Question 12.1: For the timber beam and loading shown, draw the shear and be...
For the timber beam and loading shown, draw the shear and bending-moment diagrams and determine the maximum normal stress due to bending.
STRATEGY: After using statics to find the reaction forces, identify sections to be analyzed. You should section the beam at points to the immediate left and right of each concentrated force to determine values of V and M at these points.
MODELING and ANALYSIS:
Reactions. Considering the entire beam to be a free body (Fig. 1),
\pmb{\text{R}}_B=40 \text{ kN}\uparrow \quad \quad \pmb{\text{R}}_D=14 \text{ kN}\uparrow
Shear and Bending-Moment Diagrams. Determine the internal forces just to the right of the 20-kN load at A. Considering the stub of beam to the left of section 1 as a free body and assuming V and M to be positive (according to the standard convention), write
+\uparrow \sum{F_y=0:} \quad \quad -20 \text{ kN}-V_2=0 \quad \quad \quad \quad \quad \quad V_2=-20 \text{ kN} \\ +\circlearrowleft \sum M_2=0: \quad \quad (20 \text{ kN})(2.5 \text{ m})+M_2=0) \quad \quad M_2=-50 \text{ kN}\cdot \text{m}The shear and bending moment at sections 3, 4, 5, and 6 are determined in a similar way from the free-body diagrams shown in Fig. 1:
V_3=+26 \text{ kN} \quad \quad M_3=-50 \text{ kN}\cdot \text{m} \\ V_4=+26 \text{ kN} \quad \quad M_4=+28\text{ kN}\cdot \text{m} \\ V_5=-14 \text{ kN} \quad \quad M_5=+28 \text{ kN}\cdot \text{m} \\ V_6=-14 \text{ kN} \quad \quad M_6=0
For several of the latter sections, the results may be obtained more easily by considering the portion to the right of the section to be a free body. For example, for the portion of beam to the right of section 4,
+\uparrow \sum F_y=0: \quad \quad V_4-40 \text{ kN}+14 \text{ kN}=0 \quad \quad \quad V_4=+26 \text{ kN} \\ +\circlearrowleft \sum{M_4=0:} \quad \quad -M_4+(14 \text{ kN})(2 \text{ m})=0 \quad \quad M_4=+28 \text{ kN}\cdot \text{m}Now plot the six points shown on the shear and bending-moment diagrams. As indicated earlier, the shear is of constant value between concentrated loads, and the bending moment varies linearly.
Maximum Normal Stress. This occurs at B, where |M| is largest. Use Eq. (12.4) to determine the section modulus of the beam:
S=\frac{1}{6}bh^2=\frac{1}{6}(0.080 \text{ m})(0.250 \text{ m})^2=833.33 \times 10^{-6} \text{ m}^3
Substituting this value and |M| = |M_B| = 50 × 10³ N.m into Eq. (12.3) gives
\sigma_m=\frac{|M_B|}{S}=\frac{(50 \times 10^3 \text{ N.m})}{833.33 \times 10^{-6}} =60.00 \times 10^6 \text{Pa}
Maximum normal stress in the beam = 60.0 MPa
