Question 3.7: A beam 12 in long is to support a load of 488 lbf acting 3 i...

The loading, shear-force, and bending-moment diagrams are shown in Fig. 3–19b. If the direct shear force is included in the analysis, the maximum stresses at the top and bottom of the beam will be the same as if only bending were considered. The maximum bending stresses are

σ = ±\frac {M_{c}}{I} = ±\frac {1098(1.5)}{1.66} = ±992 psi

However, the maximum stress due to the combined bending and direct shear stresses may be maximum at the point (3−, 1.227) that is just to the left of the applied load, where the web joins the flange. To simplify the calculations we assume a cross section with square corners (Fig. 3–19c). The normal stress at section ab, with x = 3 in, is

σ = -\frac {M_{y}}{I} = ±\frac {1098(1.227)}{1.66} = -812 psi

For the shear stress at section ab, considering the area above ab and using Eq. (3–30) gives

Q =\int_{y_{1}}^{c}{y dA} =\overline{y}′A′           (3–30)

Q=\overline{y}′A′=(1.227 +\frac {0.273}{2})(1.410)(0.273) =0.525 in^{3}

Using Eq. (3–31) with V = 366 lbf, I = 1.66 in^{4}, Q = 0.525 in^{3}, and b = 0.170 in yields

τ = \frac {V Q}{I b }          (3.13)

τ_{xy} = −\frac {V Q}{I b }= −\frac {366(0.525)}{1.66(0.170)}= −681 psi

The negative sign comes from recognizing that the shear stress is down on an x face of a dx dy element at the location being considered.
The principal stresses at the point can now be determined. Using Eq. (3–13), we find that at x = 3− in, y = 1.227 in,

σ_{1}, σ_{2} =\frac {σ_{x} + σ_{y}}{2} ±\sqrt {(\frac {σ_{x} − σ_{y}}{2})^{2}+ τ^{2}_{xy}}                    (3–13)

= \frac {−812 + 0}{2} ±\sqrt {(\frac {−812 − 0}{2})^{2}+ (−681)^{2}} = 387, −1200 psi

For a point at x = 3^{−} in, y = −1.227 in, the principal stresses are σ_{1}, σ_{2} = 1200, −387 psi. Thus we see that the maximum principal stresses are ±1200 psi, 21 percent higher than thought by the designer.