Question 5.15: A beam having a T-shaped cross section (Fig. 5-40a) is subje...
A beam having a T-shaped cross section (Fig. 5-40a) is subjected to a vertical shear force V = 10,000 lb. The cross-sectional dimensions are b = 4 in., t = 1.0 in., h = 8.0 in., and h_{1} = 7.0 in.
Determine the shear stress \tau_{1} at the top of the web (level nn) and the maximum shear stress \tau_{max}. (Disregard the areas of the fillets.)
Location of neutral axis. The neutral axis of the T-beam is located by calculating the distances c_{1} and c_{2} from the top and bottom of the beam to the centroid of the cross section (Fig. 5-40a). First, we divide the cross section into two rectangles, the flange and the web (see the dashed line in Fig. 5-40a). Then we calculate the first moment Q_{aa} of these two rectangles with respect to line aa at the bottom of the beam. The distance c_{2} is equal to Q_{aa} divided by the area A of the entire cross section (see Chapter 12, Section 12.3, for methods for locating centroids of composite areas). The calculations are as follows:
A=\sum A_{i}=b\left(h-h_{1}\right)+t h_{1}=11.0 \text { in.}^{2}Q_{a a}=\sum y_{i} A_{i}=\left(\frac{h+h_{1}}{2}\right)(b)\left(h-h_{1}\right)+\frac{h_{1}}{2}\left(t h_{1}\right)=54.5 \text { in.}^{3}
c_{2}=\frac{Q_{a a}}{A}=\frac{54.5 in. ^{3}}{11.0 in. ^{2}}=4.955 \text { in. } \quad c_{1}=h-c_{2}=3.045 in.
Moment of inertia. The moment of inertia I of the entire cross-sectional area (with respect to the neutral axis) can be found by determining the moment of inertia I_{aa} about line aa at the bottom of the beam and then using the parallel-axis theorem (see Section 12.5):
I=I_{a a}-A c_{2}^{2}The calculations are as follows:
I_{a a}=\frac{b h^{3}}{3}-\frac{(b-t) h_{1}^{3}}{3}=339.67 \text {in.}^{4} \quad A c_{2}^{2}=270.02 \text {in.}^{4} \quad I=69.65 \text {in.}^{4}Shear stress at top of web. To find the shear stress \tau_{1} at the top of the web (along line nn) we need to calculate the first moment Q_{1} of the area above level nn. This first moment is equal to the area of the flange times the distance from the neutral axis to the centroid of the flange:
Q_{1}=b\left(h-h_{1}\right)\left(c_{1}-\frac{h-h_{1}}{2}\right)
= (4 in.)(1 in.)(3.045 in. – 0.5 in.) = 10.18 in.³
Of course, we get the same result if we calculate the first moment of the area below level nn:
Q_{1}=t h_{1}\left(c_{2}-\frac{h_{1}}{2}\right)=(1 \text { in.})(7 \text { in.})(4.955 \text {in.}-3.5 \text { in.})=10.18 \text { in.}^{3}Substituting into the shear formula, we find
\tau_{1}=\frac{V Q_{1}}{I t}=\frac{(10,000 lb )\left(10.18 in. ^{3}\right)}{\left(69.65 in. ^{4}\right)(1 in .)}=1460 psiThis stress exists both as a vertical shear stress acting on the cross section and as a horizontal shear stress acting on the horizontal plane between the flange and the web.
Maximum shear stress. The maximum shear stress occurs in the web at the neutral axis. Therefore, we calculate the first moment Q_{max} of the cross-sectional area below the neutral axis:
As previously indicated, we would get the same result if we calculated the first moment of the area above the neutral axis, but those calculations would be slighter longer.
Substituting into the shear formula, we obtain
which is the maximum shear stress in the beam.
The parabolic distribution of shear stresses in the web is shown in Fig. 5-40b.