Question 5.9: A metal beam with span L=1 m is simply supported at points A...
A metal beam with span L=1 m is simply supported at points A and B (Fig. 5-29a). The uniform load on the beam (including its own weight) is q = 28 kN/m. The cross section of the beam is rectangular (Fig. 5-29b) with width b = 25 mm and height h = 100 mm. 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 25 mm below the top of the beam and 200 mm 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 = 2.22 kN·m V_C=-8.4 kN
The 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. 5-29b) is
I=\frac{bh^3}{12} =\frac{1}{12} (25 mm)(100 mm)^3 = 2083 \times 10^3 mm^4 = 5.333 in.^4
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 25 mm; thus,
\sigma _C=-\frac{My}{I} (5-13)
=-\frac{(2.24 \times 10^6 N\cdot mm)(25 mm)}{2083 \times 10^3 mm^4} =-26.9 MPa
The 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. 5-29b). This first moment is equal to the product of the area and its centroidal distance (denoted y_C) from the z axis; thus,
A_C = (25 mm)(25 mm) = 625 mm² y_C = 37.5 mm
Q_C = A_C y_C = 23,440 mm³
Now we substitute numerical values into the shear formula (Eq. 5-35) and obtain the magnitude of the shear stress:
\tau =\frac{VQ}{Ib} (5-35)
\tau _C=\frac{V_CQ_C}{Ib}=\frac{(8400 N)(23,440 mm^3)}{(2083 \times 10^3 mm^4)(25 mm)} = 3.8 MPa
The 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. 5-29c, is cut from the side of the beam at point C (Fig. 5-29a). Compressive stresses \sigma_C = 26.9 MPa act on the cross-sectional faces of the element and shear stresses \tau_C = 3.8 MPa act on the top and bottom faces as well as the cross sectional faces.
