Question 10.4.2: The cross section of the cantilever beam of Example 10.3.3 (...
CALCULATING BEAM NORMAL STRESS
The cross section of the cantilever beam of Example 10.3.3 (Figure 1) is a rectangle 0.4 m high and 0.1 m wide (Figure 2). Determine the largest stress in the beam.
Goal Determine the maximum stress in beam ABC based on the moment diagram developed in Example 10.3.3 (Figure 3).
Given The beam cross section and the shear and bending moment diagrams developed in Example 10.3.3.
Assume Since no other information is given, assume the cross section of the beam is constant throughout its length.
Formulate Equations and Solve To calculate stress, we use (10.7)
normal stress = \sigma = \frac{- M_{bz}y}{I} (10.7)
We can examine the variation of stress throughout the beam. Since the cross section is assumed constant throughout the beam, the moment of inertia I and the distance from the centroid y in (10.7) are the same everywhere along the length of the beam. Thus the maximum stress will occur where \left\|M_{bz} (x)\right\| is maximum. Examining the moment diagram in Figure 3, we see that the maximum moment of −100 kN.m occurs at C.
According to Appendix C, the area moment of inertia for a rectangular beam is bh³/12, where b is the beam width and h is the height.
I=\frac{0.1 m(0.4 m)^{3}}{12} =0.533 \times 10^{-3} m^{4}
at the top of the beam
\sigma _{max}=\frac{(- 100 kN.m)(0.2 m)}{0.533 \times 10^{-3} m^{4}}
=37.5 \times 10^{3}\frac{N}{m^{2}} =37.5 MPa
at the bottom of the beam
\sigma _{max}=\frac{(- 100 kN.m)( – 0.2 m)}{0.533 \times 10^{-3} m^{4}}
=37.5 \times 10^{3}\frac{N}{m^{2}} = – 37.5 MPa
Answer \sigma _{max}= = 37.5 MPa and occurs at x = 6 m. because the beam is curving downward, the top of the beam experiences tensile stresses and the bottom of the beam experiences compressive stresses, as illustrated in Figure 4.
