A beam having the cross-section shown in Fig. 15.13 is subjected to a bending moment of 1,500 Nm in a vertical plane. Calculate the maximum direct stress due to bending stating the point at which it acts.

Question

Accepted Answer

The position of the centroid of the section may be found by taking moments of areas about some convenient point. Thus,

[latex](120 imes 8+80 imes 8) \bar{y}=120 imes 8 imes 4+80 imes 8 imes 48[/latex]

giving

[latex]\bar{y}=21.6 mm[/latex]

and

[latex](120 imes 8+80 imes 8) \bar{x}=80 imes 8 imes 4+120 imes 8 imes 24[/latex]

giving

[latex]\bar{x}=16 mm[/latex]

The next step is to calculate the section properties referred to axes Cxy (see Section 15.4):

[latex]I_{x x}=\frac{120 imes(8)^{3}}{12}+120 imes 8 imes(17.6)^{2}+\frac{8 imes(80)^{3}}{12}+80 imes 8 imes(26.4)^{2}[/latex]

[latex]=1.09 imes 10^{6} mm ^{4}[/latex]

[latex]I_{y y}=\frac{8 imes(120)^{3}}{12}+120 imes 8 imes(8)^{2}+\frac{80 imes(8)^{3}}{12}+80 imes 8 imes(12)^{2}[/latex]

[latex]=1.31 imes 10^{6} mm ^{4}[/latex]

[latex]I_{x y}=120 imes 8 imes 8 imes 17.6+80 imes 8 imes(-12) imes(-26.4)[/latex]

[latex]=0.34 imes 10^{6} mm ^{4}[/latex]

Since [latex]M_{x}=1,500[/latex] Nm and [latex]M_{y}=0[/latex], we have, from Eq. (15.18),

[latex]\sigma_{z}=\frac{M_{x}\left(I_{y y} y-I_{x y} x ight)}{I_{x x} I_{y y}-I_{x y}^{2}}+\frac{M_{y}\left(I_{x x} x-I_{x y} y ight)}{I_{x x} I_{y y}-I_{x y}^{2}}[/latex] (15.18)

[latex]\sigma_{z}=1.5 y-0.39 x[/latex] (i)

in which the units are N and mm. By inspection of Eq. (i), we see that [latex]\sigma_{z}[/latex] is a maximum at F, where [latex]x=-8 mm , y=-66.4 mm[/latex]. Thus,

[latex]\sigma_{z, \max }=-96 N / mm ^{2} ext { (compressive) }[/latex]

In some cases the maximum value cannot be obtained by inspection, so that values of [latex]\sigma_{z}[/latex] at several points must be calculated.

Question 15.4: A beam having the cross-section shown in Fig. 15.13 is subje...

Related Answered Questions