Working Stress Method

Q1. Using the working stress method, design a rectangular section 300 mm wide and 450 mm deep carrying 30 kN/m load over an effective span of 3.6 m. Use mild steel and M20 concrete. [4] (2071 Chaitra, Q1a)

Assumptions: interpret 450 mm as overall depth, adopt an effective cover of 50 mm, and treat 30 kN/m as the total service UDL. Then b = 300 mm, d = 400 mm, m = 13.333, σ_cbc = 7 and σ_st = 140 N/mm². If the stated load excludes self-weight, the loading calculation must be revised.

M = 30 × 3.6²/8 = 48.60 kN·m
X_c = [13.333 × 7/(13.333 × 7 + 140)] × 400 = 160 mm
M_bal = 0.5 × 7 × 300 × 160 × (400 − 160/3) = 58.24 kN·m

Since M < M_bal, a steel-controlled solution exists. Set f_st = 140 and use strain compatibility f_c = 140x/[m(d − x)] together with M = 0.5f_cbx(d − x/3). Solving gives x = 148.745 mm, z = 350.418 mm, and A_st = M/(140z) = 990.65 mm². Concrete stress is 6.216 N/mm², below 7.

A practical choice is four 20 mm bars, A_st = 1256.64 mm². Re-analysis of the provided section gives x = 162.781 mm, f_c = 5.757 N/mm² and f_st = 111.861 N/mm² at 48.60 kN·m. Both permissible-stress checks pass. Its WSM classification changes slightly after rounding up the steel; ultimate ductility requires its own LSM check. Bar fit, shear and development length remain separate design checks.

Q1. What are the different types of RCC section of a rectangular beam? Explain with neat sketches. [4] (2082 Bhadra, Q1a)

By reinforcement arrangement, a rectangular section is singly reinforced when the flexural tension is carried by reinforcement without a designed compression-steel contribution, and doubly reinforced when both tension and compression steel contribute to the resisting couple.

In working stress design, the critical neutral-axis depth is X_c = dmσ_cbc/(mσ_cbc + σ_st). If the actual x is smaller than X_c, steel reaches its permissible stress first; if equal, both permissible stresses are reached together; if larger, concrete controls the permissible moment. These are elastic design limits, not physical crushing or yielding.

At ultimate flexure, a tension-controlled section develops substantial steel strain before concrete crushing. A compression-controlled section can crush with limited warning. The IS limiting neutral-axis depth enforces a minimum tensile strain; a section beyond that limit must be redesigned, rather than accepted by simply ignoring excess reinforcement.

Sketch: draw the rectangular concrete section, compression face, tension bars and neutral axis; show the neutral axis successively deeper for the steel-controlled, limiting and compression-controlled cases. The strain line is straight, with tensile and compressive strains on opposite sides of zero. Ductility also requires adequate shear resistance, anchorage and confinement.

Q2. Discuss different types of flexural sections according to the amount of reinforcement in the tension zone. Describe the different failure modes in the above-mentioned sections. [8] (2080 Bhadra, Q1a)

By reinforcement arrangement, a rectangular section is singly reinforced when the flexural tension is carried by reinforcement without a designed compression-steel contribution, and doubly reinforced when both tension and compression steel contribute to the resisting couple.

In working stress design, the critical neutral-axis depth is X_c = dmσ_cbc/(mσ_cbc + σ_st). If the actual x is smaller than X_c, steel reaches its permissible stress first; if equal, both permissible stresses are reached together; if larger, concrete controls the permissible moment. These are elastic design limits, not physical crushing or yielding.

At ultimate flexure, a tension-controlled section develops substantial steel strain before concrete crushing. A compression-controlled section can crush with limited warning. The IS limiting neutral-axis depth enforces a minimum tensile strain; a section beyond that limit must be redesigned, rather than accepted by simply ignoring excess reinforcement.

Sketch: draw the rectangular concrete section, compression face, tension bars and neutral axis; show the neutral axis successively deeper for the steel-controlled, limiting and compression-controlled cases. The strain line is straight, with tensile and compressive strains on opposite sides of zero. Ductility also requires adequate shear resistance, anchorage and confinement.

Q1. An RC beam 230 mm wide and 400 mm deep (effective depth 380 mm) with area of steel 580 mm² has permissible stresses in concrete and steel of 5 N/mm² and 140 N/mm² respectively. Find the moment of resistance of the section and the actual stresses in concrete and steel. [8] (2080 Baishakh, Q1b)

Use m = 280/(3 × 5) = 18.667, not the M20 value of 13.333. The equilibrium equation is 115x² = 18.667 × 580(380 − x), giving x = 147.840 mm. The critical depth is X_c = 152 mm, so steel governs. Thus M_R = 140 × 580 × (380 − 147.840/3) = 26.8545 × 10⁶ N·mm. The accompanying concrete stress is 2M/(230 × 147.840 × 330.720) = 4.776 MPa. The paper supplies no separate applied moment, so these are stresses at the permissible moment of resistance.

Q2. A rectangular beam is 350 mm wide and has an effective depth of 650 mm. It is reinforced with five tension bars of 25 mm diameter. Due to an accidental loading condition, the beam has to resist a service-load bending moment of 150 kN·m. Determine whether the beam is safe. Considering the safety of the beam, also determine the stresses developed in the concrete and steel at the state of failure according to the working stress method. Use M25 concrete and Fe415 steel. [8] (2082 Bhadra, Q1b)

A_st = 5π(25)²/4 = 2454.37 mm²; m = 280/(3 × 8.5) = 10.980. Solving the neutral-axis equation gives x = 248.621 mm and z = 567.126 mm. Concrete controls because x > X_c = 187.629 mm. The working-stress capacity 209.74 kN·m exceeds 150 kN·m, and the actual stresses 6.079 MPa and 107.763 MPa are below 8.5 MPa and 230 MPa.

The question also asks for stresses "at failure" in WSM. Interpreting this as the first permissible-stress limit, f_c = 8.5 MPa and f_st = 150.680 MPa. Do not call this actual crushing or a true ultimate failure analysis.

Q3. A rectangular beam 300 mm wide and 550 mm effective depth is reinforced with 4 bars of 12 mm diameter. Determine the moment of resistance using the working stress method. Use M20 concrete and Fe415 steel. [8] (2081 Bhadra, Q1c)

1. Geometry and working-stress data. Use b = 300 mm and d = 550 mm. The tension area is A_st = 4π(12)²/4 = 452.39 mm². The permissible stresses are 7 MPa in concrete and 230 MPa in steel; m = 13.3333. No ultimate load factor is applied in this working-stress analysis.

2. Neutral axis. From equal first moments of the cracked transformed section, bx²/2 = mA_st(d − x). The positive root gives x = 129.964 mm. The critical depth is X_c = dmσ_cbc/(mσ_cbc + σ_st) = 158.763 mm. The lever arm is z = d − x/3 = 506.679 mm.

3. Independent permissible moments. Concrete permits M_c = 0.5σ_cbcbxz = 69.142 kN·m; steel permits M_s = σ_stA_stz = 52.720 kN·m. Therefore M_R = min(M_c, M_s) = 52.720 kN·m.

4. Stresses at the permissible moment of resistance. With M = 52.720 kN·m, f_c = 2M/(bxz) = 5.337 MPa, and f_st = M/(A_stz) = 230.000 MPa. Use M in N·mm in the stress equations.

Q4. A rectangular beam is 300 mm wide and has an effective depth of 550 mm. It is reinforced with 4 tension bars of 16 mm diameter. Determine the stresses induced in the top compression fibre of the concrete and in the tension steel when the beam is subjected to a bending moment of 50 kN·m. Use M20 concrete and Fe415 steel, with permissible stresses in concrete and steel not exceeding 5 MPa and 140 MPa, respectively. [9] (2081 Baishakh, Q1b)

1. Geometry and working-stress data. The paper supplies the 5 MPa concrete limit, so use m = 280/(3 × 5), even though the stated concrete grade is M20. Use b = 300 mm and d = 550 mm. The tension area is A_st = 4π(16)²/4 = 804.25 mm². The permissible stresses are 5 MPa in concrete and 140 MPa in steel; m = 18.6667. No ultimate load factor is applied in this working-stress analysis.

2. Neutral axis. From equal first moments of the cracked transformed section, bx²/2 = mA_st(d − x). The positive root gives x = 189.855 mm. The critical depth is X_c = dmσ_cbc/(mσ_cbc + σ_st) = 220.000 mm. The lever arm is z = d − x/3 = 486.715 mm.

3. Independent permissible moments. Concrete permits M_c = 0.5σ_cbcbxz = 69.304 kN·m; steel permits M_s = σ_stA_stz = 54.802 kN·m. Therefore M_R = min(M_c, M_s) = 54.802 kN·m.

4. Stresses at the applied moment. With M = 50.000 kN·m, f_c = 2M/(bxz) = 3.607 MPa, and f_st = M/(A_stz) = 127.734 MPa. Use M in N·mm in the stress equations.

Conclusion: the stated service moment satisfies both permissible-stress limits. This is a working-stress conclusion, not a claim of physical ultimate collapse or completion of the shear/anchorage checks.

Q5. Determine the area of steel required and the moment of resistance for a RC beam 250 mm × 450 mm. Consider M15 concrete and mild steel. Use the working stress method. [8] (2080 Bhadra, Q1b)

No applied moment or cover is specified in this question, so the conventional study interpretation is a balanced section. If 450 mm is overall depth, assume an effective cover of 50 mm: d = 400 mm, m = 18.667, X_c = 0.4d = 160 mm. Then A_st,bal = 0.5 × 5 × 250 × 160/140 = 714.29 mm², and M_bal = 34.67 kN·m. If the examiner instead intends d = 450 mm, the corresponding results are 803.57 mm² and 43.875 kN·m. State the interpretation rather than presenting one of these as uniquely determined.

Q6. An RCC beam 25 cm wide and 60 cm deep has 4 bars of 20 mm diameter as tension reinforcement, the centre of bars being 5 cm above the bottom. Determine the uniformly distributed load the beam can carry over a simply supported effective span of 6.10 m. Permissible stresses in concrete and steel are 7 MPa and 230 MPa respectively; use the modular ratio. [10] (2078 Bhadra, Q1b)

1. Geometry and working-stress data. Convert 25 cm to 250 mm. The overall depth is 600 mm and the bar centre is 50 mm above the bottom, giving d = 550 mm. Use b = 250 mm and d = 550 mm. The tension area is A_st = 4π(20)²/4 = 1256.64 mm². The permissible stresses are 7 MPa in concrete and 230 MPa in steel; m = 13.3333. No ultimate load factor is applied in this working-stress analysis.

2. Neutral axis. From equal first moments of the cracked transformed section, bx²/2 = mA_st(d − x). The positive root gives x = 212.648 mm. The critical depth is X_c = dmσ_cbc/(mσ_cbc + σ_st) = 158.763 mm. The lever arm is z = d − x/3 = 479.117 mm.

3. Independent permissible moments. Concrete permits M_c = 0.5σ_cbcbxz = 89.148 kN·m; steel permits M_s = σ_stA_stz = 138.478 kN·m. Therefore M_R = min(M_c, M_s) = 89.148 kN·m.

4. Stresses at the permissible moment of resistance. With M = 89.148 kN·m, f_c = 2M/(bxz) = 7.000 MPa, and f_st = M/(A_stz) = 148.067 MPa. Use M in N·mm in the stress equations.

5. Allowable uniformly distributed load. For the stated simply supported span, w_total = 8M_R/L² = 8(89.14788)/6.1² = 19.166 kN/m. This is total service load, including self-weight; subtract the actual beam self-weight only when a superimposed-load capacity is requested.

Q7. Calculate the tensile reinforcement required for a rectangular RC beam of size 230 mm × 425 mm (overall) if it has to carry a moment of 64 kNm at service condition. Use M20 concrete and Fe500 steel in the working stress method. [8] (2076 Chaitra, Q1b)

The overall size is 230 × 425 mm, but cover is unspecified. Adopt an effective cover of 45 mm for this study check, giving d = 380 mm. For M20/Fe500, take σ_cbc = 7 MPa, σ_st = 275 MPa and m = 13.333. The previously quoted area of about 648 mm² is not a valid completed design: satisfying a steel-force equation does not establish acceptable concrete stress.

Even at A_st = 0.04bD = 3910 mm², the ordinary maximum beam tension-steel area, elastic analysis gives x = 246.244 mm and a concrete-governed capacity of only 59.06 kN·m. At 64 kN·m the concrete stress would be 7.586 MPa > 7 MPa. Thus this assumed singly reinforced section is inadequate; increasing steel alone within that limit cannot solve it. A deeper section or a properly designed doubly reinforced section is required. Cover, bar fit and the precise interpretation of the problem must be stated before revising the design.

Q8. A rectangular reinforced-concrete beam has overall dimensions of 300 mm × 500 mm and is reinforced with 4 tension bars of 25 mm diameter at an effective cover of 50 mm. The beam is simply supported over an effective span of 5 m. Using the working stress method, calculate the total uniformly distributed load that the beam can carry, including its self-weight. Adopt M20 concrete and Fe415 (TOR) steel. [8] (2076 Ashwin, Q1a)

1. Geometry and working-stress data. The supplied effective cover is 50 mm, so d = 500 − 50 = 450 mm; do not subtract the link diameter again. Use b = 300 mm and d = 450 mm. The tension area is A_st = 4π(25)²/4 = 1963.50 mm². The permissible stresses are 7 MPa in concrete and 230 MPa in steel; m = 13.3333. No ultimate load factor is applied in this working-stress analysis.

2. Neutral axis. From equal first moments of the cracked transformed section, bx²/2 = mA_st(d − x). The positive root gives x = 206.256 mm. The critical depth is X_c = dmσ_cbc/(mσ_cbc + σ_st) = 129.897 mm. The lever arm is z = d − x/3 = 381.248 mm.

3. Independent permissible moments. Concrete permits M_c = 0.5σ_cbcbxz = 82.566 kN·m; steel permits M_s = σ_stA_stz = 172.173 kN·m. Therefore M_R = min(M_c, M_s) = 82.566 kN·m.

4. Stresses at the permissible moment of resistance. With M = 82.566 kN·m, f_c = 2M/(bxz) = 7.000 MPa, and f_st = M/(A_stz) = 110.297 MPa. Use M in N·mm in the stress equations.

5. Allowable uniformly distributed load. For the stated simply supported span, w_total = 8M_R/L² = 8(82.56632)/5² = 26.421 kN/m. This is total service load, including self-weight; subtract the actual beam self-weight only when a superimposed-load capacity is requested.

Q9. A rectangular reinforced-concrete beam has overall dimensions of 250 mm × 450 mm. It is reinforced with 4 tension bars of 16 mm diameter at an effective cover of 40 mm. Calculate the moment of resistance of the beam using the working stress method. Adopt M20 concrete and Fe415 steel. [8] (2075 Chaitra, Q1c)

1. Geometry and working-stress data. The supplied effective cover is 40 mm, giving d = 450 − 40 = 410 mm. Use b = 250 mm and d = 410 mm. The tension area is A_st = 4π(16)²/4 = 804.25 mm². The permissible stresses are 7 MPa in concrete and 230 MPa in steel; m = 13.3333. No ultimate load factor is applied in this working-stress analysis.

2. Neutral axis. From equal first moments of the cracked transformed section, bx²/2 = mA_st(d − x). The positive root gives x = 149.492 mm. The critical depth is X_c = dmσ_cbc/(mσ_cbc + σ_st) = 118.351 mm. The lever arm is z = d − x/3 = 360.169 mm.

3. Independent permissible moments. Concrete permits M_c = 0.5σ_cbcbxz = 47.112 kN·m; steel permits M_s = σ_stA_stz = 66.623 kN·m. Therefore M_R = min(M_c, M_s) = 47.112 kN·m.

4. Stresses at the permissible moment of resistance. With M = 47.112 kN·m, f_c = 2M/(bxz) = 7.000 MPa, and f_st = M/(A_stz) = 162.644 MPa. Use M in N·mm in the stress equations.

Q10. A rectangular reinforced-concrete beam has overall dimensions of 230 mm × 350 mm and is reinforced with 4 bars of 16 mm diameter in its bottom tension zone. Determine the moment of resistance of the beam if the permissible stresses in concrete and steel must not exceed 7.0 MPa and 140 MPa, respectively. Take a nominal cover of 25 mm and a modular ratio of 13.33. [7] (2075 Ashwin, Q1a)

1. Geometry and working-stress data. Assume 8 mm links because their diameter is not supplied. With 25 mm nominal cover, d = 350 − 25 − 8 − 16/2 = 309 mm. Another link diameter changes this answer. Use b = 230 mm and d = 309 mm. The tension area is A_st = 4π(16)²/4 = 804.25 mm². The permissible stresses are 7 MPa in concrete and 140 MPa in steel; m = 13.3300. No ultimate load factor is applied in this working-stress analysis.

2. Neutral axis. From equal first moments of the cracked transformed section, bx²/2 = mA_st(d − x). The positive root gives x = 129.396 mm. The critical depth is X_c = dmσ_cbc/(mσ_cbc + σ_st) = 123.581 mm. The lever arm is z = d − x/3 = 265.868 mm.

3. Independent permissible moments. Concrete permits M_c = 0.5σ_cbcbxz = 27.694 kN·m; steel permits M_s = σ_stA_stz = 29.935 kN·m. Therefore M_R = min(M_c, M_s) = 27.694 kN·m.

4. Stresses at the permissible moment of resistance. With M = 27.694 kN·m, f_c = 2M/(bxz) = 7.000 MPa, and f_st = M/(A_stz) = 129.517 MPa. Use M in N·mm in the stress equations.

Q11. Find the moment of resistance of a reinforced-concrete beam that is 250 mm wide, has an effective depth of 500 mm and is reinforced with 3 bars of 16 mm diameter. The permissible stresses in concrete and steel are 7 MPa and 230 MPa, respectively. Take the modular ratio as 13.33. [6] (2073 Shrawan, Q1a)

1. Geometry and working-stress data. Both the effective depth and modular ratio are supplied. Use b = 250 mm and d = 500 mm. The tension area is A_st = 3π(16)²/4 = 603.19 mm². The permissible stresses are 7 MPa in concrete and 230 MPa in steel; m = 13.3300. No ultimate load factor is applied in this working-stress analysis.

2. Neutral axis. From equal first moments of the cracked transformed section, bx²/2 = mA_st(d − x). The positive root gives x = 150.037 mm. The critical depth is X_c = dmσ_cbc/(mσ_cbc + σ_st) = 144.304 mm. The lever arm is z = d − x/3 = 449.988 mm.

3. Independent permissible moments. Concrete permits M_c = 0.5σ_cbcbxz = 59.075 kN·m; steel permits M_s = σ_stA_stz = 62.428 kN·m. Therefore M_R = min(M_c, M_s) = 59.075 kN·m.

4. Stresses at the permissible moment of resistance. With M = 59.075 kN·m, f_c = 2M/(bxz) = 7.000 MPa, and f_st = M/(A_stz) = 217.648 MPa. Use M in N·mm in the stress equations.

Q12. A reinforced concrete beam of size 300 mm × 640 mm overall is reinforced with 4 bars of 20 mm diameter at the bottom. The beam carries a superimposed load of 50 kN/m, excluding its self-weight, over a span of 4 m. Find the actual stresses developed in steel and concrete. Take effective cover of 40 mm and modular ratio of 13.33. Also calculate the compressive stress in concrete at 50 mm from the top of the beam. [10] (2082 Baishakh, Q1c)

Use b = 300 mm, D = 640 mm, d = 600 mm, A_st = 4π(20)²/4 = 1256.64 mm², m = 13.33. Assume a simply supported beam and concrete unit weight 25 kN/m³. Self-weight = 0.300 × 0.640 × 25 = 4.80 kN/m; total service load = 54.80 kN/m; M = 54.80 × 4²/8 = 109.60 kN·m.

The neutral-axis equation gives x = 208.968 mm, and z = 530.344 mm. Therefore f_c,top = 2 × 109.60 × 10⁶/(300 × 208.968 × 530.344) = 6.593 MPa, while f_st = 109.60 × 10⁶/(1256.64 × 530.344) = 164.454 MPa.

Because compression stress decreases linearly from the top to zero at x, the stress 50 mm below the top is f_c,50 = 6.593(208.968 − 50)/208.968 = 5.015 MPa. The requested output is actual stress; a safety verdict would additionally need the applicable permissible limits.

Q13. A simply supported rectangular RC beam of effective span 4.2 m and overall dimensions 230 mm × 450 mm is reinforced with four 20 mm diameter bars in tension. Determine the moment of resistance. Take permissible stresses for M20 concrete and Fe415 steel. [6] (2074 Ashwin, Q1b)

1. Geometry and working-stress data. The paper omits cover. Adopt an explicitly assumed effective cover of 45 mm, giving d = 450 − 45 = 405 mm. Use the M20/Fe415 permissible bending stresses. Use b = 230 mm and d = 405 mm. The tension area is A_st = 4π(20)²/4 = 1256.64 mm². The permissible stresses are 7 MPa in concrete and 230 MPa in steel; m = 13.3333. No ultimate load factor is applied in this working-stress analysis.

2. Neutral axis. From equal first moments of the cracked transformed section, bx²/2 = mA_st(d − x). The positive root gives x = 180.754 mm. The critical depth is X_c = dmσ_cbc/(mσ_cbc + σ_st) = 116.907 mm. The lever arm is z = d − x/3 = 344.749 mm.

3. Independent permissible moments. Concrete permits M_c = 0.5σ_cbcbxz = 50.163 kN·m; steel permits M_s = σ_stA_stz = 99.642 kN·m. Therefore M_R = min(M_c, M_s) = 50.163 kN·m.

4. Stresses at the permissible moment of resistance. With M = 50.163 kN·m, f_c = 2M/(bxz) = 7.000 MPa, and f_st = M/(A_stz) = 115.791 MPa. Use M in N·mm in the stress equations.

Q14. The moment of resistance of a rectangular reinforced-concrete beam of width b and effective depth d is 0.85bd², using N and mm. The extreme concrete and steel stresses must not exceed 7 N/mm² and 140 N/mm², respectively, and the modular ratio is 18.33. Determine the ratio of neutral-axis depth from the compression face to effective depth. The beam is reinforced on the tension side only. [12] (2067 Ashadh, Q1b)

1. Write the dimensionless moment equation. Let k = x/d. Since M = 0.85bd² and the concrete compression diagram is triangular, 0.85 = 0.5f_ck(1 − k/3). Compatibility also gives f_st = mf_c(1 − k)/k.

2. Find the controlling stress. Taking f_c = 7 MPa in the first equation would give a shallow neutral axis and a steel stress greater than 140 MPa. That root cannot be the permissible moment of resistance. Instead set f_st = 140 MPa, and substitute f_c = 140k/[18.33(1 − k)].

0.85 = [140/(2 × 18.33)]k²(1 − k/3)/(1 − k)

3. Solve and verify. The physical root between zero and one is x/d = 0.39404. The accompanying concrete stress is 4.9666 MPa, less than 7 MPa, while the steel is at 140 MPa. Thus steel governs this working-stress resistance. The statement must not be interpreted as requiring both permissible stresses to occur simultaneously, because that would conflict with the supplied moment coefficient.