Paper I — Q5
(a) A circular log of weight 1000 N and radius 18 cm is supported by a pair of brackets, one of which is shown in the figure…
A circular log of weight 1000 N and radius 18 cm is supported by a pair of brackets, one of which is shown in the figure below. Bar PN is hinged at 'P' and held by a string MN 67 cm long. To induce minimum tension at MN, determine the value of 2θ, as shown for equilibrium. Consider all contact surfaces smooth. Also find the value of minimum tension. 10 marks
Determine the moment of resistance of the T-beam as shown in the figure. Use limit state method. Take M 20 grade of concrete and Fe 415 HYSD bars. 10 marks
The fixed-end bar ABCD consists of three prismatic segments as shown in the figure. The end segments AB and CD have cross-sectional area 800 mm² and length 300 mm. The middle segment has cross-sectional area 1200 mm² and length 500 mm. Two axial loads P_B = 28·5 kN and P_C = 20 kN are acting in the bar as shown in the figure. Young modulus E is same for all three segments. Determine the reaction forces at fixed supports.
Determine the compressive axial force in the middle segment. 10 marks
A flow meter when tested in a laboratory gives a pressure drop of 100 kN/m² for a discharge of 0·10 m³/s in a 150 mm diameter pipe. If a geometrically similar model is tested in 600 mm diameter pipe at identical conditions of fluid, determine the corresponding discharge and pressure drop in the model. 10 marks
A sample of dry cohesionless soil whose angle of internal friction is 35°, is subjected to a triaxial test. If the minor principal stress (σ₃) is 105 kPa, at what values of deviator stress (Δσ) and major principal stress (σ₁) will the test specimen fail? 10 marks
हिंदी में प्रश्न पढ़ें
1000 N भार और 18 cm त्रिज्या वाला एक वृत्ताकार लट्ठा, ब्रैकेट के एक जोड़े द्वारा आलंबित है जिनमें से एक नीचे चित्र में दर्शाया गया है । छड़ PN, 'P' पर हिंज है और 67 cm लंबी डोरी MN द्वारा बंधी है । MN में न्यूनतम तनाव उत्पन्न करने के लिए, दर्शाई गई साम्यावस्था के लिए, 2θ के मान को निर्धारित कीजिए । सभी संपर्क सतहों को चिकना मान लीजिए । न्यूनतम तनाव का मान भी ज्ञात कीजिए । (10 अंक)
चित्र में दर्शाई गई T-धन के आघूर्ण प्रतिरोध का निर्धारण कीजिए । सीमंत अवस्था विधि का उपयोग कीजिए । M 20 ग्रेड की कंक्रीट और Fe 415 एच.वाई.एस.डी. छड़ें लीजिए । (10 अंक)
चित्र में दर्शाए अनुसार आबद्ध-सिरा छड़ ABCD तीन समपार्श्वीय खंडों से बनी है । अंत खंडों AB और CD का अनुप्रस्थ-परिच्छेद क्षेत्रफल 800 mm² और लंबाई 300 mm है । मध्य खंड का अनुप्रस्थ-परिच्छेद क्षेत्रफल 1200 mm² और लंबाई 500 mm है । दो अक्षीय भार P_B = 28·5 kN और P_C = 20 kN चित्र में दर्शाए अनुसार छड़ में लगे हैं । यांग मापांक E सभी तीन खंडों के लिए समान है । आबद्ध आलम्बों पर प्रतिक्रिया बलों का निर्धारण कीजिए ।
मध्य खंड में संपीडन अक्षीय बल का निर्धारण कीजिए । (10 अंक)
एक प्रवाह मापी, एक प्रयोगशाला में परीक्षण करने पर, एक 150 mm व्यास की पाइप में एक 0·10 m³/s के निस्सरण के लिए 100 kN/m² का दाब पात देता है । तरल की समान अवस्था पर यदि एक ज्यामितियतः समरूप निदर्श का परीक्षण 600 mm व्यास की पाइप में किया जाता है, तो निदर्श में संगत निस्सरण और दाब पात का निर्धारण कीजिए । (10 अंक)
एक शुष्क संसजनहीन मृदा, जिसका आंतरिक घर्षण कोण 35° है, के प्रतिदर्श पर एक त्रिअक्षीय परीक्षण किया जाता है । विचलक प्रतिबल (Δσ) और उच्च मुख्य प्रतिबल (σ₁) के किन मानों पर परीक्षण प्रतिदर्श भंग होगा, यदि निम्न मुख्य प्रतिबल (σ₃) 105 kPa है ? (10 अंक)
The figure this question refers to, in words
The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.
(a) A 2D diagram showing a circular log resting in a corner formed by a vertical wall and an inclined bar. The vertical wall is a line segment labeled M at the top and P at the bottom. The inclined bar is a line segment labeled P at the bottom and N at the top right. A horizontal string connects point M to point N. A circle representing the log is centered at point G. The circle touches the vertical wall at point Q and the inclined bar at point J. A dashed line connects the center G to the contact point Q (horizontal) and to the contact point J. The angle between the vertical wall and the line segment PG is labeled theta. The angle between the line segment PG and the inclined bar PN is also labeled theta. The angle at the hinge P is thus 2theta.
(b) A cross-sectional diagram of a T-beam. The top flange has a width of 1000 mm and a thickness of 100 mm. The vertical web has a width of 300 mm and a height of 400 mm. The total depth of the beam is 500 mm. Reinforcement consists of 4 bars, each with a diameter of 25 mm, located at the bottom of the web. The center of these bars is positioned 50 mm from the bottom edge of the beam. The text below the figure specifies: 'Take M 20 grade of concrete and Fe 415 HYSD bars.'
(c) A horizontal bar ABCD is fixed at both ends A and D. The bar consists of three segments: AB, BC, and CD. Segment AB has a length of 300 mm. Segment BC has a length of 500 mm. Segment CD has a length of 300 mm. An axial load P_B is applied at point B, directed to the right. An axial load P_C is applied at point C, directed to the left. The supports at A and D are shown as fixed walls.
Model answer
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(a) Minimum tension in the string MN (10 marks)
Assume smooth contacts and neglect the bar weight. Two identical brackets share the log, so W = 1000/2 = 500 N per bracket. r = 0.18 m and MN = L = 0.67 m, which is horizontal.
G is at distance r from both the wall and the bar, so PG bisects the angle 2θ at P and PJ = r cot θ.
Log. The bar's reaction R is normal to the bar. Vertical equilibrium gives R sin 2θ = W, so R = W/sin 2θ.
Bar PN, moments about P. N is at horizontal distance L from the wall, so its height above P is L cot 2θ, which is the arm of T. The log's thrust R acts at J with arm PJ.
T L cot 2θ = R · r cot θ, so T = W r cot θ/(L cos 2θ).
Minimum. Set d(ln T)/dθ = 0: −2/sin 2θ + 2 tan 2θ = 0, so sin² 2θ = cos 2θ. With c = cos 2θ, c² + c − 1 = 0, so c = 0.618 and 2θ = 51.8°. T tends to infinity at both ends of the range, so this is the minimum.
Then cot²θ = (1 + c)/(1 − c) = 4.236, so cot θ = 2.058. T_min = 500 × 0.18 × 2.058/(0.67 × 0.618) = 447 N.
(b) Moment of resistance of the T-beam (10 marks)
Flange 1000 × 100 mm, web 300 mm, d = 500 − 50 = 450 mm. A_st = 4 × π/4 × 25² = 1,963.5 mm². f_ck = 20, f_y = 415 N/mm².
Neutral axis. Assume x_u ≤ D_f and use b = 1,000 mm: x_u = 0.87 f_y A_st/(0.36 f_ck b) = 0.87 × 415 × 1,963.5/(0.36 × 20 × 1,000) = 98.5 mm. This is less than D_f = 100 mm, so the assumption holds and the section acts as a rectangle. It is under-reinforced, since x_u < x_u,max = 0.48d = 216 mm.
Moment. T = 0.87 × 415 × 1,963.5 = 708.9 kN. Lever arm = d − 0.416x_u = 450 − 41.0 = 409.0 mm. M_u = 708.9 × 0.409 = 290 kN·m.
(c) Fixed-end bar ABCD (10 marks)
Take tension as positive, with R_D the reaction at D. P_B = +28.5 kN (right) and P_C = −20 kN (left). Cutting from the right:
- N_CD = R_D
- N_BC = R_D − 20
- N_AB = R_D + 8.5
(i) Both ends are fixed, so the total elongation is zero. With E the same, Σ N L/A = 0: (R_D + 8.5)(300/800) + (R_D − 20)(500/1200) + R_D(300/800) = 0 1.1667 R_D = 5.146, so R_D = 4.41 kN (to the right). Equilibrium of the bar: R_A = −(R_D + 8.5) = 12.91 kN (to the left).
(ii) N_BC = 4.41 − 20 = −15.59 kN, so the middle segment carries 15.59 kN compression. (N_AB = 12.91 kN tension and N_CD = 4.41 kN tension, and the elongations sum to zero.)
(d) Flow meter model (10 marks)
The model is geometrically similar, with the same fluid, so dynamic similarity needs equal Reynolds number: ρVD/μ is the same, so V_m/V_p = D_p/D_m = 150/600 = 1/4.
Discharge ratio: Q ∝ VD², so Q_m/Q_p = (1/4)(4)² = 4, and Q_m = 0.40 m³/s.
Pressure drop scales as ρV² (equal Euler number): Δp_m/Δp_p = (1/4)² = 1/16, so Δp_m = 100/16 = 6.25 kN/m².
(e) Triaxial failure of dry sand (10 marks)
For c = 0, the Mohr–Coulomb criterion gives σ₁ = σ₃ tan²(45° + φ/2) = σ₃(1 + sin φ)/(1 − sin φ).
With sin 35° = 0.5736: σ₁/σ₃ = 1.5736/0.4264 = 3.69. σ₁ = 105 × 3.69 = 387.5 kPa. Deviator stress Δσ = σ₁ − σ₃ = 387.5 − 105 = 282.5 kPa.
How this answer will be evaluated
Framework: null
What earns full marks: All parts show complete, correct derivations with clear sketches and unit checks.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Determine the value of 2θ for minimum tension and the value of that minimum tension. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Free body diagram of the log showing all forces
- Equilibrium equations (ΣF=0, ΣM=0) applied to the system
- Geometric relationship between θ and the string length MN
- Differentiation or optimization to find minimum tension
Loses marks
- Omitting the free body diagram
- Failing to state the assumption of smooth surfaces
Earns more
- Correct identification of reaction forces at contact points
- Clear labeling of angles and dimensions in the sketch
Extra mark
- Verification of the minimum using the second derivative test
- (b) Determine the moment of resistance of the T-beam using the limit state method. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Locate the neutral axis with the given flange width 1000 mm: x_u about 98.5 mm < D_f = 100 mm, so the section is treated as rectangular
- Determination of neutral axis depth (xu) and comparison with xu,max
- Calculation of lever arm and moment of resistance (Mu)
- Use of correct material strengths for M20 and Fe 415
Loses marks
- Using working stress method instead of limit state
- Incorrect calculation of steel area (Ast)
Earns more
- Explicit check for under-reinforced section
- Clear sketch of the stress block
Extra mark
- Reference to specific IS 456 clauses for design constants
- (c) Determine reaction forces at fixed supports and the compressive axial force in the middle segment. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Equilibrium equation for horizontal forces (ΣFx = 0)
- Compatibility equation for total deformation (δAB + δBC + δCD = 0)
- Expression for deformation in terms of force, length, area, and E
- Solving the simultaneous equations for reactions
Loses marks
- Ignoring the compatibility condition
- Unit inconsistency in force or length
Earns more
- Correct sign convention for tension and compression
- Step-by-step substitution of values
Extra mark
- Verification of the result using the force in segment AB
- (d) Determine the corresponding discharge and pressure drop in the 600 mm diameter model pipe. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Application of Reynolds number similarity (Re_model = Re_prototype)
- Calculation of velocity ratio from diameter ratio
- Calculation of discharge ratio (Q ∝ D²V)
- Application of pressure drop scaling law (ΔP ∝ ρV²)
Loses marks
- Assuming velocity is constant between model and prototype
- Incorrect scaling of pressure drop
Earns more
- Explicit statement of the geometric similarity assumption
- Correct unit conversion for diameter
Extra mark
- Mention of the Froude number if relevant to the flow regime
- (e) Determine the deviator stress (Δσ) and major principal stress (σ₁) at failure. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Use of the Mohr-Coulomb failure criterion for cohesionless soil
- Relationship between σ₁, σ₃, and angle of internal friction (φ)
- Calculation of σ₁ using the given σ₃ and φ
- Calculation of Δσ as the difference between σ₁ and σ₃
Loses marks
- Using the wrong formula for the relationship between principal stresses
- Confusing deviator stress with total stress
Earns more
- Drawing of the Mohr's circle at failure
- Clear identification of the failure envelope
Extra mark
- Reference to the specific triaxial test condition (drained/undrained)
What "Solve" is asking you to do
Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.
Structure that answers it
Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity
Where marks are lost
Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.
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