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A floor slab of thickness t, is cast monolithically transverse to a rectangular continuous beam of span, L and width, B. If the distance between two consecutive points of contraflexure is, L0, the effective width of compression flange at a continuous support is (GATE 2012)
- a)B
- b)L/3
- c)B + 12 t
- d)B + 6t + L0/6
Correct answer is option 'D'. Can you explain this answer?
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A floor slab of thickness t, is cast monolithically transverse to a re...
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Option (d)
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A floor slab of thickness t, is cast monolithically transverse to a re...
Given data:
Thickness of floor slab = t
Span of the continuous beam = L
Width of the continuous beam = B
Distance between two consecutive points of contraflexure = L0
To find: Effective width of compression flange at a continuous support
Solution:
The effective width of compression flange at a continuous support can be found using the following formula:
beff = (2/3)B + (L0/6) - t
where,
beff = Effective width of compression flange at a continuous support
B = Width of the continuous beam
L0 = Distance between two consecutive points of contraflexure
t = Thickness of floor slab
Substituting the given values in the above formula, we get:
beff = (2/3)B + (L0/6) - t
beff = (2/3)B + (L/2L0)(L0/6) - t
beff = (2/3)B + (L/12) - t
Multiplying both sides by 6, we get:
6beff = 4B + L - 12t
Dividing both sides by L, we get:
(6beff/L) = (4B/L) + 1 - (12t/L)
Since the beam is continuous, the maximum bending moment occurs at the mid-span of the beam. Therefore, the effective width of compression flange at a continuous support can be found at mid-span.
At mid-span, the bending moment equation can be written as:
M = (wL^2)/8
where,
M = Maximum bending moment
w = Load per unit length of the beam
The load per unit length of the beam can be assumed as the self-weight of the beam and the floor slab. Therefore,
w = (γc + γs)t
where,
γc = Unit weight of concrete
γs = Unit weight of floor slab
Substituting the given values, we get:
w = (25 + 20)t
w = 45t
Substituting the values of B, L, and t in the above equation, we get:
(6beff/L) = (4B/L) + 1 - (12t/L)
(6beff/L) = (4B/L) + 1 - (2.6667t)
(6beff/L) = (4/3) + 1 - (2.6667t/L)
(6beff/L) = (7/3) - (2.6667t/L)
beff = (L/6) [(7/3) - (2.6667t/L)]
At mid-span, the effective width of compression flange is maximum. Therefore, substituting L/2 in the above equation, we get:
beff = (L/6) [(7/3) - (2.6667t/(L/2))]
beff = (L/6) [(7/3) - (5.3333t/L)]
beff = (L/6) [(21 - 16t/L)/9]
beff = (L/54) [21L - 16t]
Substituting the given values, we get:
beff = (L/54) [21L - 16t]
beff = (9
Thickness of floor slab = t
Span of the continuous beam = L
Width of the continuous beam = B
Distance between two consecutive points of contraflexure = L0
To find: Effective width of compression flange at a continuous support
Solution:
The effective width of compression flange at a continuous support can be found using the following formula:
beff = (2/3)B + (L0/6) - t
where,
beff = Effective width of compression flange at a continuous support
B = Width of the continuous beam
L0 = Distance between two consecutive points of contraflexure
t = Thickness of floor slab
Substituting the given values in the above formula, we get:
beff = (2/3)B + (L0/6) - t
beff = (2/3)B + (L/2L0)(L0/6) - t
beff = (2/3)B + (L/12) - t
Multiplying both sides by 6, we get:
6beff = 4B + L - 12t
Dividing both sides by L, we get:
(6beff/L) = (4B/L) + 1 - (12t/L)
Since the beam is continuous, the maximum bending moment occurs at the mid-span of the beam. Therefore, the effective width of compression flange at a continuous support can be found at mid-span.
At mid-span, the bending moment equation can be written as:
M = (wL^2)/8
where,
M = Maximum bending moment
w = Load per unit length of the beam
The load per unit length of the beam can be assumed as the self-weight of the beam and the floor slab. Therefore,
w = (γc + γs)t
where,
γc = Unit weight of concrete
γs = Unit weight of floor slab
Substituting the given values, we get:
w = (25 + 20)t
w = 45t
Substituting the values of B, L, and t in the above equation, we get:
(6beff/L) = (4B/L) + 1 - (12t/L)
(6beff/L) = (4B/L) + 1 - (2.6667t)
(6beff/L) = (4/3) + 1 - (2.6667t/L)
(6beff/L) = (7/3) - (2.6667t/L)
beff = (L/6) [(7/3) - (2.6667t/L)]
At mid-span, the effective width of compression flange is maximum. Therefore, substituting L/2 in the above equation, we get:
beff = (L/6) [(7/3) - (2.6667t/(L/2))]
beff = (L/6) [(7/3) - (5.3333t/L)]
beff = (L/6) [(21 - 16t/L)/9]
beff = (L/54) [21L - 16t]
Substituting the given values, we get:
beff = (L/54) [21L - 16t]
beff = (9
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A floor slab of thickness t, is cast monolithically transverse to a rectangular continuous beam of span, L and width, B. If the distance between two consecutive points of contraflexure is, L0, the effective width of compression flange at a continuous support is (GATE 2012)a)Bb)L/3c)B + 12 td)B + 6t + L0/6Correct answer is option 'D'. Can you explain this answer?
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