 1.

Check stressHide stress

The distance of Point 1 from the point of application of the load is a. The angle between the axis of the load and the polar coordinate axis, r is 60^°.

Boussinesq formula is given in Figure 2.34.

σ_{r}^{1} = - \frac{2 F / t}{π a} \cos 60 ° = - \frac{2 \times 600}{π \times 1.5} \frac{1}{2} = - 127.32 \frac{k N}{m^{2}} (compression)

Step 2. Calculate radial stress from the inclined load F at point 2.

Check stressHide stress

The distance of Point 2 from the point of application of the load is 2a. The angle between the axis of the load coincide with the polar coordinate axis, r (φ = 0^°).

Use the Boussinesq formula given in Figure 2.34.

σ_{r}^{2} = - \frac{2 F / t}{π 2 a} \cos 0 ° = - \frac{600 \times 1}{π \times 1.5} = - 127.32 \frac{k N}{m^{2}} (compression)

Results

Worked out solutionHide solution

Boussinesq solution is applied. The solution is the same as a vertical force would act on inclined surface.

Boussinesq solution for inclined surface is given in Figure 2.34b.

Point 1

The distance of Point 1 from the point of application of the load is a. The angle between the axis of the load and the polar coordinate axis, r is 60^°. The radial stress is

$σ_{r}^{1} = - \frac{2 F / t}{π a} \cos 60 ° = - \frac{2 \times 600}{π \times 1.5} \frac{1}{2} = - 127.324 \frac{k N}{m^{2}} (compression)$

Point 2

The distance of Point 2 from the point of application of the load is 2a. The angle between the axis of the load coincide with the polar coordinate axis, r (φ = 0^°).
$σ_{r}^{2} = - \frac{2 F / t}{π 2 a} \cos 0 ° = - \frac{600 \times 1}{π \times 1.5} = - 127.32 \frac{k N}{m^{2}} (compression)$