Steel structures miscellaneous
- Consider the following two statements related to structural steel design, and identify whether they are TRUE or FALSE:
I. The Euler buckling load of a slender steel column depends on the yield strength of steel.
II. In the design of laced column, the maximum spacing of the lacing does not depend on the slenderness of column as a whole.
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The maximum spacing of lacing depend on slenderness of column as a whole.
Correct Option: B
The maximum spacing of lacing depend on slenderness of column as a whole.
- Identify the most efficient but joint (with double cover plates) for a plate in tension from the patterns (plan views) shown below, each comprising 6 identical bolts with the same pitch and gauge.
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In (a) we get more net area, A net
Correct Option: A
In (a) we get more net area, A net
- The members EJ and IJ of a steel truss shown in the figure below are subjected to a temperature rise of 30°C. The coefficient of thermal expansion of steel is 0.000012 per °C per unit length. The displacement (mm) of joint E relative to joint H along the direction HE of truss, is
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Expansion = 3000 √2 × 0.000012 × 30 = 1.527 mm
Expansion of EJ = 1.03 mm.Correct Option: D
Expansion = 3000 √2 × 0.000012 × 30 = 1.527 mm
Expansion of EJ = 1.03 mm.
- A bracket plate connected to a column flange transmits a load of 100 kN as shown in the following figure. The maximum force for which the bolts should be designed is _____ kN.
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FD = P = 100 = 20 kN n 5 Ft = (P.d)r Σ r2 = 100 × 600 × 75√2 = 141.42 kN 4 × (75√2)2
√400 + 19999.616 + 3999.96
= 156.204 kNCorrect Option: C
FD = P = 100 = 20 kN n 5 Ft = (P.d)r Σ r2 = 100 × 600 × 75√2 = 141.42 kN 4 × (75√2)2
√400 + 19999.616 + 3999.96
= 156.204 kN
- A symmetric I-section (with width of each flange = 50 mm, thickness of each flange = 10 mm, depth of web = 100 mm and thickness of web = 10 mm) of steel is subjected to a shear force of 100 kN. Find the magnitude of the shear stress (in N/mm2) in the web at its junction with the top flange.
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I = 50 × 1203 - 40 × 1003 12 12
= 3.866 × 106 mm4q = SAy = 100 × 103 × 50 × 10 × 55 Ib 3.866 × 106 × 10
= 71.12 N/mm2Correct Option: A
I = 50 × 1203 - 40 × 1003 12 12
= 3.866 × 106 mm4q = SAy = 100 × 103 × 50 × 10 × 55 Ib 3.866 × 106 × 10
= 71.12 N/mm2