Before we can find the maximum bending stress in the beam, we must find the maximum bending moment.
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[tex]$I=\frac{b h^{3}}{12}=\frac{0.12(0.2)^{3}}{12}=80.0 \times 10^{-6} \mathrm{~m}^{4}$[/tex]
1) The maximum bending stress in the beam occurs on the cross section that carries the largest
[tex]$\sigma_{\max }=\frac{|M|_{\max } c}{I}=\frac{\left(16 \times 10^{3}\right)(0.1)}{80.0 \times 10^{-6}}=20.0 \times 10^{6} \mathrm{~Pa}=20.0 \mathrm{MPa}$[/tex]
What is the bending stress formula?
The bending stress is computed for the rail by the equation Sb = Mc/I, where Sb is the bending stress in pounds per square inch, M is the maximum bending moment in pound-inches, I is the moment of inertia of the rail in (inches)4, and c is the distance in inches from the base of rail to its neutral axis.
2) The stress distribution on the cross section at D is shown in Fig. (d). When drawing the figure, we were guided by the following observations: (i) the bending stress varies linearly with distance from the neutral axis; (ii) because M_{max} is positive, the top half of the cross section is in compression and the bottom half is in tension; and (iii) due to symmetry of the cross section about the neutral axis, the maximum tensile and compressive stresses are equal in magnitude.
3) the bending stress at a point on section B that is 35 mm below the top of the beam.
[tex]$\sigma=-\frac{M y}{I}=-\frac{\left(9.28 \times 10^{3}\right)(0.075)}{80.0 \times 10^{-6}}=-8.70 \times 10^{6} \mathrm{~Pa}=-8.70 \mathrm{MPa}$[/tex]
To learn more about the bending stress, refer
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