Question 5.11: A metal beam with span L = 3 ft is simply supported at point...
A metal beam with span L = 3 ft is simply supported at points A and B (Fig. 5-32a). The uniform load on the beam (including its own weight) is q = 160 lb/in. The cross section of the beam is rectangular (Fig. 5-32b) with width b = 1 in. and height h = 4 in. The beam is adequately supported against sideways buckling.
Determine the normal stress \sigma_{C} and shear stress \tau_{C} at point C, which is located 1 in. below the top of the beam and 8 in. from the right-hand support. Show these stresses on a sketch of a stress element at point C.
Shear force and bending moment. The shear force V_{C} and bending moment M_{C} at the cross section through point C are found by the methods described in Chapter 4. The results are
M_{C}=17,920 lb \text {-in. } \quad V_{C}=-1,600 lbThe signs of these quantities are based upon the standard sign conventions for bending moments and shear forces (see Fig. 4-5).
Moment of inertia. The moment of inertia of the cross-sectional area about the neutral axis (the z axis in Fig. 5-32b) is
Normal stress at point C. The normal stress at point C is found from the flexure formula (Eq. 5-13) with the distance y from the neutral axis equal to 1.0 in.; thus,
\sigma_{C}=-\frac{M y}{I}=-\frac{(17,920 lb – in.)(1.0 in.)}{5.333 in.{ }^{4}}=-3360 psiThe minus sign indicates that the stress is compressive, as expected.
Shear stress at point C. To obtain the shear stress at point C, we need to evaluate the first moment Q_{C} of the cross-sectional area above point C (Fig. 5-32b). This first moment is equal to the product of the area and its centroidal distance (denoted y_{C}) from the z axis; thus,
Now we substitute numerical values into the shear formula (Eq. 5-38) and obtain the magnitude of the shear stress:
\tau=\frac{V Q}{I b} (5-38)
\tau_{C}=\frac{V_{C} Q_{C}}{I b}=\frac{(1,600 lb )\left(1.5 in .{ }^{3}\right)}{\left(5.333 in. ^{4}\right)(1.0 in.)}=450 psiThe direction of this stress can be established by inspection, because it acts in the same direction as the shear force. In this example, the shear force acts upward on the part of the beam to the left of point C and downward on the part of the beam to the right of point C. The best way to show the directions of both the normal and shear stresses is to draw a stress element, as follows.
Stress element at point C. The stress element, shown in Fig. 5-32c, is cut from the side of the beam at point C (Fig. 5-32a). Compressive stresses \sigma_{C} = 3360 psi act on the cross-sectional faces of the element and shear stresses \tau_{C} = 450 psi act on the top and bottom faces as well as the cross-sectional faces.
