Question 5.12: A beam of wide-flange shape (Fig. 5-36a) is subjected to a v...
A beam of wide-flange shape (Fig. 5-36a) is subjected to a vertical shear force V = 45 kN. The cross-sectional dimensions of the beam are b = 165 mm, t = 7.5 mm, h = 320 mm, and h_{1} = 290 mm.
Determine the maximum shear stress, minimum shear stress, and total shear force in the web. (Disregard the areas of the fillets when making calculations.)
Maximum and minimum shear stresses. The maximum and minimum shear stresses in the web of the beam are given by Eqs. (5-45a) and (5-45b).
τ_{max}=\frac{V}{8It}(bh^2 – bh_{1}^2 + th_{1}^2) τ_{min}=\frac{Vb}{8It}(h^2 – h_{1}^2) (5-45a,b)
Before substituting into those equations, we calculate the moment of inertia of the cross-sectional area from Eq. (5-44):
I=\frac{1}{12}(bh^3 – bh_{1}^{3} + th_{1}^{3})=130.45 \times 10^6 mm^4
Now we substitute this value for I, as well as the numerical values for the shear force V and the cross-sectional dimensions, into Eqs. (5-45a) and (5-45b):
τ_{max}=\frac{V}{8It}(bh^2 – bh_{1}^{2} + th_{1}^{2})=21.0 MPa
τ_{min}=\frac{Vb}{8It}(h^2 – h_{1}^{2})=17.4 MPa
In this case, the ratio of τ_{max} to τ_{min} is 1.21, that is, the maximum stress in the web is 21% larger than the minimum stress. The variation of the shear stresses over the height h_1 of the web is shown in Fig. 5-36b.
Total shear force. The shear force in the web is calculated from Eq. (5-46) as follows:
V_{web}=\frac{th_1}{3}(2τ_{max} + τ_{min})=43.0 kN
From this result we see that the web of this particular beam resists 96% of the total shear force.
Note: The average shear stress in the web of the beam (from Eq. 5-47) is
τ_{aver}=\frac{V}{th_1}=20.7 MPa
which is only 1% less than the maximum stress.