JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III DESIGN OF STEEL STRUCTURES Previous Question Paper (February/March - 2022 - R16)
Answer any five questions. All questions carry equal marks.
Note: Use of IS 800-2007, IS 875 (part 3) and Steel table book is permitted.
Main
Explain the design procedure for butt welds.
Answer
The design of butt welds is based on the effective throat thickness and the length of the weld. The procedure involves the following steps:
- Determine the effective throat thickness (t_e), which is taken as 5/8 of the thickness of the thinner plate for single-V butt welds (incomplete penetration) and equal to the thickness of the thinner plate for complete penetration welds.
- Calculate the design strength of the weld in tension or compression using the formula: P_dw = (l_w × t_e × f_y) / (γ_mw × √3) for shear, or (l_w × t_e × f_y) / γ_mw for tension, where l_w is the effective length and γ_mw is the partial safety factor (1.25 for shop welds, 1.50 for field welds).
- Ensure the weld length does not exceed the plate width and that the stress does not exceed the permissible design stress of the parent metal.
Two plates of thickness 16 mm and 14 mm are to be joined by a butt weld. The joint is subjected to a factored tensile force of 300 kN. Effective length of weld is 150 mm. Check the safety of joint. (i) Single-V butt joint (Incomplete penetration), (ii) Double-V butt joint (Complete penetration).
Answer
Given: Thickness of thinner plate t = 14 mm, Factored force P = 300 kN, Effective length l_w = 150 mm, f_y = 250 MPa, γ_mw = 1.25.
(i) Single-V butt joint (Incomplete penetration): Effective throat thickness t_e = 5/8 × 14 = 8.75 mm. Design strength P_d = (l_w × t_e × f_y) / γ_mw = (150 × 8.75 × 250) / 1.25 = 262.5 kN. Since 262.5 kN < 300 kN, the joint is unsafe.
(ii) Double-V butt joint (Complete penetration): Effective throat thickness t_e = 14 mm. Design strength P_d = (150 × 14 × 250) / 1.25 = 420 kN. Since 420 kN > 300 kN, the joint is safe.
A single unequal angle of ISA 90 × 60 × 6 mm is connected to 10 mm thick gusset plate with 5 numbers of 16 mm diameter bolts to transfer the tension. Determine the design tensile strength of angle. Assume yield stress as 250 N/mm2, ultimate stress as 410 N/mm2 and Gauge distance is 50 mm. The Gusset plate is connected to 90 mm leg.
Answer
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Design a simply supported beam of span 5m carrying a reinforced concrete slab capable of providing the laterally restrainment to top of compression flange. The DL is 10 KN/m and Imposed load is 20 KN/m. Assume Fe410 grade steel.
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Design a simply supported plate girder of span 20 m to support a DL and Live load of 70 KN/m. Design bearing stiffness and intermediate stiffness also.
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Design a symmetric truss of span 20 m and height 5 m. spacing of trusses is 4.5 m center to center. Design the channel section purlin to be placed at the suitable distances to resist the following loads. The weight of sheeting including bolts is 171 N/m2, Live load = 0.4 KN/m2, Wind load = 1.2 KN/m2. Spacing of purlins is 1.4 m.
Answer
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Design a batten steel column of carrying factored axial load of 1000 kN. Height of the column is 8 m; the column is restrained in position but not in direction of both ends.
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Design a slab base for a column ISHB 300 @ 58.8 Kg/m carrying an axial load of 1200 kN. Allowable bearing pressure of the concrete is 4 N/mm2. Base plate is connected to the column by 2 numbers of ISA 100 × 100 × 7.5 mm thick cleat angle. Safe bearing capacity of soil is 200 KN/m2.
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A roof truss shed is to be built at Lucknow for an industry. The size of shed is 24 m × 40 m. height of the building is 12 m at the eaves board. Determine the basic wind pressure.
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