Answer:
q = 61.71 KN/m
Explanation:
We know that shear force at one end of the beam is;
F = wl/2
Where;
w is the uniformly distributed load and l is the span.
Thus, in this question, q is the distributed load, so;
F = ql/2
Area of beam section = breadth x depth
In this case,
Area = 200 × 250 = 50000 mm²
We are given allowable shear stress of τa=1.8MPa. This can also be written as τa = 1.8 N/mm²
We know that formula for average shear stress is;
τ_avg = Force/Area
Thus, Force = τ_avg x Area
However, we are given maximum allowable shear stress as 1.8and we know that; τ_max = 1.5 × τ_avg
Thus, τ_avg = 1.8/1.5 = 1.2
Hence;
Force = 1.2 × 50,000 = 60000 N
We need
So from the earlier equation F = ql/2,we can get; 60000 = ql/2
ql = 120000 - - - - - (1)
Now, to the bending stress, we know that section modulus of a rectangular section is;
Z = bd²/6
So,for this question, we have;
Z = (200 × 250²)/6
Z = 2083333.33 mm²
Maximum bending moment of a simply supported beam is wl²/8
So,in this case, M = ql²/8
So,formula for maximum bending stress = M/Z
So, plugging in the values, we have ;
σ_max = (ql²/8) / 2083333.33
We are given σ= 14 MPa or 14 N/mm²
Thus;
14 = (ql²/8) / 2083333.33
ql² = 14 × 2083333.33 × 8
ql² = 233333332.96 - - - eq(2)
From equation 1,we saw that;ql = 120000.
Putting this for ql in equation 2,we will get;
120000l = 233333332.96
l = 233333332.96/120000
l = 1944.44 mm
So from eq 1,q = 120000/l
q = 120000/1944.44
q = 61.71 KN/m
