Unit 4 ( Design Of Rectangular Beam Sections ) 3w1l1i
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UNIT 4 DESIGN OF RECTANGULAR BEAM SECTIONS
OBJECTIVES
GENERAL OBJECTIVE To be able to provide steel reinforcement required in reinforced concrete beams. SPECIFIC OBJECTIVES
At the end of this unit, you should be able to: 1.
calculate the area and total number of reinforcement required in given beam sections with known bending moments.
2.
calculate the bending moment that can be carried by the beam section when the grade of steel is already provided.
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INPUT 1
4.1
Simply Reinforced Rectangular Section What does it mean? Simply reinforced section is the steel reinforcement that is provided only at the tension zone. The form of section, stress distribution and forces on a section are as shown in Figure 4.1: 0.45fcu
b
0.45x 0.9x Fcc=0.405fcubx
x d
Z=d-0.45x
As 0.87 fy
Section
Stress
Fst =0.87fyAs
Forces
Figure 4.1: The form of section, stress distribution and forces
on a section
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The meaning of symbols used are given in clause 3.4.4.3 For the section to be in balance, the sum of the forces must be equal to zero, i.e. Fst = Fcc 0.87fyAs = 0.405fcubx
∴
0.87 f y As
x = 0.405 f cu b
.......... .......... ...... equation
(4.1)
Taking moments about Fcc or Fst, the moment of resistance of the section can be calculated as follows, M = Fst. Z = (0.87fyAs) (d - 0.45x) or, M = Fcc. Z = (0.405fcubx) (d – 0.45 x) x 0.45 x 1− d = (0.405 d ) ( ) (fcubd2)
= K. fcu bd2 A section whereby the reinforcement is already known, the above equation can be used to calculate the moment of resistance or the bending moment. Equation 4.1 shows that, x will increase if the area of reinforcement, As increases. The moment of resistance of the section will also increase. It is clear that a section provided with more reinforcement will be able to with stand reinforcement load i.e. greater bending moment.
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In order to make sure that concrete will not crush or fail when in compression or when yielding of the steel occurs, BS 8110 limits the value of x to be not greater than 0.5d. Therefore, the maximum resistance moment of resistance of the section can be obtained by equating x = 0.5d. This is shown as follows: -
M = Mu =
(0.405 ×
0.5d 0. 5d 0.45 × d ) (1d ) (fcubd2)
= 0.156fcubd2 = K’fcubd2 •
Where Mu is known as the ultimate moment of resistance for singly reinforced section.
•
The equation above may be used to calculate the area of reinforcement required if the bending moment is known.
•
One important conclusion is that if the bending moment, M applied is less than Mu, the section requires tension reinforcement in the tension zone. This is called singly reinforced section.
4.2 Example Simply Reinforced Rectangular Section A rectangular beam section is required to carry a design moment of 300kNm. The dimension of the area, b = 250 mm and d = 700 mm. Determine the area and total number of reinforcement needed if the characteristic strength of reinforcement used, fy=460N/mm2 and that of concrete, fcu = 40 N/mm2.
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Solution: Calculate the ultimate moment of resistance for singly reinforced section as follows: Mu = 0.156fcubd2 = 0.156 (40) (250) (700)2 = 764.4kNm
This shows that Mu > M (=300kNm). Therefore this is a singly reinforced section Calculate the depth to neutral axis as follows: M = 0.405fcubx (d - 0.45x) 300 x 106 = 0.405 (4) (250) (x) (700 - 0.45x) x2 – 1560x + 16500 = 0 x = 114 mm @ 1441 mm x is taken as 114mm (the reasonable value). Calculate Lever arm, Z = d – 0.45x = 700 – 0.45(114) = 649 mm Calculate the area of reinforcement required as follows: -
As =
M 0.87 f y Z
300 ×10 6 = 0.87 × 460 × 649
= 1155mm2 Use 4T20 (As = 1260mm2)
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4.2.1Example Calculate the moment of resistance for a beam section shown in Figure 4.1 below: 250 mm
-
d= 700 mm
4T20 Figure 4.1: Beam Section
The section is already provided with 4T20 reinforcement whereby fcu and fy are given as 40N/mm2 and 460 N/mm2 respectively. Solution:
b=250mm
d = 700
x
Fcc = 0.405fcu bx Z = d – 0.45x
Fst = 0.87fyAs Figure 4.2: Simply Reinforced Rectangular Section
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Calculate the depth to neutral axis as follows: Fcc = Fst 0.405fcubx = 0.87fyAs 0.87 f y As
x = 0.405 f cu b 0.87 ( 460 )(1260 ) = 0.405 ( 40 )( 250 )
= 124 mm x 124 = < 0.5 Check, d 700
Therefore concrete will not be crushed during compression.
Calculate the moment of resistance as shown below: M = Fst. Z = 0.87 fy As (d - 0.45x) = 0.87 (460) (1260) (700 – 0.45 (124)) = 325 kNm Therefore, the moment of resistance of the beam section, provided with 4T20, is 325 kNm.
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INPUT 2
4.3
Doubly Reinforced Rectangular Section For singly reinforced section, steel reinforcement is only used for resisting tensile force. Compressive force is taken by concrete in the compressive zone. BS 8110 allows the maximum value of x to be not greater than 0.5d for the concrete stress block taking the compressive force. Based on this maximum limit x = 0.5d, the maximum moment that can be carried by concrete is given by, Mu = 0.156fcubd2 If the bending moment, M to be carried by the section is greater than Mu, steel reinforcement is to be provided in the compression zone to resist the resultant moment. The stress-strain distribution and the forces acting in a doubly reinforced rectangular section is shown in Figure 4.3 below: -
As’
Fsc
d’
Fcc
x Z
d As Fst Section
Strain
Stress & Forces
Figure 4.3: The stress-strain distribution and the forces acting in a doubly reinforced rectangular section
Z1
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The condition for equilibrium which must be obtained at any point in the section is that horizontal forces must equate to zero; i.e.: Fst = Fcc + Fsc 0.87fyAs = 0.405fcubx + 0.87fyAs’ (0.87 f y As −0.87 f y As ' )
x
=
0.405 f cu b
……..(4.2)
Taking moment (moment of resistance) about the tension steel, we have; M = Fcc . Z + Fsc . Z1 = (0.405fcubx) (d - 0.45x) + (0.87 fyAs’). (d - d’) = 0.156fcubd2 + 0.87fyAs’ (d - d’) = Mu + 0.87fyAs’ (d - d’) Therefore, area of compression reinforcement is:
As’ =
=
M − Mu 0.87 f y ( d −d ' )
( K − K ' ) f cu bd 2 0.87 f y ( d − d ' )
From equation 4.2, it can be shown that: 0.87fyAs = 0.405fcubx + 0.87fyAs’ 0.87fyAs.Z = 0.405fcubx. Z + 0.87fyAs’ Z When x = 0.5d, therefore z = 0.775d Then, 0.87fyAs. Z = 0.405fcub (0.5d - 0.775d) + 0.87fyAs’. Z 0.87fyAs.Z = 0.156 fcubd2 + 0.87fyAs’. Z
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Therefore the area of tension reinforcement is given by,
As =
0.156 f cu bd 0.87 f y Z
2
+ As '
Or
As
=
K ' f cu bd 2 + As '......... .......... .......... .......... .......... ....... 4.3 0.87 f y Z
4.4
Stress in the compression reinforcement. In deriving equation 4.3 above, it is assumed that the compression reinforcement has yielded, i.e. it has achieved its design strength of 0.87fy. From the strain design diagram, it can be shown that:
ε sc 0.0035 = x − d' x x−d ε sc = x 0.0035 d' ε sc =1− x 0.0035 For the design strength to achieve the value 0.87fy, therefore; εsc =
And
0.87 f y Es
0.87 f y d' =1 − x Es (0.0035 )
For fy = 460 N/mm2, d' 0.87 (460 ) =1 − x (200 ×10 3 )( 0.0035 )
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= 0.43 This implies that for compression reinforcement to achieve its yield point, the
d' ratio x must not exceed 0.43. If
d' x > 0.43, the compressive stress, fsc < 0.87
fy and the value fsc can be obtained from Figure 2.2 BS 8110. It can be shown that,
fsc = Es . εsc where εsc =
∴
3
0.0035 ( x − d ' ) x
fsc = 200 x 10 x 0.0035 (
= 700
(1 −
(1 −
d' ) x
d' ) x
Now do activity 1
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ACTIVITY 4a
Answer the following questions. 4.1 What is the limit for depth to neutral axis, x as specified by BS 8110?
_________________________________________________________________ _ 4.2 Write down the formula that is used to calculate the maximum moment that can be carried by concrete. _________________________________________________________________ _________________________________________________________________ __ 4.3
What should be done if the applied moment is greater than the ultimate moment, Mu?
_________________________________________________________________ _________________________________________________________________ __ 4.4
Write down the formula that is used to calculate the area of compression reinforcement.
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_________________________________________________________________ _________________________________________________________________ __
4.5
Write down the formula that is used to calculate the area of tension reinforcement.
_________________________________________________________________ _________________________________________________________________ __
4.6
d' Write down the formula used to that is calculate f sc when x is greater
than 0.43. _________________________________________________________________ _________________________________________________________________ __
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4a
Check your answers below: 4.1
x ≠ 0.5d
4.2 Mu = 0.156fcubd2 4.3 Compression reinforcement should be provided.
4.4 As’ =
M − Mu 0.87 f y ( d −d ' )
As’ =
4.5 As =
(k − k ' ) f cu bd 2 0.87 f y ( d − d ' )
0.156 f cu bd 0.87 f y Z
4.6 fsc = 700
or
(1 −
d' ) x
2
+ As '
or
As =
k ' f cu bd 2 + As ' 0.87 f y Z
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INPUT 3
4.3 Design Example (Rectangular Section)
A rectangular section of reinforced concrete beam is to be designed to carry a design moment of 880kNm. The width of this beam, b=250 mm and the depth, d=700 mm. Calculate the area and the total number of bars needed if fcu=40N/mm2 and fy = 460 N/mm2. Solution: Ultimate moment of singly reinforced section, Mu = 0156fcubd2 = 0.156 x 40 x 250 x (700)2 = 764.4 kNm
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Mu < M (= 880kNm). This shows that compression reinforcement is needed and is calculated as follows: -
As’ =
M − Mu 0.87 f y ( d −d ' )
Assuming that d’ = 60 mm, (880 − 764 .4) ×10 6 As’ = 0.87 × 460 ×(700 − 60 )
= 451 mm2.
d' Now, we have to check the x ratio. This is done as follows: -
x = 0.5d = 0.5 x 700 = 350 mm
d' 60 x = 350
= 0.17 < 0.43 This shows that the assumed compression steel achieved its design strength 0.87fy.
The area of tension steel is then calculated as shown below: -
As =
Mu + As ' 0.87 f y Z
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764 .4 ×10 6 + 451 = 0.87 × 460 × 0.775 × 700
= 3972 mm2
Now we have to check from the table below to determine the total number of bars and then the size. Bar size (mm)
Number of bars
Perimeter (mm)
1
2
3
4
5
6
7
8
6 8 10
28.3 50.3 78.6
56.5 101 157
84.9 151 236
113 201 314
141 251 393
170 302 471
198 352 550
226 402 629
18.9 25.1 31.4
12 16 20
113 201 314
226 402 629
339 603 943
453 805 1257
566 1006 1571
679 1207 1886
792 1408 2200
905 1609 2514
37.7 50.3 62.9
25 32 40
491 805 1257
982 1609 2514
1473 2414 3771
1964 3218 5029
2455 4023 6286
2496 4827 7543
3438 5632 8800
3929 6437 10057
78.6 100.6 125.7
Table 4.1: The Allowable Size and Numbers of Bars
Provided: 3T16 for compression reinforcement (As’ = 603 mm2) 5T32 for tension reinforcement (As = 4023 mm2)
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ACTIVITY 4b
4.7 Given the following information, calculate the area of reinforcement required for a bending moment of 650 kNm fcu = 30 N/mm2 fy = 460 N/mm2 b = 300 mm d = 618 mm d’ = 60 mm.
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4b
4.7
Solution: Step 1 Mu = 0.156fcubd2 = 0.156 x 30 x 300 x (618)2 = 536.2 kNm < M (= 650 kNm)
∴
Compression reinforcement is required.
Step 2
As’ =
M − Mu 0.87 f y ( d −d ' )
(650 − 536 .2) ×10 6 = 0.87 × 460 ×(618 − 60 )
=509.6 mm2
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Step 3 d' Check x ratio: -
x = 0.5d = 0.5 (618) = 309 mm d' 60 x = 309 = 0.19 < 0.43
∴
Compression steel has yielded.
Step 4
As =
Mu + As ' 0.87 f y Z
536 .2 ×10 6 + 509 .6 = 0.87 × 460 × 0.775 × 618
= 2797 mm2
Step 5 Provide:
5T12 (As’ = 565 mm2) for compression reinforced
9T20 (As = 2828 mm2) for tension reinforced
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INPUT 3
4.4 Design example (Determine moment capacity of a section)
Calculate the moment of resistance of a rectangular section beam which has been provided with 5T32 (tension steel) and 3T16 (compression steel). The 2 f = 460 N / mm 2 characteristic strengths are f cu = 40 N / mm and y .
250 mm The section is as shown in Figure 4.4 below: 3T16 700 mm
5T32
Figure 4.4: Rectangular Section Beam
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Solution: The stress and forces in the stress block is given below: -
Fsc x
Fcc
d
Z
Z1=d-d’
Fst Figure 4.5: The Stress and Forces in the Stress Block
Step: 1. For equilibrium of forces; Fst = Fcc + Fsc
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0.87 f y = 0.405 f cu bx +0.87 f y As '
2. The depth to neutral axis is given by, 0.87 f y ( As − As ' )
x=
=
0.405 f cu b 0.87 × 460 × ( 4023 − 603 ) 0.405 ( 40 )( 250 )
= 338.5 mm
3. Check the steel stress. x 338 .5 = = 0.483 < 0.5 d 700
d' 60 = = 0.17 < 0.43 x 338 .5
This shows that both the compression and tension steel have yielded at 0.87fy 4. Moment of resistance is calculated as shown below; M = Fsc Z1 + Fcc Z 2 = 0.87 f y As ' ( d −d ' ) +0.405 f cu bx (d −0.45 x )
= 0.87 (460 )( 603 )( 700 − 600 ) + 0.405 (40 )( 250 )[ 700 − 0.405 (338 .5)]
= 904 kNm. Therefore the resistance moment of the section is 904 kNm.
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Now, do activity 3
ACTIVITY 4c 1.8 Determine the moment of resistance for the given section: b = 400 mm2 As’ = 800 mm2 d’ = 40 mm d = 800 mm As = 5050mm2 fcu = 30N/mm2 fy = 460 N/mm2
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3
4.8
Refer to the answer as shown below: -
Step 1: Equilibrium of forces Fst = Fcc + Fsc 0.87 f y As = 0.405 f cu bx +0.87 f y As '
∴
x=
=
0.87 f y ( As − As ' ) 0.405 f cu b
0.87 × 460 × (5050 − 800 ) 0.405 × 30 × 400
= 350 mm Step 2: Check whether the steel has yielded. x 350 = = 0.4375 < 0.5 d 800 d' 40 = = 0.11 < 0.43 x 350
∴Compression and tension steel have yielded.
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Step 3: Calculate the moment of resistance. M = Fsc Z1 + Fcc Z 2 = 0.87 f y As ' ( d −d ' ) +0.405 f cu bx (d −0.45 x )
= (0.87 )( 30 )( 800 )( 800 − 40 ) + (0.405 )( 30 )( 400 )( 350 )[ 800 − 0.45 (350 )]
= 15.87 + 1092.9 kNm = 1108.77 kNm
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SUMMARY
SUMMARY
To calculate the area of reinforcement of a beam (having a size of b x d), with a bending moment of M, using concrete strength of fcu and steel strength fy , you should use the following steps: K =
1. Calculate
M bd 2 f cu
2. Calculate K ' ( = 0.156 ) 3. If K< K’, compression reinforcement is not required.
i)
ii)
k z = d{0.5 + ( .0 2 − 5 } 0.9 but z < 0.95d Calculate Calculate As =
M 0.87 f y Z
4. If K > K’, compression reinforcement is required.
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i)
z = d {0.5 + (0.25 −
Calculate
x=
d −z 0.45
ii)
Calculate
iii)
d' Check x ratio.
iv)
Calculate ( K − K ' ) f cu bd 2 0.87 f y ( d − d ' )
d' < 0.43 if x
( K − K ' ) f cu bd 2 As ' = d' 700 (1 − ) x
d' > 0.43 if x
As ' =
or v)
K' } 0.9
Calculate As =
K ' f cu bd 2 + As ' 0.87 f y Z
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SELF-ASSESSMENT
Test your understanding by solving the given problem. 1.
Calculate the area, size and total number of bars required by the given section shown in Figure 4.6. The section is to carry a design moment of 90kNm. Use concrete of grade 30 and high yield steel. 200 mm
400 mm
(4 Marks) Figure 4.6: Rectangular Bar
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2.
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Calculate the area, size and total number of bars required for the given section. The bending moment, fcu and fy to be used are the same as in question 1. Use d’ = 50 mm.
350 mm
(7 Marks)
150 mm
3.
The figure below shows a rectangular beam section which is to be designed using concrete of grade 25 and high yield steel reinforcement. Calculate the moment of resistance of the section. 300 mm
600 mm
4T25
(3 Marks)
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4.
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Calculate the moment of resistance of the given section using concrete of grade 25 and high yield steel reinforcement.
250mm 2T12
d=450mm
3T20
(4 Marks)
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ON SELF-ASSESSMENT Check your answers below and then total up your marks. The marks allocated are shown at every step: (TOTAL 100 MARKS) K =
1.
M 90 ×10 6 = bd 2 f cu 200 × (400 ) 2 × 30
= 0.094
……………………. 1
Since, K < K’ Therefore compression reinforcement is not required. Z = d {0.5 + (0.25 −
K )} 0.9
.0 0 9 4 = 4 {0.5 + 0 .0 2 − 5 ) } 0.9 = 352 .6mm
0.95 d = 0.95 ( 400 )
= 380 mm
∴Z
< 0.95 d
1
……………………..
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As =
=
C4301/UNIT4/33
M 0.87 f y Z
90 ×10 6 0.87 × 460 × 352 .6
= 638 mm 2
………………………….. 1
Use 6T12 (As = 679 mm2) K=
2. =
M bd 2 f cu
90 ×10 6 30 ×150 × (350 ) 2
……………………….. 1
= 0.163
∴K > K’
Compression reinforcement is required. Z = d {0.5 + (0.25 −
.0 1 5 6 = 3 {05.5 + 0( .0 2 − 5 ) } 0.9 =272 mm
K' )} 0.9
1
………………………….
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0.95d = 0.95(350) =332.5 mm
∴Z < 0.95d x=
d −Z 0.45
350 − 272 0.45
=
=173.3 mm
d' 50 = = 0.29 x 173 .3
∴
d' < 0.43 x
As ' =
=
( K − K ' ) f cu bd 2 0.87 f y (d − d ' ) (0.163 −0.156 )( 30 )(150 )( 350 ) 2 0.87 ×460 ×(350 −50 )
=32 .1mm
As =
1 ………………………….
2
K ' f cu bd 2 + As ' 0.87 f y Z
…………………………… 1
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=
C4301/UNIT4/35
0.156 × 30 ×150 × (350 ) 2 + 32 .1 0.87 × 460 × 272
………………………………1
= 822 mm2
Use : 2T8 (As’=101 mm2) as compression reinforcement 8T12 (As=905 mm2) as tension reinforcement
3. For equilibrium of forces ; Fcc = Fst 0.405 fcu b x = 0.87 fy As 0.87 f y As
x = 0.405 f cu b 0.87 ( 460 )(1964 ) = 0.405 ( 25 )( 300 )
= 258.8 mm x 258 .8 = d 600
= 0.43 < 0.5
Moment of resistance,
M = Fst Z = 0.87 x 460 x 1964 (600 – 0.45 x 258.8) = 380.06 kNm
1
. ……………… ………………..
1
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4. Fst = Fcc + Fsc 0.87 fy As = 0.405 fcu b x + 0.87 fy As’ 0.87 f y ( As − As ' )
x=
0.405 f cu b
0.87 × 460 (943 − 226 ) 0.405 × 25 × 250 =
= 113.4 mm x 113 .4 = = 0.25 < 0.5 d 450 d' 40 = = 0.35 < 0.43 x 113 .4
…………………………… 1
…………………………
1
1 …………...………………
Moment of resistance is, M = 0.87 f y As ' ( d −d ' ) +0.405 f cu bx ( d −0.45 x) = {0.87 x 460 x 226 x (450 - 40)} + {0.405 x 25 x 250 x 113.4 x [(450(0.45x113.4)]}
=
151.6 kNm
……………………………….
1
Total = 18 marks
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NOW TOTAL UP YOUR MARKS. FULL MARKS = 18 MARKS
Calculate your score as shown below: Score = total marks obtained 18
x 100%
You should score 80% or more to this unit. If you have scored 80% or more, you may proceed to unit 5. If you have scored less than 80%, you should go through unit 4 again until you this unit. See you in UNIT 5.
END OF UNIT 4
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UNIT 4 DESIGN OF RECTANGULAR BEAM SECTIONS
OBJECTIVES
GENERAL OBJECTIVE To be able to provide steel reinforcement required in reinforced concrete beams. SPECIFIC OBJECTIVES
At the end of this unit, you should be able to: 1.
calculate the area and total number of reinforcement required in given beam sections with known bending moments.
2.
calculate the bending moment that can be carried by the beam section when the grade of steel is already provided.
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INPUT 1
4.1
Simply Reinforced Rectangular Section What does it mean? Simply reinforced section is the steel reinforcement that is provided only at the tension zone. The form of section, stress distribution and forces on a section are as shown in Figure 4.1: 0.45fcu
b
0.45x 0.9x Fcc=0.405fcubx
x d
Z=d-0.45x
As 0.87 fy
Section
Stress
Fst =0.87fyAs
Forces
Figure 4.1: The form of section, stress distribution and forces
on a section
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The meaning of symbols used are given in clause 3.4.4.3 For the section to be in balance, the sum of the forces must be equal to zero, i.e. Fst = Fcc 0.87fyAs = 0.405fcubx
∴
0.87 f y As
x = 0.405 f cu b
.......... .......... ...... equation
(4.1)
Taking moments about Fcc or Fst, the moment of resistance of the section can be calculated as follows, M = Fst. Z = (0.87fyAs) (d - 0.45x) or, M = Fcc. Z = (0.405fcubx) (d – 0.45 x) x 0.45 x 1− d = (0.405 d ) ( ) (fcubd2)
= K. fcu bd2 A section whereby the reinforcement is already known, the above equation can be used to calculate the moment of resistance or the bending moment. Equation 4.1 shows that, x will increase if the area of reinforcement, As increases. The moment of resistance of the section will also increase. It is clear that a section provided with more reinforcement will be able to with stand reinforcement load i.e. greater bending moment.
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In order to make sure that concrete will not crush or fail when in compression or when yielding of the steel occurs, BS 8110 limits the value of x to be not greater than 0.5d. Therefore, the maximum resistance moment of resistance of the section can be obtained by equating x = 0.5d. This is shown as follows: -
M = Mu =
(0.405 ×
0.5d 0. 5d 0.45 × d ) (1d ) (fcubd2)
= 0.156fcubd2 = K’fcubd2 •
Where Mu is known as the ultimate moment of resistance for singly reinforced section.
•
The equation above may be used to calculate the area of reinforcement required if the bending moment is known.
•
One important conclusion is that if the bending moment, M applied is less than Mu, the section requires tension reinforcement in the tension zone. This is called singly reinforced section.
4.2 Example Simply Reinforced Rectangular Section A rectangular beam section is required to carry a design moment of 300kNm. The dimension of the area, b = 250 mm and d = 700 mm. Determine the area and total number of reinforcement needed if the characteristic strength of reinforcement used, fy=460N/mm2 and that of concrete, fcu = 40 N/mm2.
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/ 5
Solution: Calculate the ultimate moment of resistance for singly reinforced section as follows: Mu = 0.156fcubd2 = 0.156 (40) (250) (700)2 = 764.4kNm
This shows that Mu > M (=300kNm). Therefore this is a singly reinforced section Calculate the depth to neutral axis as follows: M = 0.405fcubx (d - 0.45x) 300 x 106 = 0.405 (4) (250) (x) (700 - 0.45x) x2 – 1560x + 16500 = 0 x = 114 mm @ 1441 mm x is taken as 114mm (the reasonable value). Calculate Lever arm, Z = d – 0.45x = 700 – 0.45(114) = 649 mm Calculate the area of reinforcement required as follows: -
As =
M 0.87 f y Z
300 ×10 6 = 0.87 × 460 × 649
= 1155mm2 Use 4T20 (As = 1260mm2)
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/ 6
4.2.1Example Calculate the moment of resistance for a beam section shown in Figure 4.1 below: 250 mm
-
d= 700 mm
4T20 Figure 4.1: Beam Section
The section is already provided with 4T20 reinforcement whereby fcu and fy are given as 40N/mm2 and 460 N/mm2 respectively. Solution:
b=250mm
d = 700
x
Fcc = 0.405fcu bx Z = d – 0.45x
Fst = 0.87fyAs Figure 4.2: Simply Reinforced Rectangular Section
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/ 7
Calculate the depth to neutral axis as follows: Fcc = Fst 0.405fcubx = 0.87fyAs 0.87 f y As
x = 0.405 f cu b 0.87 ( 460 )(1260 ) = 0.405 ( 40 )( 250 )
= 124 mm x 124 = < 0.5 Check, d 700
Therefore concrete will not be crushed during compression.
Calculate the moment of resistance as shown below: M = Fst. Z = 0.87 fy As (d - 0.45x) = 0.87 (460) (1260) (700 – 0.45 (124)) = 325 kNm Therefore, the moment of resistance of the beam section, provided with 4T20, is 325 kNm.
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/ 8
INPUT 2
4.3
Doubly Reinforced Rectangular Section For singly reinforced section, steel reinforcement is only used for resisting tensile force. Compressive force is taken by concrete in the compressive zone. BS 8110 allows the maximum value of x to be not greater than 0.5d for the concrete stress block taking the compressive force. Based on this maximum limit x = 0.5d, the maximum moment that can be carried by concrete is given by, Mu = 0.156fcubd2 If the bending moment, M to be carried by the section is greater than Mu, steel reinforcement is to be provided in the compression zone to resist the resultant moment. The stress-strain distribution and the forces acting in a doubly reinforced rectangular section is shown in Figure 4.3 below: -
As’
Fsc
d’
Fcc
x Z
d As Fst Section
Strain
Stress & Forces
Figure 4.3: The stress-strain distribution and the forces acting in a doubly reinforced rectangular section
Z1
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/ 9
The condition for equilibrium which must be obtained at any point in the section is that horizontal forces must equate to zero; i.e.: Fst = Fcc + Fsc 0.87fyAs = 0.405fcubx + 0.87fyAs’ (0.87 f y As −0.87 f y As ' )
x
=
0.405 f cu b
……..(4.2)
Taking moment (moment of resistance) about the tension steel, we have; M = Fcc . Z + Fsc . Z1 = (0.405fcubx) (d - 0.45x) + (0.87 fyAs’). (d - d’) = 0.156fcubd2 + 0.87fyAs’ (d - d’) = Mu + 0.87fyAs’ (d - d’) Therefore, area of compression reinforcement is:
As’ =
=
M − Mu 0.87 f y ( d −d ' )
( K − K ' ) f cu bd 2 0.87 f y ( d − d ' )
From equation 4.2, it can be shown that: 0.87fyAs = 0.405fcubx + 0.87fyAs’ 0.87fyAs.Z = 0.405fcubx. Z + 0.87fyAs’ Z When x = 0.5d, therefore z = 0.775d Then, 0.87fyAs. Z = 0.405fcub (0.5d - 0.775d) + 0.87fyAs’. Z 0.87fyAs.Z = 0.156 fcubd2 + 0.87fyAs’. Z
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/10
Therefore the area of tension reinforcement is given by,
As =
0.156 f cu bd 0.87 f y Z
2
+ As '
Or
As
=
K ' f cu bd 2 + As '......... .......... .......... .......... .......... ....... 4.3 0.87 f y Z
4.4
Stress in the compression reinforcement. In deriving equation 4.3 above, it is assumed that the compression reinforcement has yielded, i.e. it has achieved its design strength of 0.87fy. From the strain design diagram, it can be shown that:
ε sc 0.0035 = x − d' x x−d ε sc = x 0.0035 d' ε sc =1− x 0.0035 For the design strength to achieve the value 0.87fy, therefore; εsc =
And
0.87 f y Es
0.87 f y d' =1 − x Es (0.0035 )
For fy = 460 N/mm2, d' 0.87 (460 ) =1 − x (200 ×10 3 )( 0.0035 )
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/11
= 0.43 This implies that for compression reinforcement to achieve its yield point, the
d' ratio x must not exceed 0.43. If
d' x > 0.43, the compressive stress, fsc < 0.87
fy and the value fsc can be obtained from Figure 2.2 BS 8110. It can be shown that,
fsc = Es . εsc where εsc =
∴
3
0.0035 ( x − d ' ) x
fsc = 200 x 10 x 0.0035 (
= 700
(1 −
(1 −
d' ) x
d' ) x
Now do activity 1
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/12
ACTIVITY 4a
Answer the following questions. 4.1 What is the limit for depth to neutral axis, x as specified by BS 8110?
_________________________________________________________________ _ 4.2 Write down the formula that is used to calculate the maximum moment that can be carried by concrete. _________________________________________________________________ _________________________________________________________________ __ 4.3
What should be done if the applied moment is greater than the ultimate moment, Mu?
_________________________________________________________________ _________________________________________________________________ __ 4.4
Write down the formula that is used to calculate the area of compression reinforcement.
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/13
_________________________________________________________________ _________________________________________________________________ __
4.5
Write down the formula that is used to calculate the area of tension reinforcement.
_________________________________________________________________ _________________________________________________________________ __
4.6
d' Write down the formula used to that is calculate f sc when x is greater
than 0.43. _________________________________________________________________ _________________________________________________________________ __
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/14
4a
Check your answers below: 4.1
x ≠ 0.5d
4.2 Mu = 0.156fcubd2 4.3 Compression reinforcement should be provided.
4.4 As’ =
M − Mu 0.87 f y ( d −d ' )
As’ =
4.5 As =
(k − k ' ) f cu bd 2 0.87 f y ( d − d ' )
0.156 f cu bd 0.87 f y Z
4.6 fsc = 700
or
(1 −
d' ) x
2
+ As '
or
As =
k ' f cu bd 2 + As ' 0.87 f y Z
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/15
INPUT 3
4.3 Design Example (Rectangular Section)
A rectangular section of reinforced concrete beam is to be designed to carry a design moment of 880kNm. The width of this beam, b=250 mm and the depth, d=700 mm. Calculate the area and the total number of bars needed if fcu=40N/mm2 and fy = 460 N/mm2. Solution: Ultimate moment of singly reinforced section, Mu = 0156fcubd2 = 0.156 x 40 x 250 x (700)2 = 764.4 kNm
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/16
Mu < M (= 880kNm). This shows that compression reinforcement is needed and is calculated as follows: -
As’ =
M − Mu 0.87 f y ( d −d ' )
Assuming that d’ = 60 mm, (880 − 764 .4) ×10 6 As’ = 0.87 × 460 ×(700 − 60 )
= 451 mm2.
d' Now, we have to check the x ratio. This is done as follows: -
x = 0.5d = 0.5 x 700 = 350 mm
d' 60 x = 350
= 0.17 < 0.43 This shows that the assumed compression steel achieved its design strength 0.87fy.
The area of tension steel is then calculated as shown below: -
As =
Mu + As ' 0.87 f y Z
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/17
764 .4 ×10 6 + 451 = 0.87 × 460 × 0.775 × 700
= 3972 mm2
Now we have to check from the table below to determine the total number of bars and then the size. Bar size (mm)
Number of bars
Perimeter (mm)
1
2
3
4
5
6
7
8
6 8 10
28.3 50.3 78.6
56.5 101 157
84.9 151 236
113 201 314
141 251 393
170 302 471
198 352 550
226 402 629
18.9 25.1 31.4
12 16 20
113 201 314
226 402 629
339 603 943
453 805 1257
566 1006 1571
679 1207 1886
792 1408 2200
905 1609 2514
37.7 50.3 62.9
25 32 40
491 805 1257
982 1609 2514
1473 2414 3771
1964 3218 5029
2455 4023 6286
2496 4827 7543
3438 5632 8800
3929 6437 10057
78.6 100.6 125.7
Table 4.1: The Allowable Size and Numbers of Bars
Provided: 3T16 for compression reinforcement (As’ = 603 mm2) 5T32 for tension reinforcement (As = 4023 mm2)
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/18
ACTIVITY 4b
4.7 Given the following information, calculate the area of reinforcement required for a bending moment of 650 kNm fcu = 30 N/mm2 fy = 460 N/mm2 b = 300 mm d = 618 mm d’ = 60 mm.
REINFORCED CONCRETE STRUCTURAL DESIGN
4b
4.7
Solution: Step 1 Mu = 0.156fcubd2 = 0.156 x 30 x 300 x (618)2 = 536.2 kNm < M (= 650 kNm)
∴
Compression reinforcement is required.
Step 2
As’ =
M − Mu 0.87 f y ( d −d ' )
(650 − 536 .2) ×10 6 = 0.87 × 460 ×(618 − 60 )
=509.6 mm2
C4301/UNIT4/19
REINFORCED CONCRETE STRUCTURAL DESIGN
Step 3 d' Check x ratio: -
x = 0.5d = 0.5 (618) = 309 mm d' 60 x = 309 = 0.19 < 0.43
∴
Compression steel has yielded.
Step 4
As =
Mu + As ' 0.87 f y Z
536 .2 ×10 6 + 509 .6 = 0.87 × 460 × 0.775 × 618
= 2797 mm2
Step 5 Provide:
5T12 (As’ = 565 mm2) for compression reinforced
9T20 (As = 2828 mm2) for tension reinforced
C4301/UNIT4/20
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/21
INPUT 3
4.4 Design example (Determine moment capacity of a section)
Calculate the moment of resistance of a rectangular section beam which has been provided with 5T32 (tension steel) and 3T16 (compression steel). The 2 f = 460 N / mm 2 characteristic strengths are f cu = 40 N / mm and y .
250 mm The section is as shown in Figure 4.4 below: 3T16 700 mm
5T32
Figure 4.4: Rectangular Section Beam
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/22
Solution: The stress and forces in the stress block is given below: -
Fsc x
Fcc
d
Z
Z1=d-d’
Fst Figure 4.5: The Stress and Forces in the Stress Block
Step: 1. For equilibrium of forces; Fst = Fcc + Fsc
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/23
0.87 f y = 0.405 f cu bx +0.87 f y As '
2. The depth to neutral axis is given by, 0.87 f y ( As − As ' )
x=
=
0.405 f cu b 0.87 × 460 × ( 4023 − 603 ) 0.405 ( 40 )( 250 )
= 338.5 mm
3. Check the steel stress. x 338 .5 = = 0.483 < 0.5 d 700
d' 60 = = 0.17 < 0.43 x 338 .5
This shows that both the compression and tension steel have yielded at 0.87fy 4. Moment of resistance is calculated as shown below; M = Fsc Z1 + Fcc Z 2 = 0.87 f y As ' ( d −d ' ) +0.405 f cu bx (d −0.45 x )
= 0.87 (460 )( 603 )( 700 − 600 ) + 0.405 (40 )( 250 )[ 700 − 0.405 (338 .5)]
= 904 kNm. Therefore the resistance moment of the section is 904 kNm.
REINFORCED CONCRETE STRUCTURAL DESIGN
Now, do activity 3
ACTIVITY 4c 1.8 Determine the moment of resistance for the given section: b = 400 mm2 As’ = 800 mm2 d’ = 40 mm d = 800 mm As = 5050mm2 fcu = 30N/mm2 fy = 460 N/mm2
C4301/UNIT4/24
REINFORCED CONCRETE STRUCTURAL DESIGN
3
4.8
Refer to the answer as shown below: -
Step 1: Equilibrium of forces Fst = Fcc + Fsc 0.87 f y As = 0.405 f cu bx +0.87 f y As '
∴
x=
=
0.87 f y ( As − As ' ) 0.405 f cu b
0.87 × 460 × (5050 − 800 ) 0.405 × 30 × 400
= 350 mm Step 2: Check whether the steel has yielded. x 350 = = 0.4375 < 0.5 d 800 d' 40 = = 0.11 < 0.43 x 350
∴Compression and tension steel have yielded.
C4301/UNIT4/25
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/26
Step 3: Calculate the moment of resistance. M = Fsc Z1 + Fcc Z 2 = 0.87 f y As ' ( d −d ' ) +0.405 f cu bx (d −0.45 x )
= (0.87 )( 30 )( 800 )( 800 − 40 ) + (0.405 )( 30 )( 400 )( 350 )[ 800 − 0.45 (350 )]
= 15.87 + 1092.9 kNm = 1108.77 kNm
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/27
SUMMARY
SUMMARY
To calculate the area of reinforcement of a beam (having a size of b x d), with a bending moment of M, using concrete strength of fcu and steel strength fy , you should use the following steps: K =
1. Calculate
M bd 2 f cu
2. Calculate K ' ( = 0.156 ) 3. If K< K’, compression reinforcement is not required.
i)
ii)
k z = d{0.5 + ( .0 2 − 5 } 0.9 but z < 0.95d Calculate Calculate As =
M 0.87 f y Z
4. If K > K’, compression reinforcement is required.
REINFORCED CONCRETE STRUCTURAL DESIGN
i)
z = d {0.5 + (0.25 −
Calculate
x=
d −z 0.45
ii)
Calculate
iii)
d' Check x ratio.
iv)
Calculate ( K − K ' ) f cu bd 2 0.87 f y ( d − d ' )
d' < 0.43 if x
( K − K ' ) f cu bd 2 As ' = d' 700 (1 − ) x
d' > 0.43 if x
As ' =
or v)
K' } 0.9
Calculate As =
K ' f cu bd 2 + As ' 0.87 f y Z
C4301/UNIT4/28
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/29
SELF-ASSESSMENT
Test your understanding by solving the given problem. 1.
Calculate the area, size and total number of bars required by the given section shown in Figure 4.6. The section is to carry a design moment of 90kNm. Use concrete of grade 30 and high yield steel. 200 mm
400 mm
(4 Marks) Figure 4.6: Rectangular Bar
REINFORCED CONCRETE STRUCTURAL DESIGN
2.
C4301/UNIT4/30
Calculate the area, size and total number of bars required for the given section. The bending moment, fcu and fy to be used are the same as in question 1. Use d’ = 50 mm.
350 mm
(7 Marks)
150 mm
3.
The figure below shows a rectangular beam section which is to be designed using concrete of grade 25 and high yield steel reinforcement. Calculate the moment of resistance of the section. 300 mm
600 mm
4T25
(3 Marks)
REINFORCED CONCRETE STRUCTURAL DESIGN
4.
C4301/UNIT4/31
Calculate the moment of resistance of the given section using concrete of grade 25 and high yield steel reinforcement.
250mm 2T12
d=450mm
3T20
(4 Marks)
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/32
ON SELF-ASSESSMENT Check your answers below and then total up your marks. The marks allocated are shown at every step: (TOTAL 100 MARKS) K =
1.
M 90 ×10 6 = bd 2 f cu 200 × (400 ) 2 × 30
= 0.094
……………………. 1
Since, K < K’ Therefore compression reinforcement is not required. Z = d {0.5 + (0.25 −
K )} 0.9
.0 0 9 4 = 4 {0.5 + 0 .0 2 − 5 ) } 0.9 = 352 .6mm
0.95 d = 0.95 ( 400 )
= 380 mm
∴Z
< 0.95 d
1
……………………..
REINFORCED CONCRETE STRUCTURAL DESIGN
As =
=
C4301/UNIT4/33
M 0.87 f y Z
90 ×10 6 0.87 × 460 × 352 .6
= 638 mm 2
………………………….. 1
Use 6T12 (As = 679 mm2) K=
2. =
M bd 2 f cu
90 ×10 6 30 ×150 × (350 ) 2
……………………….. 1
= 0.163
∴K > K’
Compression reinforcement is required. Z = d {0.5 + (0.25 −
.0 1 5 6 = 3 {05.5 + 0( .0 2 − 5 ) } 0.9 =272 mm
K' )} 0.9
1
………………………….
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/34
0.95d = 0.95(350) =332.5 mm
∴Z < 0.95d x=
d −Z 0.45
350 − 272 0.45
=
=173.3 mm
d' 50 = = 0.29 x 173 .3
∴
d' < 0.43 x
As ' =
=
( K − K ' ) f cu bd 2 0.87 f y (d − d ' ) (0.163 −0.156 )( 30 )(150 )( 350 ) 2 0.87 ×460 ×(350 −50 )
=32 .1mm
As =
1 ………………………….
2
K ' f cu bd 2 + As ' 0.87 f y Z
…………………………… 1
REINFORCED CONCRETE STRUCTURAL DESIGN
=
C4301/UNIT4/35
0.156 × 30 ×150 × (350 ) 2 + 32 .1 0.87 × 460 × 272
………………………………1
= 822 mm2
Use : 2T8 (As’=101 mm2) as compression reinforcement 8T12 (As=905 mm2) as tension reinforcement
3. For equilibrium of forces ; Fcc = Fst 0.405 fcu b x = 0.87 fy As 0.87 f y As
x = 0.405 f cu b 0.87 ( 460 )(1964 ) = 0.405 ( 25 )( 300 )
= 258.8 mm x 258 .8 = d 600
= 0.43 < 0.5
Moment of resistance,
M = Fst Z = 0.87 x 460 x 1964 (600 – 0.45 x 258.8) = 380.06 kNm
1
. ……………… ………………..
1
REINFORCED CONCRETE STRUCTURAL DESIGN
C4301/UNIT4/36
4. Fst = Fcc + Fsc 0.87 fy As = 0.405 fcu b x + 0.87 fy As’ 0.87 f y ( As − As ' )
x=
0.405 f cu b
0.87 × 460 (943 − 226 ) 0.405 × 25 × 250 =
= 113.4 mm x 113 .4 = = 0.25 < 0.5 d 450 d' 40 = = 0.35 < 0.43 x 113 .4
…………………………… 1
…………………………
1
1 …………...………………
Moment of resistance is, M = 0.87 f y As ' ( d −d ' ) +0.405 f cu bx ( d −0.45 x) = {0.87 x 460 x 226 x (450 - 40)} + {0.405 x 25 x 250 x 113.4 x [(450(0.45x113.4)]}
=
151.6 kNm
……………………………….
1
Total = 18 marks
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C4301/UNIT4/37
NOW TOTAL UP YOUR MARKS. FULL MARKS = 18 MARKS
Calculate your score as shown below: Score = total marks obtained 18
x 100%
You should score 80% or more to this unit. If you have scored 80% or more, you may proceed to unit 5. If you have scored less than 80%, you should go through unit 4 again until you this unit. See you in UNIT 5.
END OF UNIT 4
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