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RCD-MODULE-03 (1)

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MODULE 03
Stage 2: Concrete Cracked – Elastic Stress Stage
When the load is increased, the tensile stress in the bottom of the beam becomes equal to the modulus of rupture, the
cracks then starts to develop. The moment at which these cracks begin to form is referred to as the cracking moment,
𝑴𝒄𝒓
This stage will continue as long as the compression stress in the top fibers is less than about one-half of the
concrete’s compression strength, 𝑓 ′ 𝑐, and as long as the steel stress is less than the yield stress.
𝒇 = 𝒇𝒓
When the bending moment is sufficiently large to cause the tensile stress in the extreme fibers to be
greater than the modulus of rupture of concrete, it is assumed that all of the concrete on the tensile side if
the beam is cracked and must be neglected in the flexure calculations.
Modular Ratio, n
ο‚·
The ratio of the steel modulus to the concrete modulus
𝒏=
ο‚·
𝑬𝒔
𝑬𝒄
Modular ratio is used to convert the area of steel with an equivalent area of concrete
Transformed-Area Method
a. Singly Reinforced Beam
1.
2.
3.
4.
Transform the steel section
Locate neutral axis
Compute moment of inertia
Compute the bending stress
b. Doubly Reinforced Beam
Double reinforced beam is a beam that has compression steel as well as tensile steel.
Compression steel is generally thought to be uneconomical, but occasionally its use is quite
advantageous. Compression steel will permit the use of appreciable smaller beams than those
that make use of tensile steel only. Reduced sizes can be very important where space or
architectural requirements limit the sizes of the beams
As a consequence if creep in the concrete, the stresses in the compression bars computed by
the transformed area method are assumed to double as time goes by.
The transformed area of the compression side equals the gross compression area of the
concrete plus 2nA’s minus the area of the holes in the concrete (1A’s), which theoretically
should not have been included in the concrete part
Sample Problem 1
a. Compute the bending stress in the beam, in MPa, shown in the figure by using the transformedarea method, f’c = 20.7 MPa, n = 9 and M = 94,90 kN-m
b. Determine the allowable resisting moment of the beam, in kN-m, if the allowable stress are fc =
9.30 MPa and fs = 137.90 MPa.
πœ‹π‘‘2
) (3)
4
πœ‹(28)2
𝐴𝑠 = (
) (3)
4
𝑨𝒔 = πŸπŸ–πŸ’πŸ•. πŸπŸ” π’Žπ’ŽπŸ
𝐴𝑠 = (
𝒏𝑨𝒔 = πŸπŸ”, πŸ”πŸπŸ“. πŸ‘πŸ π’Žπ’ŽπŸ
π‘₯
𝐴𝑐 ( ) = 𝑛𝐴𝑠 (435 − π‘₯)
2
π‘₯
305(π‘₯) ( ) = (16,625.31)(435 − π‘₯)
2
𝒙 = πŸπŸ”πŸ—. πŸ—πŸ– π’Žπ’Ž
𝐼𝑔 = 𝐼𝑔𝑐 + 𝐼𝑔𝑠
𝐼𝑔𝑐 =
π‘β„Ž3
3
(305)(169.98)3
3
π‘°π’ˆπ’„ = πŸ’πŸ—πŸ—. πŸ‘πŸπŸ π’™πŸπŸŽπŸ”π’Žπ’ŽπŸ’
𝐼𝑔𝑐 =
𝐼𝑔𝑠 = 𝐴𝑑2
𝐼𝑔𝑠 = 16,625.31 (265.02)2
π‘°π’ˆπ’” = 𝟏, πŸπŸ”πŸ•. πŸ”πŸ–πŸ— π’™πŸπŸŽπŸ”π’Žπ’ŽπŸ’
π‘°π’ˆ = 𝟏, πŸ”πŸ”πŸ•. 𝟎𝟎𝟏 π’™πŸπŸŽπŸ” π’Žπ’ŽπŸ’
Bending Stress
𝑀𝑦
𝐼𝑔
(94.90π‘₯106 𝑁. π‘šπ‘š)(169.98π‘šπ‘š)
𝑓𝑐 =
1,667.001π‘₯106 π‘šπ‘š4
𝒇𝒄 = πŸ—. πŸ”πŸ•πŸ• 𝑴𝒑𝒂
𝑓𝑐 =
𝑛𝑀𝑦
𝐼𝑔
(9)(94.90π‘₯106 𝑁. π‘šπ‘š)(265.02π‘šπ‘š)
𝑓𝑠 =
1,667.001π‘₯106 π‘šπ‘š4
𝒇𝒔 = πŸπŸ‘πŸ“. πŸ•πŸ– 𝑴𝑷𝒂
𝑓𝑠 =
Allowable Resisting Moment
𝑀𝑦
𝐼𝑔
𝑀
𝑐 (169.98π‘šπ‘š)
9.30 𝑁/π‘šπ‘š2 =
1,667.001π‘₯106 π‘šπ‘š4
𝑴𝒄 = πŸ—πŸ. 𝟐𝟏 π’Œπ‘΅. π’Ž
𝑓𝑐 =
𝑛𝑀𝑦
𝐼𝑔
(9)(𝑀
𝑠 )(265.02π‘šπ‘š)
137.91 𝑁/π‘šπ‘š2 =
1,667.001π‘₯106 π‘šπ‘š4
𝑴𝒔 = πŸ—πŸ”. πŸ‘πŸ– π’Œπ‘΅. π’Ž
𝑓𝑠 =
π‘¨π’π’π’π’˜π’‚π’ƒπ’π’† π‘Ήπ’†π’”π’Šπ’”π’•π’Šπ’π’ˆ π‘΄π’π’Žπ’†π’π’•, 𝑴 = πŸ—πŸ. 𝟐𝟏 π’Œπ‘΅. π’Ž
Sample Problem 2
a. Compute the bending stress, in psi, in the beam shown in th figure. Use n = 10 and M = 118 kipft
𝑨′𝒔 = 𝟐. 𝟎𝟎 π’Šπ’πŸ
𝑨𝒔 = πŸ’. 𝟎𝟎 π’Šπ’πŸ
(2𝑛 − 1)𝐴′𝑠 = 38.00 𝑖𝑛2
𝑛𝐴𝑠 = 40.00 𝑖𝑛2
π‘π‘’π‘’π‘‘π‘Ÿπ‘Žπ‘™ 𝐴π‘₯𝑖𝑠, π‘₯
π‘₯
𝐴𝑐 ( ) + (2𝑛 − 1)𝐴′𝑠 (π‘₯ − 2.5) = 𝑛𝐴𝑠 (17.5 − π‘₯)
2
π‘₯
14 (π‘₯) ( ) + 38(π‘₯ − 2.5) = 40(17.5 − π‘₯)
2
𝒙 = πŸ”. πŸ’πŸ“πŸ’ π’Šπ’
π‘€π‘œπ‘šπ‘’π‘›π‘‘ π‘œπ‘“ πΌπ‘›π‘’π‘Ÿπ‘‘π‘–π‘Ž
′ +𝐼
𝐼𝑔 = 𝐼𝑔𝑐 + 𝐼𝑔𝑠
𝑔𝑠
𝐼𝑔𝑐 =
π‘β„Ž3 (14)(6.454)3
=
= 1254.60 𝑖𝑛4
3
3
𝐼𝑔𝑠′ = 𝐴𝑑2 = 38(3.954)2 = 594.11 𝑖𝑛4
𝐼𝑔𝑠 = 𝐴𝑑2 = 40 (11.046)2 = 4880.56 𝑖𝑛4
π‘°π’ˆ = πŸ”πŸ•πŸπŸ—. πŸπŸ• π’Šπ’πŸ’
Bending Stress
𝑀𝑦
𝐼𝑔
1000𝑙𝑏 12𝑖𝑛
𝑀 = 118 π‘˜π‘–π‘. 𝑓𝑑 π‘₯
π‘₯
= 1,416,000 𝑙𝑏 − 𝑖𝑛
1π‘˜π‘–π‘
1𝑓𝑑
𝑓𝑐 =
(1416000)(6.454)
6729.27
𝒇𝒄 = 𝟏, πŸ‘πŸ“πŸ–. πŸŽπŸ– π’‘π’”π’Š
𝑓𝑐 =
2𝑛𝑀𝑦
𝐼𝑔
2(10)(1416000)(3.954)
137.91 𝑁/π‘šπ‘š2 =
6729.27
𝒇𝒔′ = πŸπŸ”, πŸ”πŸ’πŸŽ. πŸ‘πŸ‘ π’‘π’”π’Š
𝑛𝑀𝑦
𝑓𝑠 =
𝐼𝑔
(10)(1,416,000)(11.046π‘šπ‘š)
𝑓𝑠 =
6729.27
𝒇𝒔 = πŸπŸ‘, πŸπŸ’πŸ‘. πŸ’πŸ’ π’‘π’”π’Š
𝑓𝑠′ =
Exercise Problem
1. A reinforced concrete beam 300 mm wide has an effective depth of 600 mm. It is
reinforced with 4-32 mm diameter bars for tension. f’c = 21 MPa and fy = 275 MPa. Find
the moment capacity of the beam.
2. A 300 mm × 600 mm reinforced concrete beam section is reinforced with 4 - 28-mmdiameter tension steel at d = 536 mm and 2 - 28-mm-diameter compression steel at d' =
64 mm. The section is subjected to a bending moment of 150 kN·m. Use n = 9.
a) Find the maximum stress in concrete.
b) Determine the stress in the compression steel.
c) Calculate the stress in the tension steel.
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