The cantilevered tube shown in Fig. 5–17 is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276 MPa. We wish to select a stock-size tube from Table A–8 using a design factor nd = 4. The bending load is

Question

The cantilevered tube shown in Fig. 5–17 is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276 MPa. We wish to select a stock-size tube from Table A–8 using a design factor [latex]n_{d}[/latex] = 4. The bending load is F = 1.75 kN, the axial tension is P = 9.0 kN, and the torsion is T = 72 N · m. What is the realized factor of safety?

Table A–8 Properties of Round Tubing

[latex]w_{a}[/latex] = unit weight of aluminum tubing, lbf/ft
[latex]w_{s}[/latex] = unit weight of steel tubing, lbf/ft
m = unit mass, kg/m
A = area, [latex]in^{2} (cm^{2})[/latex]
I = second moment of area, [latex]in^{4} (cm^{4})[/latex]
J = second polar moment of area,[latex]in^{4} (cm^{4})[/latex]
k = radius of gyration, in (cm)
Z = section modulus,[latex]in^{3} (cm^{3})[/latex]
d, t = size (OD) and thickness, in (mm)

J	Z	K	l	A	[latex]w_{s}[/latex]	[latex]w_{a}[/latex]	Size, in
0.067	0.067	0.313	0.034	0.344	1.128	0.416	[latex]1 ×\frac{1}{8}[/latex]
0.092	0.092	0.280	0.046	0.589	2.003	0.713	[latex]1 ×\frac{1}{4}[/latex]
0.257	0.172	0.488	0.129	0.540	1.769	0.653	[latex]1 \frac {1}{2} ×\frac{1}{8}[/latex]
0.399	0.266	0.451	0.199	0.982	3.338	1.188	[latex]1 \frac {1}{2} ×\frac{1}{4}[/latex]
0.650	0.325	0.664	0.325	0.736	2.670	0.891	[latex]2 ×\frac{1}{8}[/latex]
1.074	0.537	0.625	0.537	1.374	4.673	1.663	[latex]2×\frac{1}{4}[/latex]
1.319	0.528	0.841	0.660	0.933	3.050	1.129	[latex]2 \frac {1}{2} ×\frac{1}{8}[/latex]
2.276	0.906	0.800	1.132	1.767	6.008	2.138	[latex]2 \frac {1}{2} ×\frac{1}{4}[/latex]
4.117	1.373	0.976	2.059	2.160	7.343	2.614	[latex]3 ×\frac{1}{4}[/latex]
5.436	1.812	0.938	2.718	3.093	10.51	3.742	[latex]3 ×\frac{3}{8}[/latex]
8.180	2.045	1.350	4.090	2.246	7.654	2.717	[latex]4 ×\frac{3}{16}[/latex]
14.180	3.544	1.289	7.090	4.271	14.52	5.167	[latex]4 ×\frac{3}{8}[/latex]

J	Z	k	l	A	m	Size, mm
0.163	0.136	0.361	0.082	0.628	0.490	12 × 2
0.440	0.275	0.500	0.220	0.879	0.687	16 × 2
0.545	0.341	0.472	0.273	1.225	0.956	16 × 3
1.367	0.684	0.583	0.684	2.010	1.569	20 × 4
3.015	1.206	0.756	1.508	2.638	2.060	25 × 4
3.338	1.336	0.729	1.669	3.140	2.452	25 × 5
5.652	1.885	0.930	2.827	3.266	2.550	30 × 4
6.381	2.128	0.901	3.192	3.925	3.065	30 × 5
17.430	4.151	1.351	8.717	4.773	3.727	42 × 4
20.255	4.825	1.320	10.130	5.809	4.536	42 × 5
30.810	6.164	1.632	15.409	5.778	4.512	50 × 4
36.226	7.247	1.601	18.118	7.065	5.517	50 × 5

Accepted Answer

Since the maximum bending moment is M = 120F, the normal stress, for an element on the top surface of the tube at the origin, is

[latex]σ_{x} =\frac {P}{A} +\frac {Mc}{I} =\frac {9}{A} +\frac {120(1.75)(d_{o}/2)}{I} =\frac {9}{A} +\frac {105d_{o}}{I}[/latex] (1)

where, if millimeters are used for the area properties, the stress is in gigapascals.The torsional stress at the same point is

[latex]τ_{zx} =\frac {Tr}{J} =\frac {72(d_{o}/2)}{J} =\frac {36d_{o}}{J}[/latex] (2)

For accuracy, we choose the distortion-energy theory as the design basis. The von Mises stress, as in the previous example, is

[latex]σ′ =\left (σ^{2}_{x} + 3τ^{2}_{zx} \right)^{1/2}[/latex] (3)

On the basis of the given design factor, the goal for σ′ is

[latex]σ′ ≤\frac {S_{y}}{n_{d}} =\frac {0.276}{4} = 0.0690 GPa[/latex] (4)

where we have used gigapascals in this relation to agree with Eqs. (1) and (2).
Programming Eqs. (1) to (3) on a spreadsheet and entering metric sizes from Table A–8 reveals that a 42- × 5-mm tube is satisfactory. The von Mises stress is found to be σ′ = 0.06043 GPa for this size. Thus the realized factor of safety is

[latex]n =\frac {S_{y}}{σ′} =\frac {0.276}{0.06043} = 4.57[/latex]

For the next size smaller, a 42- × 4-mm tube, σ′ = 0.07105 GPa giving a factor of safety of

[latex]n =\frac {S_{y}}{σ′} =\frac {0.276}{0.07105} = 43.88[/latex]

Question 5.4: The cantilevered tube shown in Fig. 5–17 is to be made of 20...

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