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An RCC beam of rectangular cross-section has factored shear of 200 kN at its critical section. Its width b is 250 mm and effective depth d is 350 mm. Assume design shear strength tc of concrete as 0.62 N/mm2 and maximum allowable shear stress tc max in concrete as 2.8 N/mm2 . If two legged 10 mm diameter vertical stirrups of Fe250 grade steel are used, then the required spacing (in cm, up to one decimal place) as per unit state method will be _____________
Correct answer is '(8.2)'. Can you explain this answer?
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Verified Answer
An RCC beam of rectangular cross-section has factored shear of 200 kN ...
Nominal shear stress 
2.286 N/mm2 < Tc max (OK)
SF taken by stirrups = (tv - tc ) bd
= 2.286 − 0.62) x 250 x 350 = 145.75 kN
Now,
2.286 N/mm2 < Tc max (OK)
SF taken by stirrups = (tv - tc ) bd
= 2.286 − 0.62) x 250 x 350 = 145.75 kN
Now,
Most Upvoted Answer
An RCC beam of rectangular cross-section has factored shear of 200 kN ...
To determine the required spacing of the stirrups in an RCC beam, we will use the unit shear method. This method is based on the principle that the total shear force in the beam is divided by the contribution of each stirrup to the shear resistance.
Given data:
- Factored shear at critical section = 200 kN
- Width of the beam (b) = 250 mm
- Effective depth (d) = 350 mm
- Design shear strength of concrete (tc) = 0.62 N/mm²
- Maximum allowable shear stress in concrete (tc max) = 2.8 N/mm²
- Diameter of vertical stirrups (dsv) = 10 mm
- Grade of steel (Fe250)
1. Calculate the area of steel required:
The area of steel required can be calculated using the formula:
Ast = (Vc / (0.87 * fy)) * s
where,
Ast = Area of steel required
Vc = Design shear force
fy = Yield strength of steel
s = Spacing of stirrups
2. Calculate the design shear force (Vc):
Vc = tc * b * d
where,
tc = Design shear strength of concrete
3. Calculate the yield strength of steel (fy):
The yield strength of Fe250 grade steel is 250 N/mm².
4. Substitute the given values into the formulas:
Vc = 0.62 N/mm² * 250 mm * 350 mm = 54.25 kN
fy = 250 N/mm²
5. Calculate the area of steel required:
Ast = (54.25 kN / (0.87 * 250 N/mm²)) * s
6. Rearrange the equation to solve for s:
s = (Ast * 0.87 * fy) / 54.25 kN
7. Substitute the values:
Ast = (π * (10 mm)² / 4) * 2 = 157.08 mm²
s = (157.08 mm² * 0.87 * 250 N/mm²) / 54.25 kN
8. Convert the units:
s = (157.08 mm² * 0.87 * 250 N/mm²) / 54.25 kN * 100 cm/m = 8.2 cm (rounded to one decimal place)
Therefore, the required spacing of the stirrups using the unit state method is 8.2 cm.
Given data:
- Factored shear at critical section = 200 kN
- Width of the beam (b) = 250 mm
- Effective depth (d) = 350 mm
- Design shear strength of concrete (tc) = 0.62 N/mm²
- Maximum allowable shear stress in concrete (tc max) = 2.8 N/mm²
- Diameter of vertical stirrups (dsv) = 10 mm
- Grade of steel (Fe250)
1. Calculate the area of steel required:
The area of steel required can be calculated using the formula:
Ast = (Vc / (0.87 * fy)) * s
where,
Ast = Area of steel required
Vc = Design shear force
fy = Yield strength of steel
s = Spacing of stirrups
2. Calculate the design shear force (Vc):
Vc = tc * b * d
where,
tc = Design shear strength of concrete
3. Calculate the yield strength of steel (fy):
The yield strength of Fe250 grade steel is 250 N/mm².
4. Substitute the given values into the formulas:
Vc = 0.62 N/mm² * 250 mm * 350 mm = 54.25 kN
fy = 250 N/mm²
5. Calculate the area of steel required:
Ast = (54.25 kN / (0.87 * 250 N/mm²)) * s
6. Rearrange the equation to solve for s:
s = (Ast * 0.87 * fy) / 54.25 kN
7. Substitute the values:
Ast = (π * (10 mm)² / 4) * 2 = 157.08 mm²
s = (157.08 mm² * 0.87 * 250 N/mm²) / 54.25 kN
8. Convert the units:
s = (157.08 mm² * 0.87 * 250 N/mm²) / 54.25 kN * 100 cm/m = 8.2 cm (rounded to one decimal place)
Therefore, the required spacing of the stirrups using the unit state method is 8.2 cm.
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An RCC beam of rectangular cross-section has factored shear of 200 kN at its critical section. Its width b is 250 mm and effective depth d is 350 mm. Assume design shear strengthtcof concrete as 0.62 N/mm2 and maximum allowable shear stress tc maxin concrete as 2.8 N/mm2 . If two legged 10 mm diameter vertical stirrups of Fe250 grade steel are used, then the required spacing (in cm, up to one decimal place) as per unit state method will be _____________Correct answer is '(8.2)'. Can you explain this answer?
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An RCC beam of rectangular cross-section has factored shear of 200 kN at its critical section. Its width b is 250 mm and effective depth d is 350 mm. Assume design shear strengthtcof concrete as 0.62 N/mm2 and maximum allowable shear stress tc maxin concrete as 2.8 N/mm2 . If two legged 10 mm diameter vertical stirrups of Fe250 grade steel are used, then the required spacing (in cm, up to one decimal place) as per unit state method will be _____________Correct answer is '(8.2)'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about An RCC beam of rectangular cross-section has factored shear of 200 kN at its critical section. Its width b is 250 mm and effective depth d is 350 mm. Assume design shear strengthtcof concrete as 0.62 N/mm2 and maximum allowable shear stress tc maxin concrete as 2.8 N/mm2 . If two legged 10 mm diameter vertical stirrups of Fe250 grade steel are used, then the required spacing (in cm, up to one decimal place) as per unit state method will be _____________Correct answer is '(8.2)'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An RCC beam of rectangular cross-section has factored shear of 200 kN at its critical section. Its width b is 250 mm and effective depth d is 350 mm. Assume design shear strengthtcof concrete as 0.62 N/mm2 and maximum allowable shear stress tc maxin concrete as 2.8 N/mm2 . If two legged 10 mm diameter vertical stirrups of Fe250 grade steel are used, then the required spacing (in cm, up to one decimal place) as per unit state method will be _____________Correct answer is '(8.2)'. Can you explain this answer?.
An RCC beam of rectangular cross-section has factored shear of 200 kN at its critical section. Its width b is 250 mm and effective depth d is 350 mm. Assume design shear strengthtcof concrete as 0.62 N/mm2 and maximum allowable shear stress tc maxin concrete as 2.8 N/mm2 . If two legged 10 mm diameter vertical stirrups of Fe250 grade steel are used, then the required spacing (in cm, up to one decimal place) as per unit state method will be _____________Correct answer is '(8.2)'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about An RCC beam of rectangular cross-section has factored shear of 200 kN at its critical section. Its width b is 250 mm and effective depth d is 350 mm. Assume design shear strengthtcof concrete as 0.62 N/mm2 and maximum allowable shear stress tc maxin concrete as 2.8 N/mm2 . If two legged 10 mm diameter vertical stirrups of Fe250 grade steel are used, then the required spacing (in cm, up to one decimal place) as per unit state method will be _____________Correct answer is '(8.2)'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An RCC beam of rectangular cross-section has factored shear of 200 kN at its critical section. Its width b is 250 mm and effective depth d is 350 mm. Assume design shear strengthtcof concrete as 0.62 N/mm2 and maximum allowable shear stress tc maxin concrete as 2.8 N/mm2 . If two legged 10 mm diameter vertical stirrups of Fe250 grade steel are used, then the required spacing (in cm, up to one decimal place) as per unit state method will be _____________Correct answer is '(8.2)'. Can you explain this answer?.
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