Question 3.2: A tensile member for a roof truss for a building is to carry...

A tensile member for a roof truss for a building is to carry a static axial tensile load of 88 kN. It has been proposed that a standard, equal-leg structural steel angle be used for this application using ASTM A36 structural steel. Use the AISC code. Specify a suitable angle from Appendix A–5(c).

A-5c
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Objective   Specify a standard equal-leg steel angle.

Given          F = 88 000 N static load.

Material: ASTM A36; s_{y} = 248 MPa; s_{u} = 400 MPa

(Data from Appendix A–12)

A–12  Properties of structural steels .^{a}
Ultimate strength, s_{u}^{a} Yield strength, s_{y}^{a}
Material ASTM No. and products ksi MPa ksi MPa Percent elongation in 2 in.
A36—carbon steel; available in shapes,plates, and bars 58 400 36 248 21
A 53—Grade B pipe 60 414 35 240 23
A242—HSLA, corrosion resistant; available in shapes, plates, and bars
\frac{3}{4} in. thick 70 483 50 345 21
1 \frac{1}{2} in. thick 67 462 46 317 21
1 \frac{1}{2} to 4 in. thick 63 434 42 290 21
A500—Cold-formed structural tubing
Round, Grade B 58 400 42 290 23
Round, Grade C 62 427 46 317 21
Round, Grade B 58 400 46 317 23
Round, Grade C 62 427 50 345 21
A501—Hot-formed structural tubing, round or shaped 58 400 36 248 23
A514—Quenched and tempered alloy steel; available in plate only
2 \frac{1}{2} in. thick 110 758 100 680 18
2 \frac{1}{2} to 6 in. thick 100 690 90 620 16
A572—HSLA columbium–vanadium steel; available in shapes, plates, and bars
Grade 42 60 414 42 290 24
Grade 50 65 448 50 345 21
Grade 60 75 517 60 414 18
Grade 65 80 552 65 448 17
A913—HSLA, grade 65; available in shapes only 80 552 65 448 17
A992—HSLA; available in W-shapes only 65 448 50 345 21

Analysis      Use Case C from Section 3–5.

Case C   To Determine the Shape and Dimensions of the Component

Given                  a. The magnitude and type of loading on the component of interest

b. The material, including its condition, from which the component is to be made

Objective   The shape and dimensions of critical geometry of the component

Method       1. Determine the yield strength, ultimate strength, and percent elongation for the selected material. Decide if the material is ductile (percent elongation > 5%) or brittle (percent elongation < 5%).

2. Specify an appropriate design factor considering the type of loading, the type of material, the conditions listed in the preceding section, and recommended guide- lines. For direct normal stresses, use Table 3–2.

3. Compute the design stress from Equation (3-2)   or (3-3)

\sigma_{d} = s_{y} /N based on yield strength

\sigma_{d} = s_{u} /N based on ultimate tensile strength

4. Write the equation for the expected maximum stress in the component. For direct normal stresses,

\sigma_{max} = F/A

5. Set \sigma_{max} = \sigma_{d} and solve for the required cross-sectional area.

\sigma_{max}  = \sigma_{d} = F/A

Required A=F/ \sigma_{d}

6. Determine the minimum required dimensions of the cross-sectional area to achieve the necessary total area. This is dependent on the shape you choose to make the component. It may be a solid circular, square, or rectangular, a hollow tube, a standard structural shape such as an angle, or some special shape of your own design.

7. Specify convenient dimensions from the list of preferred basic sizes listed in Appendix A–2.

TABLE 3–2  Design stress guidelines: Direct normal stresses.
Manner of loading Ductile material Brittle material
Static \sigma_{d} = s_{y} /2 \sigma_{d} = s_{u} /6
Repeated \sigma_{d} = s_{y} /8 \sigma_{d} = s_{u} /10
Impact or shock \sigma_{d} = s_{y} /12 \sigma_{d} = s_{u} /15

 

A–2     Preferred basic sizes.
Fractional (in.) Decimal (in.) SI metric (mm)
\frac{1}{64} 0.015 625 5 5.000 0.010 2.00 8.50 1.0 40
\frac{1}{32} 0.031 25 5 \frac{1}{4} 5.250 0.012 2.20 9.00 1.1 45
\frac{1}{16} 0.0625 5 \frac{1}{2} 5.500 0.016 2.40 9.50 1.2 50
\frac{3}{32} 0.093 75 5 \frac{3}{4} 5.750 0.020 2.60 10.00 1.4 55
\frac{1}{8} 0.1250 6 6.000 0.025 2.80 10.50 1.6 60
\frac{5}{32} 0.156 25 6 \frac{1}{2} 6.500 0.032 3.00 11.00 1.8 70
\frac{3}{16} 0.1875 7 7.000 0.040 3.20 11.50 2.0 80
\frac{1}{4} 0.2500 7 \frac{1}{2} 7.500 0.05 3.40 12.00 2.2 90
\frac{5}{16} 0.3125 8 8.000 0.06 3.60 12.50 2.5 100
\frac{3}{8} 0.3750 8 \frac{1}{2} 8.500 0.08 3.80 13.00 2.8 110
\frac{7}{16} 0.4375 9 9.000 0.10 4.00 13.50 3.0 120
\frac{1}{2} 0.5000   9 \frac{1}{2} 9.500 0.12 4.20 14.00 3.5 140
\frac{9}{16} 0.5625 10 10.000 0.16 4.40 14.50 4.0 160
\frac{5}{8} 0.6250   10 \frac{1}{2} 10.500 0.20 4.60 15,00 4.5 180
\frac{11}{16} 0.6875 11 11.000 0.24 4.80 15.50 5.0 200
\frac{3}{4} 0.7500   11 \frac{1}{2} 11.500 0.30 5.00 16.00 5.5 220
\frac{7}{8} 0.8750 12 12.000 0.40 5.20 16.50 6 250
1 1.000   12 \frac{1}{2} 12.500 0.50 5.40 17.00 7 280
1 \frac{1}{4} 1.250 13 13.000 0.60 5.60 17.50 8 300
1 \frac{1}{2} 1.500   13 \frac{1}{2} 13.500 0.80 5.80 18.00 9 350
1 \frac{3}{4} 1.750 14 14.000 1.00 6.00 18.50 10 400
2 2.000   14 \frac{1}{2} 14.500 1.20 6.50 19.00 11 450
2 \frac{1}{4} 2.250 15 15.000 1.40 7.00 19.50 12 500
2 \frac{1}{2} 2.500   15 \frac{1}{2} 15.500 1.60 7.50 20.00 14 550
2 \frac{3}{4} 2.750 16 16.000 1.80 8.00 16 600
3 3.000   16 \frac{1}{2} 16.500 18 700
3 \frac{1}{4} 3.250 17 17.000 20 800
3 \frac{1}{2} 3.500   17 \frac{1}{2} 17.500 22 900
3 \frac{3}{4} 3.750 18 18.000 25 1000
4 4.000   18 \frac{1}{2} 18.500 28
4 \frac{1}{4} 4.250 19 19.000 30
4 \frac{1}{2} 4.500   19 \frac{1}{2} 19.500 35
4 \frac{3}{4} 4.750 20 20.000

Let \sigma_= \sigma_{d} = 0.60s_{y} or \sigma_{d} = 0.50s_{u} (Table 3-3)

TABLE 3–3  Design stress from selected codes: Direct normal stresses—Static loads on building-like structures.
Structural steel (AISC): ASD
\sigma_{d} = s_{y} / 1.67 =0.60s_{y}        or      \sigma_{d}=s_{u}/2.00 = 0.50s_{u}
whichever is lower
Aluminum (AA):
\sigma_{d} = s_{y} / 1.67 =0.61s_{y}        or      \sigma_{d}=s_{u}/1.95 = 0.51s_{u}
whichever is lower

Stress analysis: \sigma F/A; then required area = A = F/ \sigma_{d}

Results        \sigma_{d} = 0.60s_{y} = 0.60(248 MPa) = 148.8 MPa

or \sigma_{d} = 0.50s_{u} = 0.50(400 MPa) = 200 MPa

Use lower value; \sigma_{d} = 148.8 MPa.

Required area: A = F/ \sigma_{d} = (88 000 N)/(148.8 N/mm²) = 591 mm²

This is the minimum allowable area.

pecify: L 60×60×6 steel angle (Appendix A–5(c); lightest equal-leg section).

A = 684 mm²; weight = 51.533 N/m.

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