# Question 9.2: Just to the left of point B in Fig. 1a (same as Fig. 1 of Ex......

Just to the left of point B in Fig. 1a (same as Fig. 1 of Example Problem 6.5) the transverse shear force is V = -28 kips, and the bending moment is M = -48 kip · ft. Ignoring any stress concentration due to the support at B, determine the principal stresses at point D in Fig. 1b.
Compare the maximum tensile stress at D with the maximum flexural stress in the beam, which occurs at x = 5 ft, where M(5 ft) = 50 kip · ft and V(5 ft) = 0. The beam is a W14 × 26. (See Example Problems 6.5 and 6.15.)

Step-by-Step
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Plan the Solution We can use the flexure formula to determine the normal stress on the cross section at B, and the shear stress distribution on this cross section was determined in Example Problem 6.15. We can use Mohr’s circle to combine these and to determine the principal stress magnitudes and directions at point D.

The shear force and bending moment just to the left of section B are shown in Fig. 2a, and the essential cross-sectional dimensions of the W14 × 26 beam are shown in Fig. 2b.
The normal stress at D is

$(σ_x)_D = – \frac{My}{I} = – \frac{(-48 kip · ft)(12 in./ft)(6.535 in.)}{245 in^4}$

or

$(σ_x)_D$ = 15.364 ksi (T)

The shear stress at point D has a magnitude

$τ_D ≡ |τ_{xy}|_D = \frac{|V|Q}{It}= \frac{28 kips(5.025 in.)(0.420 in.)(6.745 in.)}{(245 in^4)(0.255 in.)}$

or

$τ_D$ = 6.380 ksi ↑

We can construct a Mohr’s circle for the stresses in the plane of the web at point D. From the Mohr’s circle in Fig. 3,

R = $\sqrt{(7.682)^2 + (6.380)^2}$ = 9.986 ksi

$\left\{\begin{matrix}σ_{1_D} = 7.682 ksi + 9.986 ksi = 17.67 ksi \\ σ_{2_D} = 7.68 ksi – 9.98 ksi = -2.30 ksi \end{matrix} \right\}$

$2θ_{xp1} = \tan^{-1} \left(\frac{6.380}{7.682}\right)$ = 39.71°

$θ_{xp1}$ = 19.9°

At x = 5 ft, the maximum tensile stress is just the flexural stress at the bottom of the beam (since V = 0). Therefore,

$σ_x(5 ft, -6.955 in.) = \frac{-My}{I}= \frac{-(50 kip · ft)(12 in./ft)(-6.955 in.)} {245 in^4}$

or

$σ_x$(5 ft, -6.955 in.) = 17.03 ksi

Therefore, the tensile principal stress at D, 17.67 ksi, is slightly larger than the maximum flexural stress in the beam, 17.03 ksi. However, since an allowable stress of 19 ksi was used in Example Problem 6.5 in selecting the W14 × 26 beam cross section, the beam has enough stress margin to be safe, even though the principal stress at D is slightly larger than the maximum flexural stress on which the beam design was originally based.

Review the Solution The calculations in this problem are quite simple, so they can just be rechecked for accuracy. Since there is a significant value of transverse shear force at a cross section where the moment is nearly its maximum value, and since the cross section has heavy flanges and a thin web, we should not be surprised to find principal stresses, like the $σ_1$ stress at point D, that exceed the maximum flexural stress in the beam.

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