UPPSC AE Civil 13 Dec 2020 Official Paper - I (Previous Year Paper)

• "मरणासन्न – अव्ययीभाव समास" यह समास - युग्म बिलकुल गलत है।  अत: प्रश्नानुसार सही विकल्प 4 'मरणासन्न – अव्ययीभाव समास' है।

• मरणासन्न का समास विग्रह :- जो मृत्यु के आसन पर हो अर्थात जो मृत्यु के निकट हो

• अत: यहाँ उत्तर पद 'आसन' प्रधान पद है, इसलिए यह तत्पुरुष समास होगा।

• जबकि "जिस समास में पहला पद प्रधान हो और समस्त शब्द अव्यय का काम करे।" वह 'अव्ययी भाव समास' होता है।

Key Points

• सतसई – द्विगु समास ---> "���ात का समाहार अथवा समूह"
• तुलसीकृत – तत्पुरुष समास ---> "तुलसी द्वारा कृत अथवा तुलसी द्वारा लिखा"
• मंदोदरी – बहुव्रीहि समास  ---> "मंद उदर हो जिसका अर्थात मन्दोदरी"

Additional Information

• अलग होने के अर्थ में 'अपादान कारक' का प्रयोग किया जाता है।

• जैसे:- पेड़ से पत्ता गिरा, अर्थात पेड़ से पत्ते के अलग होने के भाव में यहाँ अपादान कारक का प्रयोग किया जाता है।

• करण कारक और अपादान कारक दोनों की विभक्ति 'से' होती है।

• परन्तु अपादान कारक की 'से' विभक्ति अलग होने के अर्थ में प्रयोग की जाती है और करण कारक की 'से',  "के द्वारा" के अर्थ में प्रयोग की जाती है।

• जैसे:- पृथ्वी सूर्य 'से' दूर है। (अपादान कारक)

• जैसे:- बच्चा बोतल 'से' दूध पीता है। (करण कारक)

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Additional Information  

कारक- संज्ञा या सर्वनाम का क्रिया से सम्बन्ध जिस रूप से जाना जाता है, उसे कारक कहते हैं।

हिंदी में जिनकी संख्या आठ मानी गई है। जो निम्नलिखित हैं:-

‘पवन’ शब्द का सन्धि-विच्छेद 'पो + अन' होगा। अतः सही विकल्प 'पो + अन' है।

Key Points

• 'पवन' में 'अयादि स्वर संधि' है।
• 'पवन' का विच्छेद करने पर 'पो + अन = पवन' होगा और नियम 'ओ + अ = अव्' है।

Additional Information

• 'निर्झरिणी' शब्द वर्तनी और उच्चारण की दृष्टि से शुद्ध और सार्थक है। अत: सही विकल्प 3 'निर्झरिणी' है।

Key Points 

• निर्झरिणी का अर्थ :- नदी होता है।
• नदी के पर्यायवाची शब्द:-  सरिता, तटिनी, सरि, सारंग, तरंगिणी, दरिया, निर्झरिणी होते है।

• 'नयन' मूल शब्द है, इसमें कोई उपसर्ग प्रयोग नही होता है। अत: सही विकल्प 4 'नयन' होता है।

Mistake Points

• नयन शब्द में 'न्' मूल शब्द है, और 'अन' प्रत्यय लगा हुआ है
• नयन में उपसर्ग नही बल्कि प्रत्यय का प्रयोग हुआ है।

Key Points

• सहज --> स + हज,  स उपसर्ग
• अनुभव --> अनु + भव, अनु उपसर्ग
• संचार ---> सन् + चार, सन् उपसर्ग

Additional Information

• उपरोक्त में से ‘अनघ’ का विलोम शब्द:- "अघ" होता है।

Important Points

• ‘अनघ’ का अर्थ:- निष्पाप, शुद्ध, सही, पूर्ण, निर्दोष, दोषरहित" होता है।
• ‘अघ’ का अर्थ:- पापी, नीच

Key Points

Additional Information

• 'पिशुन' पक्षी का पर्यायवाची शब्द नही होता है। अत: प्रश्नानुसार सही विकल्प 1 'पिशुन' होगा।
• पक्षी के सभी पर्यायवाची शब्द:- खग, चिडिया, गगनचर, पखेरू, विहंग, नभचर
• विहंग, शकुनि, द्विज :- यह तीनो पक्षी के अन्य समानार्थी अर्थात पर्यायवाची शब्द होते है।

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• द्विज का वैसे अन्य अर्थ :- दो जन्म लेने वाला भी होता है।
• जैसे:- हनुमान जी का अन्य नाम भी द्विज है।
• परन्तु यहाँ प्रश्नानुसार पक्षी का पर्यायवाची भी द्विज होता है।

Important Points

• 'पिशुन' भी एक पक्षी होता है, जिसके पर्यायवाची निम्न है:-
• पिशुन:- काक, वायस, काग, करठ, पिशुन।

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Additional Information 

• "उत्तराधिकार में प्राप्त सम्पत्ति – धरोहर" यह विकल्प युग्म गलत है। अत: प्रश्नानुसार सही विकल्प 1 'उत्तराधिकार में प्राप्त सम्पत्ति – धरोहर' है।
• "उत्तराधिकार में प्राप्त सम्पत्ति" के लिए सही वाक्यांश :-  "रिक्थ" कहलाता है।
• साथ ही इसे "पैतृक धन" भी कहा जाता है।

Additional Information

'विवाह' तत्सम शब्द है। अन्य विकल्प असंगत है। अत: सही विकल्प 1 'विवाह' है। 'विवाह' शब्द संस्कृत से ज्यों का त्यों ग्रहण किया गया है।

Key Points

• ईख का तत्सम शब्द ---> इक्षु
• खीर का तत्सम शब्द ---> क्षीर
• गिद्ध का तत्सम शब्द ---> गृद्ध

Additional Information

• उपरोक्त में से ‘महीसुर’ शब्द का अर्थ:- 'ब्राह्मण' है।
• ‘महीसुर’ का अर्थ:- मही के सुर अर्थात 'पृथ्वी का देवता' अर्थात ब्राह्मण सन्दर्भ में प्रयोग किया जाता है।

Additional Information

• पृथ्वी के रक्षक यहाँ सही विकल्प नहीं है।
• महिषासुर:- यह एक राक्षस का नाम है।
• राक्षस:- पृथ्वी के रक्षक कार्य करने न वाले अर्थात भक्षक को राक्षस कहा जाता है।

• ‘ने + अन’ = ‘नयन’ में 'अयादि सन्धि' है। अत: सही विकल्प 3 'अयादि संधि' है।

Key Points

Additional Information

• 'उज्ज्वल' शब्द शुद्ध वर्तनी का शब्द है। अत: सही विकल्प 1 'उज्ज्वल' है।

Key Points

• उज्ज्वल का अर्थ :- स्वच्छ, शुद्ध, साफ़
• उज्ज्वल शब्द में व्यंजन संधि है।
• उज्ज्वल का शुद्ध संधि विच्छेद है= उत् + ज्वल।

• ‘बुद्धिहीन’ शब्द व्याकरण की दृष्टी से इनमें से:- विशेषण के अंतर्गत आते है।
• जो शब्द किसी की विशेषता बताते है, वो विशेषण कहलाते है। जैसे:- राकेश बुद्धिहीन लड़का है।
• अत: बुद्धिहीन राकेश के बारे में विशेष है।

Additional Information

• "महादेवी वर्मा छायावाद की प्रसिद्ध कवयित्री हैं।" सिर्फ यह वाक्य उच्चारण और व्याकरणिक दृष्टि से शुद्ध है।

Key Points

• मेरा प्राण संकट में है  -----> मेरे प्राण संकट में हैं।
• सोमवार को रेलवे के कई कर्मचारी गिरफ़्तार हुए। --> सोमवार को कई रेलवे कर्मचारी गिरफ़्तार हुए।
• अपराधी को मृत्युदंड की सजा दी गयी है। --->अपराधी को मृत्युदंड दिया गया।

• इनमें से दन्त्य ध्वनियाँ:- "त, थ, द, ध" है। अन्य विकल्प असंगत है। अत: सही विकल्प 3 "त, थ, द, ध" है।

Key Points

Additional Information

दिए गए विकल्पों में 'उद्धार' शब्द व्यंजन संधि आधारित शब्द है। अतिरिक्त विकल्प स्वर तथा विसर्ग संधि हैं। अतः सही विकल्प 2 'उद्धार' है।

Key Points

• व्यंजन संधि में त् + ह आये तो त् का द् तथा ह का ध में रूपांतरण होगा। जैसे – उत् + हरण = उद्धरण, पद् + हती = पद्धति।

अन्य विकल्प

Additional Information

• ‘चौराहा’ का समास विग्रह :-  चार राहो का समूह होता है। अत: सही विकल्प 4 'द्विगु' समास है।
• द्विगु समास में प्रथम वर्ण संख्या वाचक होता है।

Key Points 

• ‘निवृत्ति’ शब्द का विलोम:- प्रवृत्ति है। अत: सही विकल्प 3 'प्रवृत्ति' है।
• ‘निवृत्ति’ का अर्थ:- छुटकारा, मुक्ति
• 'प्रवृत्ति' का अर्थ:- किसी चीज से जुड़े होने की अवस्था

Key Points

• सद्वृत्ति (सदाचार) का विलोम - असद्वृत्ति

Additional Information

• ‘शारदा’ का पर्यायवाची शब्द:- गिरा है।
• शारदा के अन्य पर्यायवाची:- भारती, सरस्वती, वाग्देवी, वीणावादिनी, विद्या, वागेश्वरी, वागीशा।

Key Points

• कमला के पर्यायवाची:- पद्मा, रमा, लक्ष्मी,हरिप्रिया, श्री, इंदिरा
• कौमुदी के पर्यायवाची:- चन्द्रिका, चाँदनी, ज्योत्स्ना, चन्द्रमरीचि, उजियारी, चन्द्रप्रभा, जुन्हाई।
• वारुणी के पर्यायवाची:- मदिरा, मधु, शराब, आसव, सुरा, मद्य, हाला, मद।

Additional Information

• 'मुमूर्षु' उसे कहते है जो मृत्यु के निकट हो या मृत्यु माँग रहा हो। अत: प्रश्नानुसार सही विकल्प 2 ''मुमूर्षु'' होगा। अन्य विकल्प असंगत है।

Key Points

• मुमुक्षा --> जो मोक्ष की इच्छा रखें, उसके लिए प्रयुक्त शब्द मुमुक्षा होता है।
• मुमुक्षु --> जो मोक्ष की इच्छा रखें, उस भव्यात्मा के लिए भी प्रयुक्त शब्द मुमुक्षा होता है।
• युयुत्सा --> जो युद्ध के लिए तत्पर अथवा युद्ध की कामना रखता हो।

Additional Information

• ‘चीनांशुक’ शब्द का सही अर्थ:- "रेशम वस्त्र" है। अत: सही विकल्प 3 'रेशम वस्त्र' होगा।

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• ‘चीनांशुक’ का वास्तविक अर्थ 'रेशम' से जुदा हुआ होता है, परन्तु इसका अर्थ मिट्टी से जुड़ा हुआ बिल्कुल भी नही होता है।
• ‘चीनांशुक’ को हम रेशमी कपड़ा, चीन से आया वस्त्र आदि के रूप में किताबो में पढ़ सकते हैं।
• परन्तु चीनी मिट्टी इसका उत्तर नही होगा।​

Additional Information

• तंतु --> सूत अथवा तागा, रेशा
• रेणु --> धूल , रेत
• चीनी मिट्टी :- यह एक सफेद मिट्टी होती है, जिसे पकाकर बर्तन या खिलौने बनाये जाते है।

• उपरोक्त विकल्पों में से 'तेल' तद्भव शब्द है। अत: सही विकल्प 2 'तेल' है।
• तद्भव शब्द, वे होते है जो संस्कृत से उत्पन्न शब्दों के नये रूप ���ोते है, जिनका उच्चारण समय के साथ बदल गया है।
• अत: तेल का तद्भव शब्द - तैल होता है।
• इसी प्रकार आँसू भी तद्भव है जिसका तत्सम शब्द 'अश्रु' होता है।

Important Points

• तत्सम शब्दों के तद्भव रूप 
• वानर - बन्दर
• पीत - पीला
• घोटक - घोड़ा

Additional Information

• उपरोक्त विकल्पों में से 'तलवार', 'सारंग' का कोई अर्थ नही है, दोनों के मध्य कोई सम्बन्ध नही है। अत: सही विकल्प 3 'तलवार' है।
• सारंग का अर्थ:- सुहावना, सुंदर, मन भावन होता है।
• भौंरा, कामदेव, ज्योतिषी, सभी सारंग के अनेकार्थक शब्द है।
• सारंग के एक से अधिक अर्थ –
• हाथी, कोयल, कामदेव, सिंह, धनुष, भौंरा, मृग, मयूर, स्त्री, नानावर्ण
• सुन्दर, सरस, बादल, वृक्ष, छाता, वस्त्र, बाल, शंख
• शिव, कपूर, चन्दन, आभूषण, स्वर्ण, मधुमक्खी, कमल।
• ज्योतिषी भी सारंग का अनेकार्थी शब्द होता है क्योंकि सारंग का अर्थ है विभिन्न रंगो वाला जैसे सारंग हिरन, सारंग सिंह उसी प्रकार ज्योतिषी भी के विद्वान् का प्रकार है अत: यह भी अनेकार्थी शब्द होगा।
• अनेकार्थी शब्द अर्थो से नही भावार्थ से भी निकाले जाते है।

Confusion Points

• अनेकार्थक शब्द में एक ही शब्द के भिन्न भिन्न अर्थ होते है।
• जैसे:- कनक का मतलब 'सोना (स्वर्ण)' , गेंहू, और धतूरा तीनो होता है।

Additional Information

• ‘अंदर-अंदर कड़ाही में गुड़ पगना’ – इस मुहावरे का सही अर्थ 2 'गुप्त मंत्रणा होना' है।
• वाक्य प्रयोग:- कौरवो ने पांडवो के खिलाफ ‘अंदर-अंदर कड़ाही में गुड़ पग लिये थे। अर्थात अंदर ही अंदर गुप्त चालाकी कर ली थी।

Key Points

Additional Information

सहस्र शब्द वर्तनी और उच्चारण की दृष्टि से शुद्ध और सही विकल्प है। अन्य विकल्प अशुद्ध है। 

Key Points

• सहस्र का अर्थ:-  गिनती में हजार गुना

Additional Information

• अनाधिकार --> शुद्ध रूप 'अनधिकार'  ---> जिसमे योग्यता की कमी हो
• रचइता --->  शुद्ध रूप 'रचयिता' ---> जो रचना करता है
• संग्रहीत ---> शुद्ध रूप 'संग्रहित'  ---> एकत्र किया हुआ,

Explanation:

Given that: M_{u, lim} = R_{u, lim} bd²

M_{u, lim} \( = 0.36{\rm{;}}{{\rm{f}}{{\rm{ck}}}}{\rm{;}}\frac{{{{\rm{X}}{\rm{u}}}}}{{\rm{d}}}{\rm{;}}\left( {1 - \frac{{0.42{\rm{;}}{{\rm{X}}_{\rm{u}}}}}{{\rm{d}}}} \right){\rm{;b}}{{\rm{d}}^2}\)

For Fe-415 Steel:

X_u = 0.48d

M_{u, lim} $= 0.36{\rm{;}} \times 20 \times {\rm{;}}\frac{{0.48{\rm{d}}}}{{\rm{d}}}{\rm{;}}\left( {1 - \frac{{0.42{\rm{;}} \times 0.48{\rm{d}}}}{{\rm{d}}}} \right){\rm{;b}}{{\rm{d}}^2}$

Mu, lim = 2.7592 bd²

R_{u, lim} = 2.7592

Additional Information

Explanation

Lateral ties

• Lateral ties are the transverse (horizontal) steel reinforcement bars wrapped around the vertical (longitudinal) main bars in a reinforced concrete column.
• In a rectangular column, they look like closed rectangular loops or links.
• General purposes in a standard (compression) column are: Positioning, Anti-buckling and Confinement.
• In a reinforced concrete member: Concrete's Shear Strength (v_u) depends heavily on the aggregate interlock and the "clamping" force of compression. When a column is under axial tension, the concrete develops cracks that go all the way through the section. This reduces the shear capacity of the concrete to almost zero.

Concept:

Area ratio:

It can be defined as the ratio of the maximum cross-sectional area of the cutting edge to the area of the soil sample.

The area ratio can be expressed as

${A_r} = \frac{{D_2^2 - D_1^2}}{{D_1^2}}$

where

D₁ = inner diameter of cutting edge

D₂ = outer diameter of cutting edge

Note:

For stiff formation (A_r)_max = 20%

Soft sensitive clay (Ar)max = 10%

Maximum thickness of cutting edge = $\frac{D_2 \ - \ D_1}{2}$

Calculation:

Given, D₂ = 70 mm 

Soil is stiff clay, So (Ar)max = 20%

${A_r} = \frac{{D_2^2 - D_1^2}}{{D_1^2}}$

${0.2} = \frac{{70^2 - D_1^2}}{{D_1^2}}$

D₁ = 63.9 mm

Maximum thickness of cutting edge = $\frac{D_2 \ - \ D_1}{2}$

= $\frac{70 \ - \ 63.9}{2}$ = 3.05 mm

Concept:

Flexural Bond develops due to variation in BM along the reinforcement. (or) Presence of shear force produces flexure bond.

We know Bond stress (τ _bd) is given by

${\tau _{bd}} = \frac{V}{{\left( {n \times \pi \times \phi } \right) \times L.A}}$

Where, n is number of bars

 Shear force (V ) = dM/dx

 ϕ is diameter of the bar

 L.A is lever arm

 τ_bd should be less than permissible value,

 Following are the permissible design bond strength

As, τ ∝ 1/n

τ ∝ 1/ϕ

If it is greater than (τ_bd) _permissible then best economical solution is to reduce the diameter of bar and increase in number.

∴ In case of bond failure most economical way to make structure safe in bond is by providing more number of bars of smaller diameter instead of less number of bars of larger diameter, for the same amount of steel (As, Surface area increases τ decreases which makes the structure safe.)

Explanation:

Assumptions made in arriving at the Differential Equation for Transient Flow:

Explanation:

Clause 21.1 of IS 456: 2000 informs us that a structure should be fire-resistant and it depends on the Size of the member, Cover of steel reinforcement, and aggregate type.

The minimum cover for fire resistance is 20 mm and is given in table 16A of IS 456: 2000.

• IS 456 states that the reinforcing details should reflect the changing pattern of the structural section and ensure that both individual elements and the structure as a whole contain adequate support, ties, bonds, and anchorages for the required fire resistance.
• The type of aggregates also has much influence. Such that the ‘carbonate’ aggregates [limestone, dolomite] consists of calcium carbonates or magnesium carbonates or both. During exposure to fire, these aggregates react and release CO₂. The reaction starts at the most exposed face and slowly progresses towards the opposite end. The result is that these aggregates behave somewhat better than normal aggregates in case of a fire.

Concept:

The load-carrying capacity(Friction) of a single pile driven in clay is given by

Q_F = α × C_U × A_F

⇒ Q_F = α × C_U × π × D × L

The load-carrying capacity(Friction), when piles act as a group is given by

QFG = α × CU × AFG

⇒ Q_FG = 1 × C_U × (3S + D) × 4L

Where, Q_F = capacity of a single pile, Q_FG = Capacity when piles act as a group,

A_F = Area for friction of single pile, C_U = Unconfined compressive strength of soil, 

α = Adhesion factor, D = Diameter, L = Length of the pile,

A_FG = Area under friction for a group of piles

Calculation:

Given:

α = 0.7, N = 16

For finding the optimum spacing between the pile, we can say that

Capacity of single pile × Number of piles = Group capacity of piles

⇒ N × Q_F = Q_FG

⇒ N ×  α × CU × π × D × L = 1 × CU × (3S + D) × 4L

⇒ 16 × 0.7 × π × D = (3S + D) × 4

⇒ 8.796 × D = 3S + D

⇒ S = 2.5988D ≈ 2.6D

∴ The optimum spacing between the piles is 2.6D

Explanation:

Law of conservation of momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

So from the law of conservation of momentum, we get m₁v₁ = m₂v₂

m₁ = Mass of 1^st object

m₂ = Mass of 2^nd object

v₁ = Velocity of 1^st object

v₂ = velocity of 2^nd object

From this, we can find the recoil velocity.

Let’s assume both bodies start from rest. So when the shot is fired, the bullet of mass m₁ gains a velocity of v₁ and that impacts the larger body m₂. Hence it will have recoil and will move backward.

So now the recoil velocity will be (v₂) = m₁v₁ / m₁

Important Points</sub>

In this regard, Newton’s laws of motion must be known. There are three laws, which are as follows

• Newton's first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

It gives us an idea about Inertia.

• Force is equal to the rate of change of momentum. For a constant mass, force equals mass times acceleration
• For every action, there is an equal and opposite reaction.

Concept:

Critical Height (H_c)

• It is the maximum height at which the slope will stand unsupported without failure under the given set of conditions.
• It is also the maximum height at which the slope is on the verge of failure i.e. FOS = 1.

We know,

For infinite slope when the slope is submerged,

$FOS = \frac{{C + {\gamma _{sub}}z{{\cos }^2}\beta \tan \phi }}{{{\gamma _{sub}}z\cos \beta \sin \beta }}$

Here,

C – cohesion value (kN/m²)

β – slope angle (degrees)

ϕ – the angle of internal friction (degrees)

γ_sub – the submerged unit weight of soil (kN/m³)

At FOS = 1, z = H_c

${H_c} = \frac{C}{{{\gamma _{sub}}\cos \beta \left( {\sin \beta - \cos \beta \tan \phi } \right)}}$                                   …..(1)

Calculation:

Given,

β = 25°

C = 30 kN / m², ϕ = 20°, e = 0.65 and G_S = 2.7

Submerge unit weight of soil,

${\gamma _{sub}} = \frac{{{G_s} - 1}}{{1 + e}}{\gamma _w} = \frac{{2.7 - 1}}{{1 + 0.65}} \times 9.81 = 10.107;kN/{m^2}$

γ_w = 9.81 kN/m³

Critical Height,

From (1)

${H_c} = \frac{{30}}{{10.107 \times \cos 25^\circ \left( {\sin 25^\circ - \cos 25^\circ \tan 20^\circ } \right)}} = 35.31;m$

Concept:

The bending moment will be maximum when,

$\frac{{d{M_x}}}{{dx}} = 0$

Calculation:

From the equilibrium equation,

Taking moment about A,

ΣM_A = 0

⟹ ${R_B}L - \frac{1}{2}\left( {{w_o}{L^2} \times \frac{{2L}}{3}} \right) = 0$

⟹ $R_B=w_ol^2/3$

Also,

$R_A=\frac{w_ol^2}{2} -\frac{w_ol^2}{3}=\frac{w_ol^2}{6}$

Taking moment at section x-x,

${M_x} = {R_A}x - \frac{1}{2}\left( {{w_o}{x^2} \times \frac{x}{3}} \right)$

For the maximum bending moment,

$\begin{array}{l} \frac{{d{M_x}}}{{dx}} = 0\ \Rightarrow {R_A} - \frac{1}{2}\left( {{w_o}{x^2}} \right) = 0\ \Rightarrow x = \pm \frac{L}{{\sqrt 3 }} \end{array}$

∴ Maximum bending moment,

${M_{max}} = \frac{{{w_o}{L^2}}}{6}\left( {\frac{L}{{\sqrt 3 }}} \right) - \frac{1}{2}\left( {{w_o}{{\left( {\frac{L}{{\sqrt 3 }}} \right)}^2} \times \frac{L}{{3\sqrt 3 }}} \right) = \frac{{{w_o}{L^3}}}{{9\sqrt 3 }} = \frac{{\sqrt 3 {w_o}{L^3}}}{{27}}$

Concept:

Design criteria for foundations for reciprocating type machines:

As per IS 2974 (Part 1):1982, cl.5.

The following are the design criteria for foundations for reciprocating machinery:

• When the number of similar machines to be installed side by side in close spacing and soil conditions, a mat foundation should be provided such that deformation is minor as compared to the resultant of vibration.
• The mass of the foundation should be greater than the mass of the whole machine.
• The foundation shall be so dimensioned such that resultant force due to mass of the machine and foundation passes through the CG of the base of the contact area.
• The eccentricity of the foundation system along both directions should not exceed 5% of the length of the corresponding side of the contact area.
• The highest natural frequency of foundation shall be higher than 40% of the highest disturbing frequency while the lowest natural frequency of the foundation shall be lower than 150% of the lowest disturbing frequency.
• The amplitudes of the vibrations at the upper edge shall not exceed 0.20mm in both directions.
• The soil stress below the foundation under dead loads only shall not exceed 80% of the allowable bearing pressure of the static loading.
• For preliminary design, the maximum pressure on soil due to static load alone may be taken as 0.4 times the corresponding safe bearing capacity

Additional Information

Frequency Ratio (f/f_n):

• It is the ratio of operating frequency to the natural frequency.
• For reciprocating machinery, it should be greater than 0.40.

Explanation:

Centroid – It is defined as a point on which, we can concentrate the entire area A of any arbitrary shape.

The centroid of the semi-circular lamina is given by,

$\bar x = \frac{{{M_y}}}{{A}} = \frac{2}{3}\frac{{R\sin \alpha }}{\alpha }$

Due to symmetry y – coordinate of the area, because the sector is symmetrical to X-axis.

The centroid of the semi-circular  area

For semicircle θ varies from \(\frac{{ - \pi }}{2}to\frac{\pi }{2}\), Hence its centroid is obtained by putting $\alpha = \frac{\pi }{2}$

 $\bar x = \frac{{4R}}{{3\pi }}$ and $\bar y = 0$

Concept:

The coefficient of permeability ‘k’ is given by:

$k = C \times \frac{γ }{μ } \times \frac{{{e^3}}}{{1 + e}} \times {D^2}$

where

e = void ratio

D = effective size of soil particle

γ = weight density of fluid

μ = viscosity of fluid

c = constant (depends upon the shape of particle)

Calculation:

Given,

Viscosity reduced to 80% i.e μ₂ = 0.8 μ₁

Unit Weight reduced to 90%  γ₂ = 0.9 γ₁  

As 'k' is directly proportional to γ / μ

∴ $\frac{{{k_2}}}{{{k_1}}} = \frac{{{γ _2} \times {μ _1}}}{{{γ _1} \times {μ _2}}} = \frac{{0.9{γ _1} \times {μ _1}}}{{{γ _1} \times 0.8{μ _1}}} = 1.125$

k2 = 1.125 k1

i.e k2 is increased by 12.5%.

Concept:

• Shear force is the resultant of all the transverse forces to the right or left of the beam.
• Bending moment is the resultant moment at a section due to all the transverse forces either to the left or to the right of the section.

Calculation:

Given

UDL intensity = W1 = 10 kN/M , Moment = M = 45 kN-m

∑M_A = 0 ,Moment about point  A 

= 10 × 4 × 4/2 + 45 - R_B × 10 = 0

= R_B = 12.5 kN

∑V = 0, Taking sum of all the vertical forces as 0

= R_A + R_B = 10 × 4

= R_A = 40 - 12.5 = 27.5 kN

Additional Information 

• Slope of the shear force diagram = intensity of distributed load.

         $\frac{{dV}}{{dx}} = W$

• Sope at any section of bending moment diagram = shear force at that section.

         $\frac{{dM}}{{dx}} = V$

• ​The point of contra flexure is the point where the bending moment changes its sign.
• The point of inflexion is the point where deflected shape of the beam changes its curvature.

Explanation:

• Well foundation is a type of deep foundation which is generally provided below the water level for bridges. Cassions or well have been in use for foundations of bridges and other structures since the Roman and Mughal periods.
• Depth of scour is the depth at which scouring occurs, below the high flood level.
• A well foundation should be sunk below the maximum scour depth, such that there is adequate lateral stability and this embedded length which is called the grip length.
• The grip length is required for developing sufficient passive resistance to counteract the overturning moment due to horizontal force at well top.
• Grip length for well foundation of railway bridges is generally taken as 50 %
• Grip length for well foundation of road bridges is generally taken as 30 %

Additional Information

Different shapes of well foundation are as follows:

1. ​ Circular shape is the most commonly used shape, as it has high circular strength and is convenient in sinking, and chances of tilting are also less. Maximum diameter limited to 9m.
2. Double D wells are generally used for the piers and the abutments of bridges which are too long to be accommodated on a circular well of 9 m diameter. This well of this shape can be sunk easily.
3. Double octagonal wells are better than double D well. The square corners are eliminated and bending stresses are reduced and also provide greater resistance to sinking.
4. Rectangular wells are generally used for bridge foundations having depth up to 7-8 m.

Concept:

• The analysis of trusses is usually based on the following simplifying assumptions.

1. The centroidal axis of each member coincides with the line connecting the centers of the adjacent members and the members only carry axial force.
2. All members are connected only at their ends by frictionless hinges in plane trusses.
3. All loads and support reactions are applied only at the joints.

• The reason for making these assumptions is to obtain an ideal truss, i.e., a truss whose members are subjected only to axial forces
• Types of forces in a truss are as follows

1. Primary Forces =member axial forces determined from the analysis of an ideal truss.
2. Secondary Forces = deviations from the idealized forces, i.e., shear and bending forces in a truss member.

         Our focus will be on primary forces. If large secondary forces are anticipated, the truss should be analyzed as a frame.

Calculation:

Given 

Load, P = 10 kN acts on joint B, AB = 1 m, AC = 0.5 m,

Let reactions at A in a vertical and horizontal direction as v_A and H_A  respectively

NOTE: Forces in truss as generally assumed as tensile. So, in truss analysis, positive member force means tensile force in that member and negative member force means compressive force in that member.

 ∑ M_C = 0, Taking moment about point C

= H_A×0.5 + 10×1 = 0 ( Moment of V_A  about C  will be 0 )

= H_A = -10/0.5 = -20 kN

∑ H_A = 0, Take sum of Horizontal forces as zero at joint A

= H_A + F_AB = 0

= - 20 + F_AB = 0

= F_AB = 20 kN and the nature of force is tensile as force comes out positive.

Important Points

• Method to identify zero force members in a given truss

1. If only two noncollinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both members is zero.
2. If three members, two of which are collinear, are connected to a joint that has no external loads or reactions applied to it, then the force in the member that is not collinear is zero.

Concept:

Relationship between total stress, effective stress, and pore water pressure:

$\sigma = ;\bar \sigma + u$

$\bar \sigma = effetive;stress$

σ = Total stress

u= Pore pressure

Given:

h = 5 m = 500 cm 

ρ_w = 1000 kg/m³ = 10^-3 kg/cm3 =10^-3 kg/cm³

σ = ρ_wh = 10^-3 × 500 = 0.5 kg/cm²

u = ρ_wh = 10^-3  × 500 = 0.5 kg/cm²

$\bar \sigma = \sigma - u = 0.5 - 0.5 = 0$

Concept:

Parallelogram law of forces:

This law states that if two forces are acting simultaneously on a body at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, their resultant is represented in magnitude and direction by the diagonal of the parallelogram which passes through the point of intersection of the two sides representing the forces.

Net resultant force, ${{R}} = \sqrt {{{{a}}^2} + {{{b}}^2}}$ $\% MathType!MTEF!2!1!+- \% feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn \% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr \% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 \% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x \% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 \% qacaqGsbGaeyypa0ZaaOaaa8aabaWdbiaabggapaWaaWbaaSqabeaa \% peGaaGOmaaaakiabgUcaRiaabkgapaWaaWbaaSqabeaapeGaaGOmaa \% aaaeqaaaaa!3CE6! {\rm{R}} = \sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}}$

Calculation:

Given:

$\rm Let;{\rm{\vec a}} = 800{\rm{;N}},{\rm{;\vec b}} = 200{\rm{;N}},{\rm{;\vec c}} = 400{\rm{;N}},{\rm{;\vec d}} = 600{\rm{;N}}$(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qaceqGHbWdayaalaWdbiabg2da9iaaiIdacaaIWaGaaGimaiaabcka % caqGobGaaiilaiaabckaceqGIbWdayaalaWdbiabg2da9iaaikdaca % aIWaGaaGimaiaabckacaqGobGaaiilaiaabckaceqGJbWdayaalaWd % biabg2da9iaaisdacaaIWaGaaGimaiaabckacaqGobGaaiilaiaabc % kaceqGKbWdayaalaWdbiabg2da9iaaiAdacaaIWaGaaGimaiaabcka % aaa!53CD! {\rm{\vec a}} = 800{\rm{;N}},{\rm{;\vec b}} = 200{\rm{;N}},{\rm{;\vec c}} = 400{\rm{;N}},{\rm{;\vec d}} = 600{\rm{;}}\)

ΣX = 200 – 600 = -400 N

ΣY = 400 – 800 = -400 N

${\rm{\theta }} = {\tan ^{ - 1}}\frac{{Σ {\rm{X}}}}{{Σ {\rm{Y}}}} = {\tan ^{ - 1}}\frac{{400}}{{400}} = 45^\circ$, as shown in the figure.

${\rm{R}} = \sqrt {{{\left( {Σ {\rm{X}}} \right)}^2} + {{\left( {Σ {\rm{Y}}} \right)}^2}} = \sqrt {{{400}^2} + {{400}^2}} = 400\sqrt 2 {\rm{;N}}$

ΣM_A = 400 × 1 + 600 × 1 = 1000 Nm.

Let x be the distance from A along the x-axis, where the resultant cuts AB. Then,

${\rm{x}} = \frac{{\Sigma {{\rm{M}}_{\rm{A}}}}}{{\Sigma {\rm{Y}}}} = \frac{{1000}}{{400}} = 2.5{\rm{;m}}$

Concept

There are four compounds (Called Bogue's Compounds) formed as a result of hydration of cement:

Alite: C3S, or Tricalcium Silicate

Belite: C2S, or Dicalcium Silicate

Aluminate phase: C3A, or Tricalcium Aluminate

Ferrite phase: C4AF, or Tetracalcium Aluminoferrite

Bogue Compounds

a) Dicalcium Silicate (C2S): This compound will undergo a reaction slowly. It is responsible for the progressive strength of concrete. This is also called as Belite. 

A higher percentage of C2S results in slow hardening, less heat of hydration, and great resistance to chemical attack.

b) Tricalcium silicate (C3S): This is also called as Alite. It undergoes hydration within one week and helps in the development of strength in the early stages of concrete (aka hardening).

It has the best cementitious property among all the other Bogue's Compounds. Tricalcium Silicate (C3S) hardens rapidly and is largely responsible for the initial set and early strength.

 The cement that has more C­­­­3S content is good for cold weather concreting. 

c) Tricalcium aluminate (C3A): Celite is the quickest one to react when water is added to the cement. It is responsible for the flash setting. The increase of this content will help in the manufacture of Quick Setting Cement. 

It provides weak resistance against sulphate attack and contribution to the development of strength is significantly less than the above two bogue compounds.

d) Tetra calcium Alumino ferrite (C4AF): This is called Felite. It has the poorest cementing value but it responsible for the long-term gain of strength of the cement.

Concept:

Impulse:

• When a large force works on a body for a very small time interval, it is called impulsive force.
• An impulsive force does not remain constant, but changes first from zero to maximum and then from maximum to zero. In such a case we measure the total effect of force.
• The impulse caused by a force during a specific time interval is equal to the body's change of momentum during that time interval.
• Impulse, effectively, is a measure of the change in momentum.
• Impulse of a force is a measure of total effect of force.
  

\(\vec I = \mathop \smallint \nolimits_{{t_1}}^{{t_2}} \overrightarrow {F;} \cdot dt = {\overrightarrow {F;} _{av}}{\rm{\Delta }}t\)  or

  Where, I = impulse, F = force and Δt = very small time interval

  Force = m × dv/dt

• Impulse is a vector quantity and its direction is the same as that of force.
• SI unit of impulse is Newton-second or kg-m-s-1.

Calculation:

Given,

 Mass (m) = 30 g = 0.03 kg,

 v = 500 m/s and t = 0.0018 sec

 Force = 0.03 × 500/0.018 = 8333.34 N

Concept:

​Optimum time:

• We know that Total project cost = Direct cost + Indirect cost.

• If direct and indirect cost varies with time is shown in the figure below, then total project cost versus time curve has the shape as indicated in the figure below

• Such a curve will have a point A where the tangent drawn to the curve would be horizontal.

• The total cost of the project will be minimum at this point which is called as optimum cost of the project and the time corresponding to this point A is called as optimum project duration.

• If the project duration is increased total cost will increase, while if project duration is equal to the crash time, the project cost will be the highest.

• In the process of reducing the project time, efforts should be made to achieve optimum duration such that project cost is min

Additional Information

• Normal time: It is the standard time that an estimator will assign to an activity.
• Crash time: It is the minimum possible time in which an activity can be completed by assigning extra resources.

Concept:

Simple harmonic motion:

it is a special type of periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position. It results in an oscillation which, if uninhibited by friction or any other dissipation of energy

a = ω² x and $;\omega = \frac{{2\pi }}{T}$

Where, a = acceleration, x = distance, T = time period, ω = angular velocity 

Calculation:

Given: a = 6 m/s², x = 1.5 m

6 = ω2 × 1.5

ω = 2 rad/sec

$T = {\frac{{2\pi}}{ \omega}} =\frac{2\pi}{2}=3.14 ~sec$

Explanation:

Harmful ingredient in bricks:

1. Lime:

• If lime is present in clay brick excess amount it causes a change of colour of brick from red to yellow and when lime is burn-in lamp it absorbs water and swells and brick becomes disintegrated.
• So lime should be present in bricks as finely graded material if there is present it might be removed in the beginning process.

2. Iron Pyrite:

• Iron pyrites tend to oxidize and decompose the brick during burning. The brick may split into pieces.
• Pyrites discolourises the bricks

3. Pebbles gravels And grits:

• It can not be allowed to mix with pebbles gravels and grit , because it can make brick porous and lose its strength and water-absorbing capacity is increased.

4. Alkalies:

• Excess of alkali metal causes bricks to melt and lose their shape this causes efflorescence.
• When bricks with alkalies come in contact with moisture is absorb and alkali material crystallized on drying of bricks when water is evaporated and crystallized material form white powder deposit on the bricks which destroy and spoil the appearance of bricks this phenomenon is known as effluorescence.

5. Sulphur:

• If sulphur is present in the clay or brick it provides oxidation of carbon and sulphur.
• During burning sulphur provides the spongy and swelling structure of white patches on brick and brick become discolored due to white patches.

Important Points Magnesia rarely exceeding 1 percent. In burning; it causes the clay to soften at slower rate than in most case is lime and reduces warping.

Explanation:

Coefficient of friction is the ratio of limiting friction to the normal reaction, between the two bodies, and is generally denoted by μ.

$\mu = \frac{F}{N}$

F = μ N

The frictional force is directly proportional to the normal force. So, if the normal force is doubled then the frictional force will also double. Hence their ratio will remain the same.

Explanation:

Fibre reinforced concrete is a composite material consisting of mixtures of cement, mortar or concrete and discontinuous, discrete, uniformly dispersed suitable fibres.

Fibre reinforced concrete (FRC) is concrete containing fibrous material which increases its structural integrity. It contains short discrete fibres that are uniformly distributed and randomly oriented.  Different types of fibres include steel fibres, glass fibres, synthetic fibres and natural fibres.

Following are the benefits one can obtain of a fibre reinforced concrete:

i) It increases the tensile strength, Compressive strength, flexural strength of the concrete.

ii) It reduce the air voids and water voids the inherent porosity of gel.

iii) It increases the toughness, ductility of the concrete.

iv) Fibres such as graphite and glass have excellent resistance to creep. Therefore, the orientation and volume of fibres have a significant influence on the creep performance of reinforcement.

Explanation:

D’Alembert’s principle:

• D’Alembert’s principle states that For a system of mass of particles, the sum of the difference of the force acting on the system and the time derivatives of the moment is zero when projected onto any virtual displacement.
• D’Alembert’s principle is used for analyzing the dynamic problem which can reduce it into a static equilibrium problem.
• It is also known as the Lagrange-d’Alembert principle, named after the French mathematician and physicist Jean le Rond d’Alembert.
• It is an alternative form of Newton’s second law of motion.
• According to the 2nd law of motion, F = ma while it is represented as F – ma = 0 in D’Alembert’s law.
• So it can be said that the object is in equilibrium when a real force is acting on it.
• Here, F is the real force while -ma is the fictitious force known as inertial force.

Concept: -

Pozzolana may be defined as a siliceous material that itself does not possess any cementitious properties but when reacts with lime in presence of lime forms compounds of low solubility having cementitious properties.

All pozzolanas are rich in silica and alumina and contain only a small quantity of alkalis.

The most commonly used pozzolans today are industrial by-products such as fly ash, silica fume from silicon smelting, highly reactive metakaolin, and burned organic matter residues rich in silica such as rice husk ash.

A pozzolanic material must be composed of microscopic and amorphous silica. As crystalline silica is not reactive with Portland cement which does not cause an alkali-silica reaction.

Explanation

Lame's equation is used to find the thickness of the thick cylinder subjected to internal pressure, it is given by,

$t = \frac{{{D_i}}}{2}\left[ {\sqrt {\frac{{{\sigma _t} + {P_i}}}{{{\sigma _t} - {P_i}}}} - 1} \right]$

Where,

t = thickness of thick cylinder,

Di = Internal diameter of the thick cylinder, σt = Allowable tensile stress in the material of thick cylinder, Pi = Internal pressure in the thick cylinder

Important Points

• Lame's equation is based on the maximum principal stress theory of failure, as this theory is more suitable for brittle materials, Lame's equation is also applicable to brittle materials like Cast iron or Cast Steel.

Additional Information

Clavarino's equation is used to find the thickness of the cylinder when the material is ductile and the cylinder has closed ends, it is given by

$t = \frac{{{D_i}}}{2}\left[ {\sqrt {\frac{{{\sigma _t} + \left( {1 - 2μ } \right){P_i}}}{{{\sigma _t} - \left( {1 + μ } \right){P_i}}}} - 1} \right]$

Birnie's equation is used to find the thickness of thick cylinders, which are made up of ductile material and has open ends, it is given by

$t = \frac{{{D_i}}}{2}\left[ {\sqrt {\frac{{{\sigma _t} + \left( {1 - μ } \right){P_i}}}{{{\sigma _t} - \left( {1 + μ } \right){P_i}}}} - 1} \right]$

μ is the Poisson's ratio

Explanation

(i) Blow holes are cavities on the vertical surface of the concrete. These are caused by air in concrete being trapped against the form face or sometimes due to insufficient vibration.

(ii) The occurrence of blowholes can be minimized by the use of a suitable air release agent on the surface of the formwork or by providing adequate vibrations. In addition to this concrete should have adequate workability. Blows holes does not affect the long-term performance of the concrete structures.

(iii) Improper design of shuttering may cause trapping of air against form face which can lead to blow holes.

Concept:

Elastic Modulus (E): When the body is loaded within its elastic limit, the ratio of stress and strain is constant. This constant is known as Elastic modulus

${\rm{E}} = \frac{{{\rm{Stress}}}}{{{\rm{Strain}}}} = \frac{{\rm{\sigma }}}{\epsilon}$

S.I unit: N/m²

Rigidity modulus (C): When a body is loaded within its elastic limit, the ratio of shear stress and shear strain is constant, this constant is known as the shear modulus.

${\rm{C}} = \frac{{{\rm{Shear;stress;}}}}{{{\rm{Shear;strain}}}} = \frac{{\rm{\tau }}}{\phi }$

S.I unit: N/m²

Bulk modulus (K): When a body is subjected to three mutually perpendicular like stresses of the same intensity then the ratio of direct stress and the volumetric strain of the body is known as bulk modulus

${\rm{K}} = \frac{{{\rm{Direct;stress}}}}{{{\rm{Volumetric;strain}}}} = \frac{{\rm{\sigma }}}{{\frac{{{\rm{\delta V}}}}{{\rm{V}}}}}$

S.I unit: N/m²

Explanation:

The relation between E, C, μ is

E = 2C (1 + μ)           ---(1)

Relation Between E, K, μ is

E = 3K (1 – 2μ)         ---(2)

Equating (1) & (2)

2C (1 + μ) = 3K(1 – 2μ)

2C + 2μC = 3K – 6μK

μ (2C + 6K) = 3K – 2C

$\mu = \frac{{3K; -; 2C}}{{2C; + ;6K}}$

Important Points

Poisson’s ratio: 

 When the body is loaded within its elastic limit, the ratio of lateral strain and linear strain is constant. This constant is known as Poisson's ratio.

${\rm{\mu }} = \frac{{{\rm{lateral;strain}}}}{{{\rm{linear;strain}}}}$

Other relations between elastic constants are,

1. E = 2C (1 + μ)

2. E = 3K (1 – 2μ)

3) ${\bf{E}} = \frac{{9{\bf{KC}}}}{{3{\bf{K}}; + ;{\bf{C}}}}$

Explanation

Given

Diameter of boiler, D = 1.5 m

Thickness, t = 10 mm

Steam pressure, p = 2 MPa

Since,

t / D = 1 / 150 < 1 / 10, thin cylindrical section.

For thin cylindrical section,

Hoop stress is given by - 

${\sigma _h} = \frac{{pD}}{{2t}} = \frac{{2 \times 1500}}{{2 \times 10}} = 150;MPa$