Paper I — Q3
(a) Design a floor slab to cover a room with internal dimensions of 4·5 m × 6·0 m. The slab is simply supported on all the sides…
Design a floor slab to cover a room with internal dimensions of 4·5 m × 6·0 m. The slab is simply supported on all the sides on 230 mm thick masonry walls. The slab carries a live load of 4·0 kN/m² and a dead load due to finishing work of 1·0 kN/m². The corners of the slab are prevented from lifting up. Use M 20 concrete and Fe 415 steel. Assume mild exposure conditions. 20 marks
Table : Bending Moment coefficients when four edges are discontinuous
| l_y/l_x | Short span coefficient, α_x | Long span coefficient, α_y | |||||
|---|---|---|---|---|---|---|---|
| 1·0 | 1·1 | 1·2 | 1·3 | 1·4 | 1·5 | for all values of l_y/l_x | |
| α_x | 0·056 | 0·064 | 0·072 | 0·079 | 0·085 | 0·089 | 0·056 |
Modification Factor for Tension Reinforcement Note : f_s is steel stress of service loads in N/mm² f_s = 0.58 f_y (Area of cross-section of steel required)/(Area of cross-section of steel provided)
A built-up column of effective length 10 m is designed by placing two ISMC 300 @ 363 N/m back to back at a spacing 'S' mm. The column is to carry a factored axial load of 1100 kN. Find the economical spacing 'S' of the two channel sections. Also design the batten system for the column. M 20 bolts of grade 4.6 are used for making the connections. Do not design the connections. Use E 250 grade of steel. 20 marks
For connections : Edge distance = 32 mm Gauge distance = 50 mm
Properties of ISMC 300 : A = 4630 mm² r_zz = 118 mm, r_yy = 26.0 mm I_zz = 6420 × 10⁴ mm⁴ I_yy = 313 × 10⁴ mm⁴ C_y = 23.5 t_f = 13.6 300 z z t_w=7.8 y 90 ISMC 300 (All dimensions are in mm)
Using slope deflection method, determine the final end moments for the portal frame shown in the figure. The frame is fixed at A and D, and has rigid joints at B and C. Take EI as constant. 10 marks
हिंदी में प्रश्न पढ़ें
4·5 m × 6·0 m की आंतरिक विमाओं के एक कक्ष के आच्छादन के लिए एक फर्श छतपट की अभिकल्पना कीजिए । छतपट सभी ओर पर 230 mm मोटी चिनाई की दीवारों पर शुड्डालम्बित है । छतपट 4·0 kN/m² का एक चल भार और परिष्करण कार्य के कारण 1·0 kN/m² का एक अचल भार वहन करता है । छतपट के कोनों को ऊपर उठने से रोका गया है । M 20 कंक्रीट और Fe 415 इस्पात का उपयोग कीजिए । हल्की प्रभावन अवस्थाएँ मान लीजिए । (20 अंक)
सारणी : बंकन आधुर्ण गुणांक, जब चारों कोर असतत हैं
| लघु विस्तृति गुणांक, αₓ | दीर्घ विस्तृति गुणांक, αᵧ | ||||||
|---|---|---|---|---|---|---|---|
| lᵧ/lₓ | 1·0 | 1·1 | 1·2 | 1·3 | 1·4 | 1·5 | lᵧ/lₓ के सभी मानों के लिए |
| αₓ | 0·056 | 0·064 | 0·072 | 0·079 | 0·085 | 0·089 | 0·056 |
तनन प्रबलन के लिए आशोधन गुणक नोट : f_s सेवा भारों का इस्पात प्रतिबल N/mm² में है f_s = 0.58 f_y आवश्यक इस्पात का अनुप्रस्थ-परिच्छेद क्षेत्रफल —————————————————————— प्रदत इस्पात का अनुप्रस्थ-परिच्छेद क्षेत्रफल
दो ISMC 300 @ 363 N/m को 'S' mm के अंतरण पर सहपृष्ठ रखकर, 10 m प्रभावी लंबाई के एक संधित स्तंभ की अभिकल्पना की गई है । स्तंभ को 1100 kN का एक गुणित अक्षीय भार बहन करना है । दो चैनल परिछेदों के मितव्ययी अंतरण 'S' को ज्ञात कीजिए । स्तंभ के लिए बता तंत्र की भी अभिकल्पना कीजिए । जोड़ों को बनाने के लिए 4·6 ग्रेड के M 20 बोल्टों को उपयोग किया गया है । जोड़ों की अभिकल्पना नहीं कीजिए । E 250 ग्रेड इस्पात का उपयोग कीजिए । (20 अंक)
जोड़ों के लिए : कोर दूरी = 32 mm गेज दूरी = 50 mm
ISMC 300 के गुण : A = 4630 mm² r_zz = 118 mm, r_yy = 26.0 mm I_zz = 6420 × 10⁴ mm⁴ I_yy = 313 × 10⁴ mm⁴ C_y = 23.5 t_f = 13.6 300 z z t_w=7.8 y 90 ISMC 300 (सभी विमाएँ mm में हैं)
प्रवणता विषम विधि का उपयोग करके, चित्र में दर्शाए पोर्टल फ्रेम के लिए अंतिम सिरा आघूर्णों को निर्धारित कीजिए। फ्रेम, A और D पर आबद्ध है तथा B और C पर दृढ़ जोड़ हैं। EI को नियत लीजिए। (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) Graph titled 'Modification Factor for Tension Reinforcement'. The vertical axis is 'Modification Factor' ranging from 0.4 to 2.0 with grid lines at 0.4, 0.8, 1.2, 1.6, 2.0. The horizontal axis is 'Percentage Tension Reinforcement' ranging from 0 to 3.0 with grid lines at 0, 0.4, 0.8, 1.2, 1.6, 2.0, 2.4, 2.8, 3.0. The graph contains five curves representing different values of steel stress (fs). The curves are labeled from top to bottom as: fs = 120, fs = 145, fs = 190, fs = 240, fs = 290. A note inside the graph states: 'Note: fs is steel stress of service loads in N/mm2'. Below the graph is the formula: fs = 0.58 fy (Area of cross-section of steel required / Area of cross-section of steel provided).
(b) A cross-sectional view of a built-up column section composed of two ISMC 300 channels placed back-to-back. The channels are oriented with their web plates vertical and facing each other, separated by a gap. The flanges of the channels point outwards to the left and right. The section is labeled 'ISMC 300'. Dimensions provided in millimeters: The depth of the channel is 300. The width of the flange is 90. The thickness of the flange (t_f) is 13.6. The thickness of the web (t_w) is 7.8. The centroidal axis 'z-z' is shown as a horizontal line passing through the center of the section. The centroidal axis 'y-y' is shown as a vertical line passing through the center of the gap between the two channels. The distance from the y-y axis to the centroid of one channel is labeled C_y = 23.5.
(c) A cross-sectional view of a built-up column consisting of two ISMC 300 channel sections placed back-to-back. The sections are oriented with their webs vertical and flanges facing outward. The vertical axis is labeled 'z z' and the horizontal axis is labeled 'y'. Dimensions provided: depth of channel is 300 mm, flange width is 90 mm, web thickness t_w is 7.8 mm, and flange thickness t_f is 13.6 mm. The text 'ISMC 300' is written below the section.
What "Design" is asking you to do
Produce a specification that meets the given brief and demonstrate that it does. In civil and electrical papers the design is incomplete until it is expressed in buildable numbers — diameter, spacing, section, component value — and checked back against every limit stated.
Structure that answers it
Requirements and permissible values listed → code clause or design basis adopted → proportioning calculations → the specification in final dimensions → check against each requirement → sketch
Where marks are lost
Stopping at a required area or a required value without converting it into the bar size, spacing or component actually provided. The provided-against-required comparison and the serviceability or stability check are separately marked and routinely left out.
How this answer will be evaluated
Approach
Framework: IS 456:2000 (Limit State Method) and IS 800:2007 (Limit State Method). (a) calculate: given > formula > substitution > result with units > interpretation | (b) calculate: given > formula > substitution > result with units > interpretation | (c) calculate: given > formula > substitution > result with units > interpretation Full marks: Complete design/analysis with all checks, correct units, and neat sketches.
Key points expected
- Calculate total factored load (1.5DL + 1.5LL)
- Determine bending moments using provided coefficients
- Calculate required steel area (As) for both spans
- Check deflection using modification factor chart
- Calculate required moment of inertia (I_req) from P = P_cr
- Determine spacing 'S' using parallel axis theorem
- Check buckling about the minor axis (y-y)
- Design batten spacing and thickness
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Design a two-way slab for flexure and deflection control. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Calculate total factored load (1.5DL + 1.5LL)
- Determine bending moments using provided coefficients
- Calculate required steel area (As) for both spans
- Check deflection using modification factor chart
Loses marks
- Ignoring the modification factor for deflection
- Using wrong load factors (e.g., 1.2 instead of 1.5)
Earns more
- Correct calculation of effective cover (20mm)
- Proper selection of bar diameter and spacing
- Verification of minimum and maximum reinforcement limits
Extra mark
- Neat sketch of slab reinforcement layout
- (b) Determine economical spacing 'S' and design batten system for built-up column. 20 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Calculate required moment of inertia (I_req) from P = P_cr
- Determine spacing 'S' using parallel axis theorem
- Check buckling about the minor axis (y-y)
- Design batten spacing and thickness
Loses marks
- Ignoring the self-weight of the column in load calculation
- Incorrect application of the parallel axis theorem
Earns more
- Correct calculation of effective length for battened column
- Verification of minimum spacing requirements (IS 800)
Extra mark
- Sketch of the built-up column cross-section
- (c) Determine final end moments for the portal frame using slope deflection method. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Calculate fixed-end moments (FEM) for the loaded beam
- Formulate slope deflection equations for all members
- Apply equilibrium conditions at joints B and C
- Solve for unknown rotations and final moments
Loses marks
- Sign errors in the slope deflection equations
- Incorrect calculation of fixed-end moments
Earns more
- Correct handling of the fixed supports at A and D
- Clear presentation of the moment distribution
Extra mark
- Free body diagram of the frame
Model answer coming soon
Every evaluation on this site is marked against a verified model answer. This question's answer is still being written; evaluation opens the moment it lands.
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