PYQs / GATE CE / 2015 / Set 2 / Q45 GATE CE 2015 Set 2 — Question 45 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Medium Prestressed Concrete Beams Slabs, Columns & Prestressed Structural Engineering
Structural Engineering → Slabs, Columns & Prestressed → Prestressed Concrete Beams
Last updated 5 September 2026
Question In a pre-stressed concrete beam section shown in the figure, the net loss is
10 % 10\% 10% and the final prestressing force applied at
X X X is
750 kN 750\text{ kN} 750 kN . The initial fiber stresses (in
N/mm 2 \text{N/mm}^2 N/mm 2 ) at the top and bottom of the beam were:
Correct answer (D) -4.166 and 20.833
Solution 1. Identify given data:
Final prestressing force P f i n a l = 750 kN P_{final} = 750\text{ kN} P f ina l = 750 kN Loss = 10 % = 10\% = 10% Initial prestressing force P i n i t i a l = P f i n a l 1 − 0.10 = 750 0.9 = 833.33 kN P_{initial} = \frac{P_{final}}{1 - 0.10} = \frac{750}{0.9} = 833.33\text{ kN} P ini t ia l = 1 − 0.10 P f ina l = 0.9 750 = 833.33 kN Dimensions: b = 250 mm b = 250\text{ mm} b = 250 mm d 1 = 200 mm d_1 = 200\text{ mm} d 1 = 200 mm d 2 = 100 mm d_2 = 100\text{ mm} d 2 = 100 mm D = 2 d 1 = 400 mm D = 2d_1 = 400\text{ mm} D = 2 d 1 = 400 mm Prestressing point X X X d 2 = 100 mm d_2 = 100\text{ mm} d 2 = 100 mm Eccentricity e = D 2 − 100 = 200 − 100 = 100 mm e = \frac{D}{2} - 100 = 200 - 100 = 100\text{ mm} e = 2 D − 100 = 200 − 100 = 100 mm
2.
Calculate Section Properties:
Area A = b × D = 250 × 400 = 10 5 mm 2 A = b \times D = 250 \times 400 = 10^5\text{ mm}^2 A = b × D = 250 × 400 = 1 0 5 mm 2 Moment of Inertia I = b D 3 12 = 250 × 400 3 12 = 1.333 × 10 9 mm 4 I = \frac{bD^3}{12} = \frac{250 \times 400^3}{12} = 1.333 \times 10^9\text{ mm}^4 I = 12 b D 3 = 12 250 × 40 0 3 = 1.333 × 1 0 9 mm 4 Section Modulus Z = I D / 2 = 1.333 × 10 9 200 = 6.667 × 10 6 mm 3 Z = \frac{I}{D/2} = \frac{1.333 \times 10^9}{200} = 6.667 \times 10^6\text{ mm}^3 Z = D /2 I = 200 1.333 × 1 0 9 = 6.667 × 1 0 6 mm 3
3.
Calculate Initial Stresses:
Stress at top fiber σ t o p = P i n i t i a l A − P i n i t i a l ⋅ e Z = 833.33 × 10 3 10 5 − 833.33 × 10 3 × 100 6.667 × 10 6 = 8.333 − 12.5 = − 4.167 N/mm 2 \sigma_{top} = \frac{P_{initial}}{A} - \frac{P_{initial} \cdot e}{Z} = \frac{833.33 \times 10^3}{10^5} - \frac{833.33 \times 10^3 \times 100}{6.667 \times 10^6} = 8.333 - 12.5 = -4.167\text{ N/mm}^2 σ t o p = A P ini t ia l − Z P ini t ia l ⋅ e = 1 0 5 833.33 × 1 0 3 − 6.667 × 1 0 6 833.33 × 1 0 3 × 100 = 8.333 − 12.5 = − 4.167 N/mm 2 Stress at bottom fiber σ b o t t o m = P i n i t i a l A + P i n i t i a l ⋅ e Z = 8.333 + 12.5 = 20.833 N/mm 2 \sigma_{bottom} = \frac{P_{initial}}{A} + \frac{P_{initial} \cdot e}{Z} = 8.333 + 12.5 = 20.833\text{ N/mm}^2 σ b o tt o m = A P ini t ia l + Z P ini t ia l ⋅ e = 8.333 + 12.5 = 20.833 N/mm 2
Thus, the initial stresses are
− 4.166 -4.166 − 4.166 and
20.833 N/mm 2 20.833\text{ N/mm}^2 20.833 N/mm 2 . Option D is correct.
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Correct answer (D) -4.166 and 20.833
Solution 1. Identify given data:
Final prestressing force P f i n a l = 750 kN P_{final} = 750\text{ kN} P f ina l = 750 kN Loss = 10 % = 10\% = 10% Initial prestressing force P i n i t i a l = P f i n a l 1 − 0.10 = 750 0.9 = 833.33 kN P_{initial} = \frac{P_{final}}{1 - 0.10} = \frac{750}{0.9} = 833.33\text{ kN} P ini t ia l = 1 − 0.10 P f ina l = 0.9 750 = 833.33 kN Dimensions: b = 250 mm b = 250\text{ mm} b = 250 mm d 1 = 200 mm d_1 = 200\text{ mm} d 1 = 200 mm d 2 = 100 mm d_2 = 100\text{ mm} d 2 = 100 mm D = 2 d 1 = 400 mm D = 2d_1 = 400\text{ mm} D = 2 d 1 = 400 mm Prestressing point X X X d 2 = 100 mm d_2 = 100\text{ mm} d 2 = 100 mm Eccentricity e = D 2 − 100 = 200 − 100 = 100 mm e = \frac{D}{2} - 100 = 200 - 100 = 100\text{ mm} e = 2 D − 100 = 200 − 100 = 100 mm
2.
Calculate Section Properties:
Area A = b × D = 250 × 400 = 10 5 mm 2 A = b \times D = 250 \times 400 = 10^5\text{ mm}^2 A = b × D = 250 × 400 = 1 0 5 mm 2 Moment of Inertia I = b D 3 12 = 250 × 400 3 12 = 1.333 × 10 9 mm 4 I = \frac{bD^3}{12} = \frac{250 \times 400^3}{12} = 1.333 \times 10^9\text{ mm}^4 I = 12 b D 3 = 12 250 × 40 0 3 = 1.333 × 1 0 9 mm 4 Section Modulus Z = I D / 2 = 1.333 × 10 9 200 = 6.667 × 10 6 mm 3 Z = \frac{I}{D/2} = \frac{1.333 \times 10^9}{200} = 6.667 \times 10^6\text{ mm}^3 Z = D /2 I = 200 1.333 × 1 0 9 = 6.667 × 1 0 6 mm 3
3.
Calculate Initial Stresses:
Stress at top fiber σ t o p = P i n i t i a l A − P i n i t i a l ⋅ e Z = 833.33 × 10 3 10 5 − 833.33 × 10 3 × 100 6.667 × 10 6 = 8.333 − 12.5 = − 4.167 N/mm 2 \sigma_{top} = \frac{P_{initial}}{A} - \frac{P_{initial} \cdot e}{Z} = \frac{833.33 \times 10^3}{10^5} - \frac{833.33 \times 10^3 \times 100}{6.667 \times 10^6} = 8.333 - 12.5 = -4.167\text{ N/mm}^2 σ t o p = A P ini t ia l − Z P ini t ia l ⋅ e = 1 0 5 833.33 × 1 0 3 − 6.667 × 1 0 6 833.33 × 1 0 3 × 100 = 8.333 − 12.5 = − 4.167 N/mm 2 Stress at bottom fiber σ b o t t o m = P i n i t i a l A + P i n i t i a l ⋅ e Z = 8.333 + 12.5 = 20.833 N/mm 2 \sigma_{bottom} = \frac{P_{initial}}{A} + \frac{P_{initial} \cdot e}{Z} = 8.333 + 12.5 = 20.833\text{ N/mm}^2 σ b o tt o m = A P ini t ia l + Z P ini t ia l ⋅ e = 8.333 + 12.5 = 20.833 N/mm 2
Thus, the initial stresses are
− 4.166 -4.166 − 4.166 and
20.833 N/mm 2 20.833\text{ N/mm}^2 20.833 N/mm 2 . Option D is correct.
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More questions on Slabs, Columns & Prestressed ← Q44 Full paper Q46 → 
