The overhanging beam carries two concentrated loads, $ W $, and a uniformly distributed load of magnitude $ 4W $. The working stresses are 5000 psi in tension, 9000 psi in compression, and 6000 psi in shear. Determine the largest allowable value of $ W $ in lb. Use three decimal places.

The 12-ft long walkway of a scaffold is constructed by screwing two 12-in by 0.5-in sheets of plywood to 1.5-in by 3.5-in timbers, as shown. The screws have a 3-in spacing along the length of the walkway. The working stress in bending is 700 psi for both the plywood and the timbers, and the allowable shear force in each screw is 300 lb. What limit should be placed on the weight $ W $ of a person who walks across the plank? Use three decimal places.

Since there are two concentrated loads of magnitude W, the maximum bending moment occurs at the center of the beam, where the loads are applied. The maximum bending moment for each concentrated load is given by:

M = W * L/4

Uniformly distributed load (4W):

The maximum bending moment due to the uniformly distributed load occurs at the center of the beam. The maximum bending moment for a uniformly distributed load is given by:

M = (w * L^2) / 8

Where w is the load per unit length and is equal to 4W/L.

To determine the largest allowable value of W, we need to consider the maximum bending moment caused by either the concentrated loads or the uniformly distributed load.

The total bending moment is the sum of the bending moments due to the concentrated loads and the uniformly distributed load:

M_total = 2 * (W * L/4) + ((4W/L) * L^2) / 8

M_total = (WL/2) + W * L^2 / 8

To ensure that the working stress limits are not exceeded, we need to equate the maximum bending moment to the moment of resistance of the beam. Assuming the beam is rectangular in shape, the moment of resistance (M_r) is given by:

M_r = (b * h^2) / 6

Where b is the width of the beam (assumed to be constant) and h is the height of the beam.

We can equate the maximum bending moment to the moment of resistance and solve for W:

(WL/2) + (W * L^2 / 8) = (b * h^2) / 6

Now, substitute the given working stress values for tension, compression, and shear:

For tension: (WL/2) + (W * L^2 / 8) = (5000 * b * h^2) / 6

For compression: (WL/2) + (W * L^2 / 8) = (9000 * b * h^2) / 6

For shear: (WL/2) + (W * L^2 / 8) = (6000 * b * h^2) / 6

Solve these equations simultaneously to find the largest allowable value of W.

Moving on to the second problem regarding the scaffold walkway:

To determine the weight limit W for a person walking across the plank, we need to consider the bending stress and the shear stress on the screws.

Bending stress:

The maximum bending stress occurs at the midpoint between screws due to the distributed load of the person's weight. The maximum bending stress is given by:

σ = (M * c) / I

Where σ is the bending stress, M is the bending moment, c is the distance from the neutral axis to the outer fiber (assumed to be half the thickness of the plank), and I is the moment of inertia of the plank.

Shear stress:

The maximum shear stress occurs in the screws due to the shear force caused by the person's weight. The maximum shear stress is given by:

τ = (V * A) / (n * d)

Where τ is the shear stress, V is the shear force, A is the cross-sectional area of the screw, n is the number of screws, and d is the spacing between screws.To ensure that the working stress limits are not exceeded, we need to equate the maximum bending stress and the maximum shear stress to their respective working stress limits and solve for W.

Substitute the given working stress values for bending and shear:

For bending: σ = (M * c) / I = 700 psi

For shear: τ = (V * A) / (n * d) = 300 lb

Solve these equations simultaneously to find the limit on the weight W of a person who walks across the plank.

To learn more about beam

https://brainly.com/question/31681764

#SPJ11