UPSC Previous Year Question Paper-Engineering Services (Main) Examination, 2026 -CIVIL ENGINEERING PAPER I 

 

 

 

1. (a) AB, a vertical post of a crane as shown in figure, is hinged at A and B. The horizontal arm CD is hinged to AB at C and supported by the strut EF which is hinged at E and F. Draw the bending moment diagram for AB and CD.

[Figure: Crane diagram]

1. (b) Explain Fibre Reinforced Plastics (FRPs). Draw a comparison between different type of FRPs based on their composition, engineering properties and their civil engineering applications. 12

1. (c) There are two circular beams of same length, same material and of same weight. One of these is solid with diameter D and the other one is hollow with outer diameter Do and inner diameter Di. Show that the ratio of section modulus of solid beam to that of the hollow beam is (D × Do) / (Do² + Di²). 12

1. (d) What are the different types of failure of black bolts? Explain with sketches. 12

1. (e) A short column of a 4 m effective height is subjected to an ultimate load of 1500 kN. The column is of 400 mm diameter. The column is reinforced by 6 bars of 20 mm φ of Fe-415 grade. If M-30 concrete is used, obtain the ultimate moment carrying capacity of the column. The nominal cover to the longitudinal bar is 40 mm.

[Figure: Circular cross-section of 400mm dia]

UPSC Previous year question paper Geo scientist exam

2. (a) Determine the horizontal displacement of support D of the frame shown in figure. Take E = 200 GPa and I = 300 × 10⁻⁶ m⁴.

[Figure: Frame diagram]

2. (b)(i) Discuss the classification of building lime based on IS 712: 1973 specifically mentioning different classes, their names and uses in construction.

Also enlist the field tests to be carried out on building lime as recommended by IS 1624: 1986. 10

2. (b)(ii) What is cold weather concreting? Discuss the effects of cold weather on concreting. Explain in brief, various measures recommended by IS 7861 (Part- II) : 1981 to deal with the problems of cold weather concreting. 10

2. (c) A T-shaped retaining wall for a height 4 m above ground level is as shown in figure below. The angle of repose of earth is 30° and density 18 kN/m³. The safe bearing capacity of soil is 100 kN/m² and coefficient of friction between concrete and soil is 0.55. Apply all the stability checks. Also design the stem if M-25 concrete and Fe-415 bars are used. Show details of reinforcement in stem. Consider nominal cover to steel as 75 mm.

[Figure: Retaining wall diagram]

UPSC Previous year question paper Geo scientist exam

3. (a) A bracket plate is welded to the flange of a column of ISMB 600 as shown in figure below. Determine the load carrying capacity of the joint for 10 mm fillet weld. Assume E250 grade steel and shop welding. Use limit state method. γmθ = 1.25.

[Figure: Bracket plate diagram]

3. (b) Determine the member forces in the truss as shown in the figure. The members in tension have a cross- sectional area a and the members in compression have a cross- sectional area 2a. Take E = constant.

[Figure: Truss diagram]

3. (c) The network of a certain project is shown in figure given below, with the estimated duration of various activities. Determine the following :

(i) Earliest event time and latest event time. (ii) Earliest and latest start and finish times of each activity. (iii) Total and free floats for each activity. (iv) Critical path for the network.

[Figure: Network diagram]

UPSC Previous year question paper Geo scientist exam

4. (a) The floor plan of a building is as shown in figure below. The floor load consists of a live load of 5 kN/m² and dead load of 2 kN/m² (excluding self weight). Design the slab thickness and reinforcement required at the various critical sections. Use Fe-500 steel and M-20 concrete. Assume that the structure is subjected to mild exposure conditions. 20

[Figure: Floor plan]

Table 12 Bending Moment Coefficients (Clause 22.5.1)

Type of Load | Span Moments | Support Moments
--- | --- | ---
| Near Middle of End Span | At Middle of Interior Span | At Support Next to the End Support | At Other Interior Supports
(1) | (2) | (3) | (4) | (5)
Dead load and imposed load (fixed) | + 1/12 | + 1/16 | - 1/10 | - 1/12
Imposed load (not fixed) | + 1/10 | + 1/12 | - 1/9 | - 1/9

NOTE - For obtaining the bending moment, the coefficient shall be multiplied by the total design load and effective span.

Table 13 Shear for Coefficients (Clauses 22.5.1 and 22.5.2)

Type of Load | At End Support | At Support Next to the End Support | At All Other Interior Supports
--- | --- | --- | ---
| | Outer Side | Inner Side |
(1) | (2) | (3) | (4) | (5)
Dead load and imposed load (fixed) | 0.4 | 0.6 | 0.55 | 0.5
Imposed load (not fixed) | 0.45 | 0.6 | 0.6 | 0.6

NOTE - For obtaining the shear force, the coefficient shall be multiplied by the total design load.

22.6 Critical Sections for Moment and Shear 22.6.1 For monolithic construction, the moments computed at the face of the supports shall be used in the design of the members at those sections. For non- monolithic construction the design of the member shall be done keeping in view 22.2.

## 22.6.2 Critical Section for Shear

The shears computed at the face of the support shall be used in the design of the member at that section except as in 22.6.2.1.

22.6.2.1 When the reaction in the direction of the applied shear introduces compression into the end region of the member, sections located at a distance less than d from the face of the support may be designed for the same shear as that computed at distance d see Fig.2.

NOTE- The above clauses are applicable for beams generally carrying uniformly distributed load or where the principal load is located farther than 2d from the face of the support.

## 22.7 Redistribution of Moments

Redistribution of moments may be done in accordance with 37.1.1 for limit state method and in accordance with B- 1.2 for working stress method. However, where simplified analysis using coefficients is adopted, redistribution of moments shall not be done.

## 23 BEAMS

## 23.0 Effective Depth

Effective depth of a beam is the distance between the centroid of the area of tension reinforcement and the maximum compression fibre, excluding the thickness of finishing material not placed monolithically with the member and the thickness of any concrete provided to allow for wear. This will not apply to deep beams.

## 23.1 T-Beams and L-Beams

## 23.1.1 General

A slab which is assumed to act as a compression flange of a T- beam or L- beam shall satisfy the following:

a) The slab shall be cast integrally with the web, or the web and the slab shall be effectively bonded together in any other manner; and 
b) If the main reinforcement of the slab is parallel to the beam, transverse reinforcement shall be provided as in Fig. 3; such reinforcement shall not be less than 60 percent of the main reinforcement at mid span of the slab.

## 23.1.2 Effective Width of Flange

In the absence of more accurate determination, the effective width of flange may be taken as the following

[Figure: T-beam diagram]

FIG.3 TRANSVERSE REINFORCEMENT IN FLANGE OF T-BEAM WHEN MAIN REINFORCEMENT OF SLAB IS PARALLEL TO THE BEAM

UPSC Previous year question paper Geo scientist exam

1) For flanged beams, the values of (a) or (b) be modified as per Fig. 6 and the reinforcement percentage for use in Fig. 4 and 5 should be based

on area of section equal to bf d.

NOTE- When deflections are required to be calculated, the method given in Annex C may be used.

[Figure: Modification factor graph]

FIG. 4 MODIFICATION FACTOR FOR TENSION REINFORCEMENT

4. (b) Determine the plastic moment capacity for the frame of uniform cross-section shown in figure below under applied factored load.

[Figure: Frame diagram]

4. (c) A circle of 60 mm is marked on a mild steel plate prior to the application of stresses on the plate. The plate is stressed as shown in figure. Find the lengths of the major and minor axes of an ellipse formed as a result of the deformation of the circle marked. Take E = 2 × 10⁵ N/mm² and Poisson's ratio, μ = 0.25

[Figure: Plate stress diagram]

## SECTION 'B'

5. (a) Design a bolted connection for double angle member ISA 100×100×10 in a truss connected on either side of a 10 mm thick gusset plate. The length of the member is 2.5 m. Use 16 mm dia bolt of grade 4-6. Assume net area at threads is 0.85 times area at shunk. γmb = 1.25, Kb = 0.6. The factored loads in the member are 175 kN (Compression) and 150 kN (Tension). 12

5. (b) What is preservation of timber? Explain various methods of preservation of timber. 12

5. (c) A beam of square section is used with one diagonal horizontal. The beam is subjected to a shear force F at a section. Find the maximum shear stress in the cross-section of the beam and draw the shear stress distribution diagram for the section. 12

5. (d) A post tensioned beam of span 8 m is pre-stressed by a parabolic cable, concentric at the support and having an eccentricity of 240 mm at midspan. The cross-sectional area of high tensile wire of the cable is 600 mm². The wires are stressed by using a jack at the left end so that the initial force in the cable at the right end is 260 kN. Calculate the jacking force at the left end and the total losses of stress in wires.

UPSC Previous year question paper Geo scientist exam

5. (e) Discuss various safety measures to be taken for workers while working at height. 12

6. (a) A cantilever has a quadratal shape as shown in figure with a radius of curvature of R. It carries a vertical downward load P at its free end A, while its other end is fixed. If EI is constant, find horizontal and vertical deflection of the free end A.

[Figure: Cantilever diagram]

6. (b) A factory shed is to be constructed using pitched roof trusses supported on R.C. columns. The spacing of the truss is 5 m. Purlins are placed at the nodes (Refer figure below). Design the purlins with circular hollow section for the unfactored loading as follows:

Dead load = 180 N/m Live load = 920 N/m Wind load = 2080 N/m (suction)

Assume E250 grade steel. Apply limit state method of design. Assume sections are compact. γmo = 1.1. Draw also the connection with the rafter (100 mm Nominal bore dia).

[Figure: Factory shed diagram]

6. (c) Design a circular water tank of capacity 350 m³ resting on ground. The base of wall is fixed with base slab. Height of Tank is restricted to 3.4 m including free board of 300 mm. The materials used are M-30 concrete and Fe-500 steel. Use IS-3370 (Part IV) method. Adopt working stress design approach. Use wall thickness of 200 mm. Show details of reinforcement.

Nominal bore dia. (mm) | Outer dia. (mm) | Thickness (mm) | Moment of Inertia (cm⁴) | Plastic section Modulus (cm³)
--- | --- | --- | --- | ---
50 | 60.3 | 3.6 | 25.9 | 11.6
65 | 76.1 | 3.6 | 54.0 | 19.0
80 | 88.9 | 4.0 | 96.4 | 28.8
90 | 101.6 | 4.0 | 146.3 | 38.1
100 | 114.3 | 4.5 | 234.3 | 54.3

Design a circular water tank of capacity 350 m³ resting on ground. The base of wall is fixed with base slab. Height of Tank is restricted to 3.4 m including free board of 300 mm. The materials used are M- 30 concrete and Fe- 500 steel. Use IS- 3370 (Part IV) method. Adopt working stress design approach. Use wall thickness of 200 mm. Show details of reinforcement.

Shear V = coefficient × [γw H² / MH]

H²/DT | Shear at the Base of Cylindrical Wall | Moment at edge
--- | --- | --- | ---
| Triangular load, fixed base | Rectangular load, fixed base | Triangular or rectangular load, hinged base
0.4 | +0.436 | +0.755 | +0.245 | -1.58
0.8 | +0.374 | +0.552 | +0.234 | -1.75
1.2 | +0.339 | +0.460 | +0.220 | -2.00
1.6 | +0.317 | +0.407 | +0.204 | -2.28
2.0 | +0.299 | +0.370 | +0.189 | -2.57
3.0 | +0.262 | +0.310 | +0.158 | -3.18
4.0 | +0.236 | +0.271 | +0.137 | -3.68
5.0 | +0.213 | +0.243 | +0.121 | -4.10
6.0 | +0.197 | +0.222 | +0.110 | -4.49
8.0 | +0.174 | +0.193 | +0.096 | -5.18
10.0 | +0.158 | +0.172 | +0.087 | -5.81
12.0 | +0.145 | +0.158 | +0.079 | -6.38
14.0 | +0.135 | +0.147 | +0.073 | -6.88
16.0 | +0.127 | +0.137 | +0.068 | -7.36

Notes: 1. γw Density of the liquid 2. Positive sign indicates shear acting inward.

UPSC Previous year question paper Geo scientist exam

Coefficients at point | 0.0H | 0.1H | 0.2H | 0.3H | 0.4H | 0.5H | 0.6H | 0.7H | 0.8H | 0.9H
--- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
0.4 | +0.149 | +0.134 | +0.120 | +0.101 | +0.082 | +0.066 | +0.049 | +0.029 | +0.014 | +0.004
0.8 | +0.263 | +0.239 | +0.215 | +0.190 | +0.160 | +0.130 | +0.096 | +0.063 | +0.034 | +0.010
1.2 | +0.283 | +0.271 | +0.254 | +0.234 | +0.209 | +0.180 | +0.142 | +0.099 | +0.054 | +0.016
1.6 | +0.265 | +0.268 | +0.268 | +0.266 | +0.250 | +0.226 | +0.185 | +0.134 | +0.075 | +0.023
2.0 | +0.234 | +0.251 | +0.273 | +0.285 | +0.285 | +0.274 | +0.232 | +0.172 | +0.104 | +0.031
3.0 | +0.134 | +0.203 | +0.267 | +0.322 | +0.357 | +0.362 | +0.330 | +0.262 | +0.157 | +0.052
4.0 | +0.067 | +0.164 | +0.256 | +0.339 | +0.403 | +0.429 | +0.409 | +0.334 | +0.210 | +0.073
5.0 | +0.025 | +0.137 | +0.245 | +0.346 | +0.428 | +0.477 | +0.469 | +0.398 | +0.259 | +0.092
6.0 | +0.018 | +0.119 | +0.234 | +0.344 | +0.441 | +0.504 | +0.514 | +0.447 | +0.301 | +0.112
8.0 | -0.011 | +0.104 | +0.218 | +0.335 | +0.443 | +0.534 | +0.575 | +0.530 | +0.381 | +0.151
10.0 | -0.011 | +0.098 | +0.208 | +0.323 | +0.437 | +0.542 | +0.608 | +0.589 | +0.440 | +0.179
12.0 | -0.005 | +0.097 | +0.202 | +0.312 | +0.429 | +0.543 | +0.628 | +0.633 | +0.494 | +0.211
14.0 | -0.002 | +0.098 | +0.200 | +0.306 | +0.420 | +0.539 | +0.639 | +0.666 | +0.541 | +0.241
16.0 | -0.000 | +0.099 | +0.199 | +0.304 | +0.412 | +0.531 | +0.641 | +0.687 | +0.582 | +0.265

Table Hoop Tension in Cylindrical Walls Fixed at Base and Free at Top (IS:3370-1967,Part IV) Hoop tension T = Coefficient × (γw H)

Coefficients at point | 0.0H | 0.1H | 0.2H | 0.3H | 0.4H | 0.5H | 0.6H | 0.7H | 0.8H | 0.9H
--- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
0.4 | +0.149 | +0.134 | +0.120 | +0.101 | +0.082 | +0.066 | +0.049 | +0.029 | +0.014 | +0.004
0.8 | +0.263 | +0.239 | +0.215 | +0.190 | +0.160 | +0.130 | +0.096 | +0.063 | +0.034 | +0.010
1.2 | +0.283 | +0.271 | +0.254 | +0.234 | +0.209 | +0.180 | +0.142 | +0.099 | +0.054 | +0.016
1.6 | +0.265 | +0.268 | +0.268 | +0.266 | +0.250 | +0.226 | +0.185 | +0.134 | +0.075 | +0.023
2.0 | +0.234 | +0.251 | +0.273 | +0.285 | +0.285 | +0.274 | +0.232 | +0.172 | +0.104 | +0.031
3.0 | +0.134 | +0.203 | +0.267 | +0.322 | +0.357 | +0.362 | +0.330 | +0.262 | +0.157 | +0.052
4.0 | +0.067 | +0.164 | +0.256 | +0.339 | +0.403 | +0.429 | +0.409 | +0.334 | +0.210 | +0.073
5.0 | +0.025 | +0.137 | +0.245 | +0.346 | +0.428 | +0.477 | +0.469 | +0.398 | +0.259 | +0.092
6.0 | +0.018 | +0.119 | +0.234 | +0.344 | +0.441 | +0.504 | +0.514 | +0.447 | +0.301 | +0.112
8.0 | -0.011 | +0.104 | +0.218 | +0.335 | +0.443 | +0.534 | +0.575 | +0.530 | +0.381 | +0.151
10.0 | -0.011 | +0.098 | +0.208 | +0.323 | +0.437 | +0.542 | +0.608 | +0.589 | +0.440 | +0.179
12.0 | -0.005 | +0.097 | +0.202 | +0.312 | +0.429 | +0.543 | +0.628 | +0.633 | +0.494 | +0.211
14.0 | -0.002 | +0.098 | +0.200 | +0.306 | +0.420 | +0.539 | +0.639 | +0.666 | +0.541 | +0.241
16.0 | -0.000 | +0.099 | +0.199 | +0.304 | +0.412 | +0.531 | +0.641 | +0.687 | +0.582 | +0.265

Note: Positive sign indicates tension.

Table Tension in Circular Ring Wall, Hinged Base, Free Top and Subjected to Triangular Load

Coefficients at point | 0.0H | 0.1H | 0.2H | 0.3H | 0.4H | 0.5H | 0.6H | 0.7H | 0.8H | 0.9H
--- | --- | --- | --- | --- | --- | --- | --- | --- | --- | ---
0.4 | +0.474 | +0.440 | +0.395 | +0.352 | +0.308 | +0.264 | +0.215 | +0.165 | +0.111 | +0.057
0.8 | +0.423 | +0.402 | +0.381 | +0.358 | +0.330 | +0.297 | +0.249 | +0.202 | +0.145 | +0.076
1.2 | +0.350 | +0.355 | +0.361 | +0.362 | +0.358 | +0.342 | +0.309 | +0.256 | +0.186 | +0.098
1.6 | +0.271 | +0.303 | +0.341 | +0.369 | +0.385 | +0.385 | +0.362 | +0.314 | +0.233 | +0.124
2.0 | +0.205 | +0.260 | +0.321 | +0.373 | +0.411 | +0.434 | +0.419 | +0.369 | +0.280 | +0.151
3.0 | +0.074 | +0.179 | +0.281 | +0.375 | +0.449 | +0.506 | +0.519 | +0.479 | +0.375 | +0.210
4.0 | +0.017 | +0.137 | +0.253 | +0.367 | +0.469 | +0.545 | +0.579 | +0.553 | +0.447 | +0.256
5.0 | -0.008 | +0.114 | +0.235 | +0.356 | +0.469 | +0.562 | +0.617 | +0.606 | +0.503 | +0.294
6.0 | -0.011 | +0.103 | +0.223 | +0.343 | +0.463 | +0.566 | +0.639 | +0.643 | +0.547 | +0.327
8.0 | -0.015 | +0.096 | +0.208 | +0.324 | +0.443 | +0.564 | +0.661 | +0.697 | +0.621 | +0.386
10.0 | -0.008 | +0.095 | +0.200 | +0.311 | +0.428 | +0.552 | +0.666 | +0.730 | +0.678 | +0.433
12.0 | -0.002 | +0.097 | +0.197 | +0.302 | +0.417 | +0.541 | +0.664 | +0.750 | +0.720 | +0.477
14.0 | 0.000 | +0.098 | +0.197 | +0.299 | +0.408 | +0.531 | +0.659 | +0.761 | +0.752 | +0.513
16.0 | +0.002 | +0.100 | +0.198 | +0.299 | +0.403 | +0.521 | +0.650 | +0.764 | +0.776 | +0.536

Notes: 1. γw is the density of the liquid. 2. Positive sign indicates tension.

UPSC Previous year question paper Geo scientist exam

7. (a)(i) A beam of 200 mm × 200 mm section is 4.0 m long and is simply supported at its ends. The beam fails under bending when a point load of 200 kN is applied at the centre of the beam. Find the intensity of the uniformly distributed load which will cause failure of 5.0 m long cantilever of same material having cross- section of 400 mm wide and 500 mm deep. 10

7. (a)(i) A beam of 200 mm × 200 mm section is 4.0 m long and is simply supported at its ends. The beam fails under bending when a point load of 200 kN is applied at the centre of the beam. Find the intensity of the uniformly distributed load which will cause failure of 5.0 m long cantilever of same material having cross- section of 400 mm wide and 500 mm deep. 10

7. (a)(ii) Determine the output of an angle dozer with a blade width = 3.35 m, blade height = 0.855 m, forward speed = 9.4 kmph on a haul road 90 m long consisting of earth and gravel (swell factor = 0.83). The efficiency of dozer is 80% and its working time is 50 min/hr. The job conditions are good and management conditions are fair (Factor for these conditions may be taken as 0.69). 10

7. (b) An unstiffened suspension cable carries a uniformly distributed load of 65 kN/m over a span of 30 m as shown in the figure. The suspension cable is supported on frictionless rollers, fixed to the piers. The anchor cables are inclined at 30° to horizontal. One pier is 4.5 m below the other and the maximum dip of the cable at the lowest point (C) is 3 m below the lower pier. Calculate

(i) the maximum and minimum tension in the cable.

(ii) the horizontal and vertical forces at support A.

[Figure: Suspension cable diagram]

7. (c) Design a combined footing with a strap beam for two reinforced concrete columns 400 mm × 400 mm size spaced 5 m apart and each supporting a factored load of 1000 kN. Assume the safe bearing capacity of soil as 100 kN/m². Maximum length of footing is restricted to 8 × 0 M. Adopt M- 25 concrete and Fe- 500 Bars. Nominal cover for foundation shall be considered as 50 mm. Load factor considered is 1.5. Show details of reinforcement. 20

Design a combined footing with a strap beam for two reinforced concrete columns 400 mm × 400 mm size spaced 5 m apart and each supporting a factored load of 1000 kN. Assume the safe bearing capacity of soil as 100 kN/m². Maximum length of footing is restricted to 8 × 0 M. Adopt M- 25 concrete and Fe- 500 Bars. Nominal cover for foundation shall be considered as 50 mm. Load factor considered is 1.5. Show details of reinforcement. 20

TABLE 3 FLEXURE-REINFORCEMENT PERCENTAGE, p1 FOR SINGLY REINFORCED SECTIONS

M, lb/ft³, N/mm³ | f, N/mm³ | f, N/mm³ | f, N/mm³ | M, lb/ft³, N/mm³ | f, N/mm³ | f, N/mm³ | f, N/mm³

UPSC Previous year question paper Geo scientist exam

8. (a) A built up column of effective length 5 m consists of two ISMC 250 (back to back). The sections are laced. Find the spacing of the sections for maximum axial compressive load capacity. Also design the single lacing system using bars connected to the channels by weld (shop). Draw the lacing system also. For ISMC 250, A = 38.67 cm², Ixx = 3817 cm⁴, Iyy = 219 cm⁴, Cyy = 2.3 cm. Assume lacing bars are at 50° with the vertical.

Given:

KL / Γ | 40 | 50 | 60 | 70 | 80 | 90 | 100 | 110 | 120
--- | --- | --- | --- | --- | --- | --- | --- | --- | ---
fcd (MPa) | 198 | 183 | 168 | 152 | 136 | 121 | 107 | 95 | 84

8. (b)(i) Discuss the important features of common post tensioned systems in prestressed concrete. 10

8. (b)(ii) Calculate the moment of resistance of a T-beam as shown in figure given below. The grade of concrete is M-25 and Grade of steel is Fe-500.

[Figure: T-beam diagram]

(All Dimensions are in mm)

UPSC Previous year question paper Geo scientist exam

IS 456 : 2000

ANNEX G

(Clause 38.1)

MOMENTS OF RESISTANCE FOR RECTANGULAR AND T-SECTIONS

G-0 The moments of resistance of rectangular and T-sections based on the assumptions of 38.1 are given in this annex.

G-1 RECTANGULAR SECTIONS

G-1.1 Sections Without Compression Reinforcement

The moment of resistance of rectangular sections without compression reinforcement should be obtained as follows :

a) Determine the depth of neutral axis from the following equation :

xu = (0.87 fy Ast) / (0.36 fck b d)

b) If the value of xu/d is less than the limiting value (see Note below 38.1), calculate the moment of resistance by the following expression :

Mu = 0.87 fy Ast d (1 - (Ast fy) / (b d fck))

c) If the value of xu/d is equal to the limiting value, the moment of resistance of the section is given by the following expression :

Mu,lim = 0.36 (xu,max / d) (1 - 0.42 (xu,max / d)) fck b d²

d) If xu/d is greater than the limiting value, the section should be redesigned.

In the above equations,

xu = depth of neutral axis, 
d = effective depth, 
fy = characteristic strength of reinforcement, 
Ast = area of tension reinforcement, 
fck = characteristic compressive strength of concrete, 
b = width of the compression face, 
Mu,lim = limiting moment of resistance of a section without compression reinforcement, and 
xu,max = limiting value of xu from 39.1.

G-1.2 Section with Compression Reinforcement

Where the ultimate moment of resistance of section exceeds the limiting value, Mu,lim compression reinforcement may be obtained from the following equation :

Mu - Mu,lim = fsc Asc (d - d')

where

Mu, Mu,lim, d are same as in G-1.1, 
fsc = design stress in compression reinforcement corresponding to a strain of

0.0035 ((xu,max - d') / xu,max)

where

xu,max = the limiting value of xu from 38.1, 
Asc = area of compression reinforcement, and 
d' = depth of compression reinforcement from compression face.

The total area of tension reinforcement shall be obtained from the following equation :

Ast = Ast1 + Ast2

where

Ast = area of the total tension reinforcement, 
Ast1 = area of the tension reinforcement for a singly reinforced section for Mu,lim, and 
Ast2 = Asc fsc / 0.87 fy

G-2 FLANGED SECTION

G-2.1 For xu < Df the moment of resistance may be calculated from the equation given in G-1.1.

G-2.2 The limiting value of the moment of resistance of the section may be obtained by the following equation when the ratio Df / d does not exceed 0.2 :

Mu = 0.36 (xu,max / d) (1 - 0.42 (xu,max / d)) fck bw d²

+ 0.45 fck (bf - bw) Df (d - Df/2)

where

Mu, xu,max, d and fck are same as in G-1.1, 
bf = breadth of the compression face/flange, 
bw = breadth of the web, and 
Df = thickness of the flange.

G-2.2.1 When the ratio Df/d exceeds 0.2, the moment of resistance of the section may be calculated by the following equation :

Mu = 0.36 (xu,max / d) (1 - 0.42 (xu,max / d)) fck bw d²

+ 0.45 fck (bf - bw) yf (d - yf/2)

where yf = (0.15 xu + 0.65 Df), but not greater than Df, and the other symbols are same as in G-1.1 and G-2.2.

G-2.3 For xu,max > xu > Df, the moment of resistance may be calculated by the equations given in G-2.2 when Df/d does not exceed 0.43 and G-2.2.1 when Df/d exceeds 0.43; in both cases substituting xu,max by xu.

8.(c) A hollow marine propeller shaft turning at 120 rpm is required to propel a vessel at 16 m/s at the expense of 6500 kW of shaft power. The efficiency of the propeller is 70%. The ratio of the outer diameter to inner diameter of the shaft is 2/3. If the direct stress due to thrust is not to exceed 10 N/mm², calculate the following :

(i) Outer and inner diameter of the shaft

(ii) Maximum shearing stress due to torque

 

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