इन्जिनियरिङ सेवा सिभिल इन्जिनियर (सातौँ तह) – प्राविधिक पत्र
इन्जिनियरिङ सेवा – सिभिल इन्जिनियर (सातौँ तह)
Civil Engineer – 7th Level · Engineering Service
१,९४० प्रश्न उपलब्ध। मोड छान्नुहोस्:
प्राविधिक पत्र: प्रश्न र उत्तर
सही लागेको विकल्पमा थिच्नुहोस्। सही भए हरियो, गलत भए रातो देखिन्छ; ५ सेकेन्डपछि सही उत्तर र व्याख्या देखिन्छ।
सही: ० / ०
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As per IS 456:2000, a compression member is treated as a short column when the ratio of effective length to least lateral dimension is less than:
सही उत्तर: ख) 12
A column with le/D below 12 is classified as short.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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As per IS 456:2000, the minimum number of longitudinal bars in a circular RCC column is:
सही उत्तर: ख) 6
Circular columns need at least six longitudinal bars; rectangular columns need four.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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As per IS 456:2000, the minimum longitudinal reinforcement in an RCC column is:
सही उत्तर: घ) 0.8% of gross cross-sectional area
IS 456 requires column longitudinal steel of at least 0.8% of gross area.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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As per IS 456:2000, the flexural strength (modulus of rupture) of M25 concrete is:
सही उत्तर: ग) 3.5 N/mm²
fcr = 0.7√fck = 0.7 × 5 = 3.5 N/mm².
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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As per IS 456:2000, the short-term modulus of elasticity of M25 concrete is:
सही उत्तर: घ) 25 000 N/mm²
Ec = 5000√fck = 5000 × √25 = 25 000 N/mm².
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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As per IS 456:2000, the limiting ratio xu,max/d for Fe 415 steel is:
सही उत्तर: क) 0.48
IS 456 gives xu,max/d = 0.53, 0.48 and 0.46 for Fe 250, Fe 415 and Fe 500 respectively.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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As per IS 456:2000, the maximum compressive strain in concrete in the outermost fibre for flexure is:
सही उत्तर: क) 0.0035
IS 456 limits the extreme fibre compressive strain in flexure to 0.0035.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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As per IS 456:2000 limit state design, the design yield strength of Fe 415 steel is approximately:
सही उत्तर: ख) 361 N/mm²
Design strength = 0.87 fy = 0.87 × 415 ≈ 361 N/mm².
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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As per IS 456:2000, the partial safety factor for reinforcing steel in the limit state of collapse is:
सही उत्तर: क) 1.15
IS 456 specifies γm = 1.15 for steel and 1.5 for concrete.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The principle used to obtain influence line diagrams by applying a unit displacement at the point of the required reaction or force is:
सही उत्तर: ग) Müller-Breslau principle
Müller-Breslau principle states the influence line has the shape of the deflected structure under a unit displacement of that response.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
यी अभ्यास प्रश्न हुन्, आधिकारिक प्रश्न होइनन्। पाठ्यक्रम, विज्ञापन र नतिजाका लागि सम्बन्धित निकायको आधिकारिक वेबसाइट हेर्नुहोस्। उत्तरमा गल्ती भेटे प्रश्नमुनिको "टिप्पणी / गल्ती रिपोर्ट" मा लेख्नुहोस्।
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