A uniform beam with the cross-section shown in Fig. P.20.5(a) is supported and loaded as shown in Fig. P.20.5(b). If the direct and shear stresses are given by the basic theory of bending, the direct stresses being carried by the booms and the shear stresses by the walls, calculate the vertical

Question

A uniform beam with the cross-section shown in Fig. P.20.5(a) is supported and loaded as shown in Fig. P.20.5(b). If the direct and shear stresses are given by the basic theory of bending, the direct stresses being carried by the booms and the shear stresses by the walls, calculate the vertical deflection at the ends of the beam when the loads act through the shear centres of the end cross-sections, allowing for the effect of shear strains.

Take E = 69 000 N/mm² and G = 26 700 N/mm². Boom areas: 1, 3, 4, 6 = 650 mm², 2, 5 = 1300 mm².

Accepted Answer

Referring to Fig. S.20.5(a) the x axis of the beam cross-section is an axis of symmetry so that [latex]I_{xy}[/latex] = 0. Further, [latex]S_y[/latex] at the end A is equal to −4450 N and [latex]S_x[/latex] = 0. The total deflection, Δ , at one end of the beam is then, from Eqs (20.17) and (20.19)

[latex]\Delta_M=\frac{1}{E} \int_L\left(\frac{M_{y, 1} M_{y, 0}}{I_{y y}}+\frac{M_{x, 1} M_{x, 0}}{I_{x x}} ight) \mathrm{d} z[/latex] (20.17)

[latex]\Delta_S=\int_L\left(\int_{ ext {sect }} \frac{q_0 q_1}{G t} \mathrm{~d} s ight) \mathrm{d} z[/latex] (20.19)

[latex]\Delta=\int_L \frac{M_{x, 1} M_{x, 0}}{E I_{x x}} \mathrm{~d} z+\int_L\left(\int_{\operatorname{sect}} \frac{q_0 q_1}{G t} \mathrm{~d} s ight) \mathrm{d} z[/latex] (i)

in which [latex]q_0[/latex], from Eqs (20.20) and (20.11) is given by

[latex]\begin{aligned}q_0=&-\left(\frac{S_{x, 0} I_{x x}-S_{y, 0} I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} x \mathrm{~d} s+\sum_{r=1}^n B_r x_r ight) \&-\left(\frac{S_{y, 0} I_{y y}-S_{x, 0} I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r ight)\end{aligned}[/latex] (20.20)

[latex]\begin{aligned}q_s=&-\left(\frac{S_x I_{x x}-S_y I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} x \mathrm{~d} s+\sum_{r=1}^n B_r x_r ight) \&-\left(\frac{S_y I_{y y}-S_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r ight)+q_{s, 0}\end{aligned}[/latex] (20.11)

[latex]q_0=-\frac{S_{y, 0}}{I_{x x}} \sum_{r=1}^n B_r y_r+q_{s, 0}[/latex] (ii)

and

[latex]q_1=\frac{q_0}{4450}[/latex]

Since the booms carrying all the direct stresses, [latex]I_{xx}[/latex] in Eq. (i) is, from Fig. S.20.5(a)
[latex]I_{x x}=2 imes 650 imes 100^2+2 imes 650 imes 75^2+2 imes 1300 imes 100^2=46.3 imes 10^6 \mathrm{~mm}^4[/latex]
Also, from Fig. S.20.5(b) and taking moments about C
[latex]R_B[/latex] × 500 − 4450 × 1750 − 4450 × 1250 = 0

from which
[latex]R_B[/latex] = 26 700 N
Therefore in AB
[latex]M_{x, 0}=4450 z \quad M_{x, 1}=z[/latex]
and in BC
[latex]M_{x, 0}=33.4 imes 10^6-22250 z \quad M_{x, 1}=7500-5 z[/latex]

Thus the deflection, [latex]Δ_M[/latex], due to bending at the end A of the beam is, from the first term on the right-hand side of Eq. (i)

[latex]\Delta_{\mathrm{M}}=\frac{1}{E I_{x x}}\left\{\int_0^{1250} 4450 z^2 \mathrm{~d} z+\int_{1250}^{1500} 4450(7500-5 z)^2 \mathrm{~d} z ight\}[/latex]

i.e.

[latex]\Delta_{\mathrm{M}}=\frac{4450}{69000 imes 46.3 imes 10^6}\left\{\left[\frac{z^3}{3} ight]_0^{1250}-\frac{1}{15}\left[(7500-5 z)^3 ight]_{1250}^{1500} ight\}[/latex]

from which
[latex]Δ_M[/latex] = 1.09 mm
Now ‘cut’ the beam section in the wall 12. From Eq. (20.11), i.e.

[latex]q_s=-\frac{S_y}{I_{x x}} \sum_{r=1}^n B_r y_r+q_{s, 0}[/latex] (iii)

[latex]\begin{aligned}&q_{\mathrm{b}, 12}=0 \&q_{\mathrm{b}, 23}=-\frac{S_y}{I_{x x}} imes 1300 imes 100=-130000 \frac{S_y}{I_{x x}} \&q_{\mathrm{b}, 34}=-130000 \frac{S_y}{I_{x x}}-\frac{S_y}{I_{x x}} imes 650 imes 100=-195000 \frac{S_y}{I_{x x}} \&q_{\mathrm{b}, 16}=-\frac{S_y}{I_{x x}} imes 650 imes 75=-48750 \frac{S_y}{I_{x x}}\end{aligned}[/latex]

The remaining distribution follows from symmetry. The shear load is applied through the shear centre of the cross-section so that [latex]\mathrm{d} heta / \mathrm{d} z=0 ext { and } q_{s, 0}[/latex] is given by Eq. (17.28), i.e.

[latex]q_{s, 0}=-\frac{\oint q_{\mathrm{b}} \mathrm{d} s}{\oint \mathrm{d} s}[/latex] (17.28)

[latex]q_{s, 0}=-\frac{\oint q_{\mathrm{b}} \mathrm{d} s}{\oint \mathrm{d} s} \quad(t= ext { constant })[/latex]

in which

[latex]\oint \mathrm{d} s-2 imes 300+2 imes 250+2 imes 100+2 imes 75-1450 \mathrm{~mm}[/latex]

i.e.

[latex]q_{s, 0}=-\frac{2 S_y}{1450 I_{x x}}(-130000 imes 250-195000 imes 100+48750 imes 75)[/latex]

from which
[latex]q_{s, 0}=66681 S_y / I_{x x}[/latex]
Then

[latex]\begin{aligned}&q_{12}=66681 S_y / I_{x x} \&q_{23}=-63319 S_y / I_{x x} \&q_{34}=-128319 S_y / I_{x x} \&q_{16}=-115431 S_y / I_{x x}\end{aligned}[/latex]

Therefore the deflection, [latex]Δ_S[/latex], due to shear is, from the second term in Eq. (i)

[latex]\Delta_{\mathrm{S}}=\int_L\left(\int_{\mathrm{sect}} \frac{q_0 q_1}{G t} \mathrm{~d} s ight) \mathrm{d} z[/latex]

i.e.

[latex]\begin{aligned}\Delta_{\mathrm{S}}-\int_L\{2& \frac{S_{y, 0} S_{y, 1}}{G t I_{x x}^2}\left(115431^2 imes 75+66681^2 imes 300+63319^2 imes 250 ight.\&\left.\left.+128319^2 imes 100 ight) ight\} \mathrm{d} z\end{aligned}[/latex]

Thus

[latex]\Delta_{\mathrm{S}}=\int_L 2 \frac{S_{y, 0} S_{y, 1} imes 4.98 imes 10^{12}}{26700 imes 2.5 imes\left(46.3 imes 10^6 ight)^2} \mathrm{~d} z=\int_L 6.96 imes 10^{-8} S_{y, 0} S_{y, 1} \mathrm{~d} z[/latex]

Then

[latex]\Delta \mathrm{s}=\int_0^{1250} 6.96 imes 10^{-8} imes 4450 imes 1 \mathrm{~d} z+\int_{1250}^{1500} 6.96 imes 10^{-8} imes 22250 imes 5 \mathrm{~d} z[/latex]

from which
[latex]\Delta_{\mathrm{S}}=2.32 \mathrm{~mm}[/latex]
The total deflection, Δ , is then
[latex]\Delta=\Delta_{\mathrm{M}}+\Delta_{\mathrm{S}}=1.09+2.32=3.41 \mathrm{~mm}[/latex]

Question p.20.5: A uniform beam with the cross-section shown in Fig. P.20.5(a...

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