Question

a timber beam of rectangular section 10 cm wide and 25 cm deep is simply 16 supported over a span of 4m. what uniformly distributed load in kn/m should the beam carry to produce a central deflection of 0.6 cm?

Answer 1

The beam should carry a uniformly distributed load of approximately 0.1875 kN/m to produce a central deflection of 0.6 cm.

Step-by-step explanation:

To find the uniformly distributed load that the beam should carry, we can use the formula for the deflection of a simply supported beam:

d = (5 * w * L^4) / (384 * E * I)

Where:

• d is the deflection
• w is the load per unit length
• L is the span
• E is the modulus of elasticity for the material
• I is the moment of inertia of the cross-sectional area

From the given information, we have:

• d = 0.006 m
• L = 4 m
• b = 0.1 m (width)
• h = 0.25 m (depth)

To determine the moment of inertia, we can use the formula:

I = (b * h^3) / 12

Plugging in the values, we can substitute the known values into the formula and solve for w:

0.006 = (5 * w * 4^4) / (384 * E * ((0.1 * 0.25^3) / 12))

Solving for w, we find:

w ≈ 0.1875 kN/m