Steel structures miscellaneous
- Rivets and bolts subjected to both shear stress (τvf, cal) and axial tensile stress (σtf, cal) shall be so proportioned that the stresses do not exceed the respective allowable stresses τvf and σtf and the value of
τvf, cal + σtf, cal τvf σtf
does not exceed
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1.4
Correct Option: C
1.4
- A steel strip of length, L = 200 mm is fixed at end A and rests at B on a vertical spring of stiffness, k = 2 N/mm. The steel strip is 5 mm wide and 10 mm thick. A vertical load, P = 50 N is applied at B, as shown in the figure. Considering E = 200 GPa, the force (in N) developed in the spring is____________
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K = F δ K = R δ ⇒ δ = R K
For cantilever,δ = (p - R)L3 = R 3EI k
⇒ R = 3NCorrect Option: A
K = F δ K = R δ ⇒ δ = R K
For cantilever,δ = (p - R)L3 = R 3EI k
⇒ R = 3N
- Cross-section of a column consisting of two steel strips, each of thickness t and width b is shown in the figure below. The critical loads of the column with perfect bond and without bond between the strips are P and P0, respectively. The ration P/P0 is
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P α 2 E I
α z × 1 × bt3 12
P α E l1α E × 1 × b(2t)3 12 ∴ P = 4 P0 Correct Option: B
P α 2 E I
α z × 1 × bt3 12
P α E l1α E × 1 × b(2t)3 12 ∴ P = 4 P0
- A mild steel specimen is under uni-axial tensile stress. Young’s modulus and yield stress for mild steel are 2 × 105 MPa respectively. The maximum amount of strain energy per unit volume that can be stored in this specimen without permanent set is
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Max strain energy unit volume
= σy2 2E = 250 × 106 = 156250 Nm/m3 2 × 2 × 1011
= 0.156 Nmm/mm3Correct Option: D
Max strain energy unit volume
= σy2 2E = 250 × 106 = 156250 Nm/m3 2 × 2 × 1011
= 0.156 Nmm/mm3
- The plastic collapse load Wp for the propped cantilever supporting two point loads as shown in figure terms of plastic moment capacity, Mp is given by
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Max positive moment under load = PL/3
∴ PL = mp + mp = 4mp 3 3 3 ∴ P = 4mp L Correct Option: B
Max positive moment under load = PL/3
∴ PL = mp + mp = 4mp 3 3 3 ∴ P = 4mp L