Question 5.13: A beam having a T-shaped cross section (Fig. 5-37a) is subje...

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. 5-37a). First, we divide the cross section into two rectangles, the flange and the web (see the dashed line in Fig. 5-37a). 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 10, Section 10.3 available online, for methods for locating centroids of composite areas). The calculations are as follows:

A=\sum{A_i}=b(h-h_1)+th_1 =11.0  in.^{2}

Q_{aa}=\sum{y_iA_i}=(\frac{h  +  h_1}{2})(b)(h  –  h_1)  +  \frac{h_1}{2}(th_1)=54.5  in.^3

c_2=\frac{Q_{aa}}{A}=\frac{54.5  in.^3}{11.0  in.^2}=4.955  in.                    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 10.5 available online):

I=I_{aa}  –  Ac_{2}^{2}

The calculations are as follows:

I_{aa}=\frac{bh^{3}}{3}  –  \frac{(b  –  t)h_{1}^{3}}{3}=339.67  in.^4                    Ac_{2}^{2}=270.02  in.^4                   I=69.65  in.^4

Shear stress at top of web. To find the shear stress  τ_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(h  –  h_1)(c_1  –  \frac{h  –  h_1}{2})

=(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=th_1(c_2  –  \frac{h_1}{2})=(1  in.)(7  in.)(4.955  in.  –  3.5  in.)=10.18  in.^3

Substituting into the shear formula, we find

τ_1=\frac{VQ_1}{It}=\frac{(10,000  lb)(10.18  in.^3)}{(69.65  in.^4)(1  in.)}=1460  psi

This 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:

Q_{max}=tc_2(\frac{c_2}{2})=(1  in.)(4.955  in.)(\frac{4.955  in.}{2})=12.28  in.^3

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 slightly longer.

Substituting into the shear formula, we obtain

τ_{max}=\frac{VQ_{max}}{It}=\frac{(10,000  lb)(12.28  in.^3)}{(69.65  in.^4)(1  in.)}=1760  psi

which is the maximum shear stress in the beam.

The parabolic distribution of shear stresses in the web is shown in Fig. 5-37b.