The gantry shown in Fig. 12.15 is constructed from 100 mm × 50 mm rectangular crosssection and, under service conditions, supports a maximum load P of 20 kN. Determine the maximum distance d at which P can be safely applied if the maximum tensile and compressive stresses for the material used are limited to 30 MN/m² and 100 MN/m² respectively.

How would this value change if the cross-section were circular, but of the same crosssectional area?

Question

The gantry shown in Fig. 12.15 is constructed from 100 mm × 50 mm rectangular crosssection and, under service conditions, supports a maximum load P of 20 kN. Determine the maximum distance d at which P can be safely applied if the maximum tensile and compressive stresses for the material used are limited to 30 MN/m² and 100 MN/m² respectively.

How would this value change if the cross-section were circular, but of the same crosssectional area?

Accepted Answer

For the gantry and cross-section of Fig. 12.15 the following values are obtained by inspection:

[latex]R_{c} = 150 mm [/latex] [latex]R_{i} = 100 mm [/latex] [latex]R_{0} = 200 mm[/latex] [latex] b = 50 mm.[/latex]

:. From Table 12.1(a)

Table 12.1. Values of [latex]\int{\frac{dA}{r} } [/latex] for curved bars
Cross-section	[latex]\int{\frac{dA}{r} } [/latex]
(a) Rectangle	[latex]b\log _{e}(\frac{R_{0}}{R_{i}} )[/latex] (N.B. The two following cross-sections are simply produced by the addition of terms of this form for each rectangular portion)
(b )T-section	[latex]b_{1}\log _{e}(\frac{R_{i}+d_{1}}{R_{i}} )+b_{2}\log _{e}(\frac{R_{0}}{R_{i}+d_{1}} )[/latex]
(c) I-beam	[latex]b_{1}\log _{e}(\frac{R_{i}+d_{1}}{R_{i}} )+b_{2}\log _{e}(\frac{R_{0}-d_{3}}{R_{i}+d_{1}} )+b_{3}\log _{e}(\frac{R_{0}}{R_{0}+d_{3}} )[/latex]
(d) trapezoid	[latex][\frac{(b_{1}R_{0}-b_{2}R_{i})}{d} \log _{e}(\frac{R_{0}}{R_{i}} )]-b_{1}+b{2}[/latex]
(e ) triangle	As above (d) with [latex]b_{2} = 0[/latex] As above (d) with [latex]b_{1} = 0[/latex]
	[latex]2\pi \left\{(R_{i}+R)-[(R_{i}+R)^{2}-R_{2}]^{1/2} ight\}[/latex]

[latex]\int{\frac{dA}{r} } =b\log _{e}(\frac{R_{o}}{R_{i}} )=50 \log _{e}(\frac{200}{100} )[/latex]
[latex]=34.6574 mm[/latex]
[latex]R_{1}=\frac{A}{\int{\frac{dA}{r} }} =\frac{50 imes 100}{34.6574} =144.269 mm[/latex]
[latex]h=R_{c}-R_{1}=150-144.269=5.731 mm[/latex]

Direct stress (Compressive) due to [latex] P=\frac{P}{A} =\frac{20 imes 10^{3}}{(100 imes 50)10^{-6}} =4 MN/m^{2}[/latex]

Thus, for maximum tensile stress of 30 MN/m² to be reached at B the bending stress (tensile) must be 30 + 4 = 34 MN/m².

Now [latex]y_{max}=50+5.731 [/latex]
[latex]=55.731 [/latex] at B

and bending stress at [latex]B=\frac{My}{hA(R_{1}+y)} =34 MN/m^{2} [/latex]
[latex]\frac{(40 imes 10^{3} d) imes 55.731 imes 10^{-3}}{(5.731 imes 10^{-3})(50 imes 100 imes 10^{-6})(200 imes 10^{-3})} =34 imes 10^{6} [/latex]
[latex]d=174.69 mm [/latex]

For maximum compressive stress of 100 MN/m² at A the compressive bending stress must be limited to 100 − 4 = 96 MN/m² in order to account for the additional direct load effect

At A, with [latex] y_{min}=50-5.731=44.269[/latex]

bending stress [latex]=\frac{(20 imes 10^{3})44.269 imes 10^{-3}}{(5.731 imes 10^{-3})(50 imes 100 imes 10^{-6})(100 imes 10^{-3})} =96 imes 10^{6}[/latex]

[latex]d = 310.7 mm[/latex].

The critical condition is therefore on the tensile stress at B and the required maximum value of d is 174.69 mm

If a circular section were used of radius R and of equal cross-sectional area to the rectangular section then πR² = 100 × 50 and R = 39.89 mm.
:. From Table 12.1 assuming [latex]R_{c}[/latex] remains at 150 mm

[latex]\int{\frac{dA}{r} }=2\pi \left\{(R_{i}+R)-\sqrt{(R_{1}+R)^{2}-R^{2}} ight\}[/latex]
[latex]=2\pi \left\{150-\sqrt{150^{2}-39.894^{2}} ight\}[/latex]
[latex]=2\pi 5.4024=33.944 mm[/latex]
[latex]R_{1}=\frac{A}{\int{\frac{dA}{r} }}=\frac{50 imes 100}{33.944} =147.301 mm[/latex]

with [latex]h = R_{c} - R_{1} = 150 - 147.3 = 2.699 mm.[/latex]

For critical tensile stress at B with y = 39.894 + 2.699 = 42.593.

[latex]\frac{My}{hA(R_{1}+y)} =34 MN/m^{2} [/latex]

[latex]\frac{(20 imes 10^{3} d) imes(42.593 imes 10^{-3})}{2.699 imes 10^{-3}(\pi imes 39.894^{2} imes 10^{-6})(150+39.894)} =34 imes 10^{6} [/latex]
[latex]d=102.3 [/latex]

i.e. Use of the circular section reduces the limit of d within which the load P can be applied

Question 12.1: The gantry shown in Fig. 12.15 is constructed from 100 mm × ...

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