Question 6.2: The cross section of a beam is a T with the dimensions shown......

Plan the Solution To use the flexure formula, Eq. 6.13, we need first to locate the centroid of the cross section and then compute the moment of inertia about an axis through the centroid (Appendix C). Since M is positive, the maximum compressive stress occurs in the top fibers, and the maximum tensile stress occurs in the bottom fibers.

σ_x = \frac{-My}{I}        Flexure Formula      (6.13)

(a) Locate the neutral axis. As indicated in Fig. 2, we can pick an arbitrary origin at the bottom and let a coordinate in the y-direction be called η. Then, by summing area contributions to the first moment, we have

or

\bar{η}A = \bar{η}_1A_1 + \bar{η}_2A_2 = (5.5 in.)(5 in²) + (2.5 in.)(5 in²) = 40 in³

where

A = A_1 + A_2 = (5 in²) + (5 in²) = 10 in²

Then,

\bar{η} = \frac{40  in^3} {10  in^2} = 4.0 in.     (a)

(b) Determine the moment of inertia with respect to the neutral axis. The moment of inertia of a rectangle about an axis through its own centroid is

and, from the parallel-axis theorem, the moment of inertia about an axis through C′ parallel to the axis through the centroid C is

Therefore,

= \left[\frac{1} {12} (5  in.)(1  in.)^3 + (5  in.)(1  in.)(1.5  in.)^2\right]

+ \left[ \frac{1}{12}(1  in.)(5  in.)^3 + (1  in.)(5  in.)(1.5  in.)^2\right]

so,

I = 33.3 in^4    (b)

(c) Compute σ_{max  T} and σ_{max  C}. The maximum compression occurs at the top of the beam, and the maximum tension occurs at the bottom of the beam. From the flexure formula, Eq. 6.13,

σ_x = \frac{-My}{I}:

σ_{max  C} = σ(x, 2 in.) = \frac{-(4  kip · ft)(12  in./ft)(2  in.)} {33.3  in^4}

= -2.88 ksi (2.88 ksi C)

σ_{max  T} = σ(x, -4 in.) = \frac{-(4  kip · ft)(12  in./ft)(-4  in.)} {33.3  in^4}

= 5.76 ksi (5.76 ksi T)

σ_{max  T} = 5.76 ksi, σ_{maxC} = -2.88 ksi   (c)

Review the Solution The centroid must lie between the centroids of the two areas A_1 and A_2. Furthermore, since the areas are equal, the combined centroid C lies midway between the individual centroids, as we obtained above.
To see if the order of magnitude of I is reasonable, we can compare our answer with the moment of inertia of a 1 in. × 6 in. web about its own centroid (\frac{1}{12}(1)(6)^3 = 18  in^4). Thus, the value of I = 33.3 in^4 appears reasonable.
Finally, since the T-section is not symmetric about the neutral axis, the maximum tension and maximum compression are not equal in magnitude, but, since σ is linear in y, their magnitudes are in the ratio of the distances to the top and bottom fibers.
(See Homework Problem 6.6-6, where the effect of misalignment of the load is examined.)