When selecting a standard rolled steel section to carry a specified load, the procedure is as follows:
1. Calculate the effective length of your strut from its actual length and the type of end fixings. State its value in mm.
2. Choose a trial column from Table 2.2 for universal columns.
Structural engineers know from experience roughly the size of section required but for a first trial choose one from somewhere near the centre of the table. Make a note of its cross-sectional area, A and radius of gyration, k about the y–y axis of bending.
Convert their units to mm2and mm. Make a note also of the flange thickness.
3. Calculate the slenderness ratio of your trial column.
4. From Table 2.4 read off the maximum allowable compressive stress, for your calculated value of slenderness ratio from the column appropriate to your flange thickness. You should note that the units used for stress are N mm 2 which is why your cross-sectional area needs to be in mm2.
5. Calculate the safe axial load that your trial column can carry using the formula,
safe axial load¼ allowable stress cross-sectional area
F= sA ð2:8Þ
6. Compare this with the load which the member has to carry. If it is less or appreciably greater, choose another trial column and repeat the calculations. Eventually, after no more than two or three trials, you should be able to narrow the choice down to a section with the least weight that can carry slightly more than the required load.
Example 2.5
A stanchion of length 10 m with direction-fixed ends is required to carry a load of 500 kN. Select a suitable universal steel column from standard section tables which has the least mass.
Finding effective length:
l ¼ 0:7L ¼ 0:7 10 l = 7 m or 7 · 103mm 1st trial:
Serial size 203 203 mm2of mass 86 kg m 1 Cross-sectional area, A ¼ 110.1 cm2¼ 110.1 102mm Least radius of gyration, k ¼ 5.32 cm ¼ 53.2 mm
Flange thickness, T ¼ 20.5 mm Finding slenderness ratio:
S:R: ¼ l
k ¼7 103 53:2 S:R: = 132
Finding allowable compressive stress from Table 2.4:
s = 89 N mm–2
Finding maximum load which the member can carry:
F ¼ A ¼ 89 110:1 102 F = 980 · 103N or 980 kN
This is much higher than the 500 kN load and so another trial is necessary.
2nd trial:
Serial size 152 152 mm2of mass 37 kg m 1 Cross-sectional area, A ¼ 47.4 cm2¼ 47.4 102mm Least radius of gyration, k ¼ 3.87 cm ¼ 38.7 mm Flange thickness, T ¼ 11.5 mm
Finding slenderness ratio:
S:R: ¼ l
k ¼7 103 38:7 S:R: = 181
Finding allowable compressive stress from Table 2.4:
s = 53 N mm–2
Finding maximum load which the member can carry:
F ¼ A ¼ 53 47:4 102 F = 251 · 103N or 251 kN
This is below the 500 kN load and so another trial is necessary.
3rd trial:
Serial size 203 203 mm2 of mass 46 kg m 1 Cross-sectional area, A ¼ 58:8 cm2¼ 58:8 102mm Least radius of gyration, k ¼ 51:1 cm ¼ 51:1 mm Finding slenderness ratio:
S:R: ¼ l
k ¼7 103 51:1 S:R: = 137
Finding allowable compressive stress from Table 2.4:
s = 87 N mm 2
Finding maximum load which the member can carry:
F ¼ A ¼ 87 58:8 102 F = 512 · 103N or 512 kN
This can carry slightly more than the 500 kN load and is the most suitable section.
Example 2.6
A strut of length 8 m with one end direction-fixed and one end pin-jointed is required to carry a compressive load of 750 kN. Select a suitable universal steel column from standard section tables which has the least mass.
Finding effective length:
l ¼ 0:85 L ¼ 0:85 8 l = 6:8 m or 6:8 · 103mm 1st trial:
Serial size 254 254 mm2of mass 167 kgm 1 Cross-sectional area, A ¼ 212.4 cm2¼ 212.4 102mm Least radius of gyration, K ¼ 6.79 cm ¼ 67.9 mm
Flange thickness, T ¼ 31.7 mm Finding slenderness ratio:
S:R: ¼ l
k ¼6:8 103 67:9 S:R: = 100
Finding allowable compressive stress from Table 2.4:
s = 138 N mm–2
Finding maximum load which the member can carry:
F ¼ A ¼ 138 212:4 102 F = 2:93 · 106N or 2:93 MN
This is much higher than the 750 kN load and so another trial is necessary.
2nd trial:
Serial size 203 203 mm2of mass 86 kg m 1 Cross-sectional area, A ¼ 110.1 cm2¼ 110.1 102mm Least radius of gyration, k ¼ 5.32 cm ¼ 53.2 mm
Flange thickness, T ¼ 20.5 mm Finding slenderness ratio:
S:R: ¼ l
k¼6:8 103 53:2 S:R: = 128
Finding allowable compressive stress from Table 2.4:
s = 96 N mm–2
Finding maximum load which the member can carry:
F ¼ A ¼ 96 110:1 102 F = 1:06 · 106N or 1:06 MN
This is still greater than the required load of 750 kN and is unsatisfactory. A further trial is needed on a smaller section.
3rd trial:
Serial size 203 203 mm2of mass 60 kg m 1 Cross-sectional area, A ¼ 7.58 cm2¼ 75.8 102mm Least radius of gyration, k ¼ 5.19 cm ¼ 51.9 mm Flange thickness, T ¼ 11.0 mm
Finding slenderness ratio:
S:R: ¼ l
k ¼6:8 103 51:9 S:R: = 131
Finding allowable compressive stress from Table 2.4:
s = 93 N mm–2
Finding maximum load which the member can carry:
F ¼ A ¼ 93 575:8 102 F = 705 · 103N or 705 kN
This is slightly below the 750 kN load and so another trial is necessary:
4th trial:
Serial size 203 503 mm2of mass 71 kg m 1 Cross-sectional area, A ¼ 91.1cm2¼ 91.1 102mm Least radius of gyration, k ¼ 5.28 cm ¼ 52.8 mm Flange thickness, T ¼ 17.3 mm
Finding slenderness ratio:
S:R: ¼ l
k ¼6:8 103 52:8 S:R: = 129
Finding allowable compressive stress from Table 2.4:
s = 95 N mm–2
Finding maximum load which the member can carry:
F ¼ A ¼ 95 91:1 102 F = 865 · 103N or 865 kN
This can carry slightly more than the 750 kN load and is the most suitable section.
Activity 2.3
A strut of length 7.5 m is pin-jointed at both ends and is required to support an axial load of 800 kN. Select a suitable universal column made from 43 grade steel which has the least possible weight. If the ultimate tensile strength of the steel is 500 MPa, what will be the factor of safety in operation for the selected section?
Test your knowledge 2.3 1. What does axial loading in struts and stanchions mean?
2. What is meant by the effective length of a strut?
3. What is the effective length of a strut with one end direction-fixed and one end pin-jointed?
4. What is the slenderness ratio of a strut?
5. How do you convert cross-sectional area in cm2to mm2?
Problems 2.2
1. A stanchion 3.6 m high has one end direction-fixed and one end pin-jointed. The standard section selected for use is Grade 43 steel serial size 202 203 mm2and mass 60 kg m 1. Determine (a) the slenderness ratio of the stanchion, (b) the maximum compressive load that it can be allowed to carry.
[59, 1683 kN]
2. A strut of length 4 m is pin-jointed at both ends and is required to carry a compressive load of 2 MN. Select a suitable I-section with the lowest possible weight using standard section tables for Grade 43 steel.
[Serial size 254 254 mm2, mass 73 kg m 1] 3. A stanchion is 4.3 m long and direction-fixed at both ends.
Using standard section tables, select a suitable I-section made from Grade 43 steel which can safely support a compressive load of 2000 kN.
[Serial size 203 203 mm2, mass 71 kg m 1] 4. Calculate the maximum allowable compressive loads that the following stanchions can carry. (a) Length 6 m with both ends pin-jointed. (b) Length 9 m with one end direction-fixed, one end pin-jointed. (c) Length 12 m with both ends direction-fixed.
The material used is Grade 43 steel of serial size 254 254 mm1 and mass 89 kg m 1.
[1.78 MN, 1.25 MN, 1.09 MN]
5. A strut of length 5 m has one end pin-jointed and one end free.
The strut is to carry an axial compressive load of 1500 kN.
Select a suitable I-section from standard section tables for Grade 43 steel which has (a) the lowest possible weight, (b) the lowest possible values of depth and breadth.
[(a) Serial size 305 305 mm2, mass 137 kg m 1] [(b) Serial size 254 254 mm2, mass 167 kg m 1]