Question 12.S-P.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.
Reactions. Considering the entire beam as a free body, we find
R_{B} = 46 kN ↑ R_{D} = 14 kN ↑
Shear and Bending-Moment Diagrams. We first 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), we write
+↑\sum{F_{y}} = 0 : -20 kN – V_{1} = 0 V_{1} = -20 kN+\curvearrowleft \sum{M_{1}} = 0 : (20 kN) (0 m)+M_{1} = 0 M_{1} = 0
We next consider as a free body the portion of beam to the left of section 2 and write
+↑\sum{F_{y}} = 0 : -20 kN – V_{2} = 0 V_{2} = -20 kN+\curvearrowleft \sum{M_{2}} = 0 : (20 kN) (2.5 m)+M_{2} = 0 M_{2} = -50 kN \cdot 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. We obtain
V_{3} = +26 kN M_{3} = -50 kN \cdot m
V_{4} = +26 kN M_{4} = +28 kN \cdot m
V_{5} = -14 kN M_{5} = +28 kN \cdot m
V_{6} = -14 kN M_{6} =0
For several of the latter sections, the results may be more easily obtained by considering as a free body the portion of the beam to the right of the section. For example, for the portion of the beam to the right of section 4, we have
+↑\sum{F_{y}} = 0 : V_{4} – 40 kN + 14 kN = 0 V_{4} = +26 kN+\curvearrowleft \sum{M_{4}} = 0 : -M_{4} + (14 kN) (2 m) = 0 M_{4} = +28 kN \cdot m
We can now plot the six points shown on the shear and bendingmoment diagrams. As indicated earlier in this section, the shear is of constant value between concentrated loads, and the bending moment varies linearly; we obtain therefore the shear and bending-moment diagrams shown.
Maximum Normal Stress. It occurs at B, where |M| is largest. We use Eq. (12.4) to determine the section modulus of the beam:
S = \frac{1}{6} bh^{2} (12.4)
S = \frac{1}{6} bh^{2} = \frac{1}{6} (0.080 m) (0.250 m)^{2} = 833.33 × 10^{-6} m^{3}
Substituting this value and |M| = |M_{B}| = 50 × 10^{3} N \cdot m into Eq. (12.3):
σ_{m} =\frac{|M|}{S} (12.3)
σ_{m} =\frac{|M_{B}|}{S} = \frac{(50 × 10^{3} N \cdot m)}{833.33 × 10^{-6}} = 60.00 × 10^{6} Pa
Maximum normal stress in the beam = 60.0 MPa
