सिभिल इन्जिनियरिङ
सिभिल इन्जिनियरिङ
Civil Engineering · ५,१७१ प्रश्न
५,१७१ प्रश्न उपलब्ध। मोड छान्नुहोस्:
सिभिल इन्जिनियरिङ: प्रश्न र उत्तर
सही लागेको विकल्पमा थिच्नुहोस्। सही भए हरियो, गलत भए रातो देखिन्छ; ३ सेकेन्डपछि सही उत्तर र व्याख्या देखिन्छ।
सही: ० / ०
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In IS 456 limit state design, the strain in Fe 415 steel at the design yield point (0.87fy/Es + 0.002) is about:
सही उत्तर: ख) 0.0038
ε = 0.87 × 415/(2 × 10⁵) + 0.002 = 0.0018 + 0.002 ≈ 0.0038 (cold-worked bars have no sharp yield, hence the 0.002 offset).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Using IS 456 (τbd = 1.2 N/mm² for M20, increased by 60% for deformed bars), the development length of an Fe 415 deformed bar in tension in M20 concrete is about:
सही उत्तर: क) 47φ
Ld = 0.87fy·φ/(4τbd) = 0.87 × 415 φ/(4 × 1.2 × 1.6) = 361φ/7.68 ≈ 47φ.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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As per IS 456:2000, the maximum shear stress τc,max in a beam of M20 concrete (limit state method) is:
सही उत्तर: ग) 2.8 N/mm²
IS 456 Table 20 gives τc,max = 2.5, 2.8, 3.1, 3.5 N/mm² for M15, M20, M25, M30; if τv exceeds it the section must be redesigned.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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As per IS 456:2000, the partial safety factor for loads at the limit state of serviceability (DL + LL) is:
सही उत्तर: ख) 1.0
For checking deflection and cracking (serviceability), loads are taken at their characteristic values, i.e. factor 1.0 for DL + LL.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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As per IS 456:2000, the partial safety factor for loads in the limit state of collapse for the combination DL + LL + EL (or WL) is:
सही उत्तर: क) 1.2
IS 456 Table 18 gives 1.5 for DL + LL but 1.2 for DL + LL + WL/EL at the limit state of collapse.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A plane truss has 11 members, 7 joints and 3 external reaction components. The truss is:
सही उत्तर: ख) Statically determinate and stable (m + r = 2j)
m + r = 11 + 3 = 14 = 2j = 14, so the truss is statically determinate (assuming proper arrangement).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Influence line diagrams for reactions and internal forces of statically determinate beams consist of:
सही उत्तर: ख) Straight line segments
For determinate structures ILDs are made up of straight lines; indeterminate structures give curved ILDs.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The influence line diagram for shear force at a section of a simply supported beam consists of:
सही उत्तर: ग) Two parallel lines with a sudden jump of unity at the section
Shear ILD runs from 0 to −a/L on the left of the section and from +b/L to 0 on the right, the two parallel lines differing by 1 at the section.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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For a simply supported beam of 8 m span, the maximum ordinate of the influence line for bending moment at a section 2 m from the left support is:
सही उत्तर: घ) 1.5 m
The ILD for moment at a section is a triangle with peak ordinate a·b/L = 2 × 6/8 = 1.5 (units of length) under the section.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A cable of span 40 m with central sag 4 m carries a UDL of 2 kN/m (horizontal). The maximum tension in the cable is about:
सही उत्तर: घ) 107.7 kN
H = wL²/(8d) = 2 × 40²/32 = 100 kN; V = wL/2 = 40 kN; Tmax = √(100² + 40²) ≈ 107.7 kN at the supports.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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