Shear Stresses in Homogeneous Beams With Example Diagram and Formulas

Shear Stresses in Homogeneous Beams 

The beams which are made of the same material are called homogeneous beams. Since in the cross section of such beams the same materials is present, so their withstanding of shear force is same throughout. To find the shear force at any point the shear force acting there is found out. The shear stress is calculated by dividing the shear force by the area of cross section of the beam at that point.
Average shear stress = Shear Force
                                        Area
              Q (avr) = Q        = Q 
                                A           bd
Maximum Shear Stress = qmax = 3 . Q 
                                                        2   bd

Example of Shear Stresses in Homogeneous Beams:-

Cantilever timber beam 30*20 cm is cross section is 2 meter long. This beam is loaded at the rate of 400 kg per meter including its own weight. Determine shear stress at each half meter length.

• Given Data:

Breadth of the beam = b = 20cm
Depth of the beam = d = 30cm
Length of the beam = l = 2m
Uniformly distributed load = 400Kg/m

• Solution:

Shear force at free end = Q0 =0
Shear Force at 0.5m from free end = Q1 = 400 x 0.5 = 200kg.
Shear force ate 1m form free end = Q2 = 400 x 1 = 400kg.
Shear force at 1.5m from free end = Q3 = 400 x 1.5 =600kg.
Shear force at fixed end = Q4 = 400 x 2 =800kg.
We know that average shear stress = Q (avr) = Q/A
And maximum shear stress = Q(max) = 3/2 . Q/A
So average shear stress and maximum shear stress at free end = 0
Average shear stress at 0.5 m from free end =q(avr) = Q1/bd = 200/20x30 = 0.33kg/cm2
Maximum shear stress at 0.5 m from free end= q(avr) =3/2 xq(avr)
                                  =3/2 x q(avr)
                                   =3/2 x 0.33 = 0.495 Kg/cm Square
Average shear stress at 1 m from end =q2(avr) = Q2/bd
                                = 400/20x30 = 0.67 Kg/ cm square

Maximum shear stress at 1 m from end= q(max) = 3/2 x q(avr)
                               = 3/2 x 0.67 = 1 Kg/cm square

Average shear stress at 1.5 m from free end =  q3(avr) = Q3/bd
                             = 600/20 x 30 = 1 Kg/cm square

Maximum shear stress at 1.5 from free end = 3/2 x q(avr)
                        = 3/2 x 1 = 1.50 Kg/ cm square

Average shear stress at 2 m from free end = q4(avr) = Q4/bd
                         = 800/20 x 30 = 1.33 Kg/cm square

Maximum shear stress at 2 m from free end
                                                              = 3  x q(avr) =  3   x 1.33 = 1.99 Kg/cm square
                                                                   2                     2

Last Words!
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