Analysis and Design of Roof Beams
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Transcript of Analysis and Design of Roof Beams
A B C
ANALYSIS AND DESIGN OF BEAMS
Roof beam
Roof beam: RB 1 and RB2
BS 8110 Reference
Calculations Output
Table 3.3
Specification:
Self weight of concrete = 24 KN/m3
f cu = 30mm
f y = 460 N/mm2
Fire resistance = 1 hours
Concrete exposure = mild
Cover = 25mm
Ceiling = 0.10KN/m2
Roof = 0.75 KN/m2
Tension φ = 12mm
Link φ = 8mm
f yv = 250 N/mm2
Case 1 : Adverse + adverse
2m 4m
Self weight of the concrete = 24 × 0.3 × 0.2 = 1.44KN/m
Dead load of roof + ceiling = (0.75 + 0.1 ) × 1m width = 0.85 KN/m
Live load = (0.5KN/m/m) 1m width = 0.5 KN/m
A AB C
Design load = 1.4Gk+1.6Qk
= 1.4( 1.44 + 0.85) + 1.6( 0.5) = 4.00KN/m
Gk= 0.85 KN/mQk= 0.5 KN/m
Design load= 4.00kN/m
4 KN/m
2m 4m
i) Fixed end momentFEMAB = FEMCB = 0
FEMBA = wL2
8 =
(4.00 )(2)2
8 = 2.00KN. m
FEMBC = -wL2
8 = -
(4.00 )(2)2
8 = - 8.00KN.m
ii) Stiffness factor for each member (K)
Kab = Kba = 3EIL
= 3EI
2 = 1.5EI
Kbc = Kcb = 3EIL
= 3EI
4 = 0.75EI
iii) Joint stiffness factor (∑K)∑ Kb = Kba + Kbc
= 1.5EI + 0.75EI = 2.25EI
iv) Distribution factor, DF = K∑K
DFab = 1
DFba = 1.5EI
2.25EI = o.67
DFbc = 0.75 EI2.25 EI
= 0.33
DFcb = 1
v) Carry over
0.5 0 0 0.5
A
2m 4m
Calculations
vi) Moment distribution table
Joint a COF b COF cSpan ab ba bc cbDF 1 0.67 0.33 1
FEM 0 2.00 -8.00 0Balance 0 6.00 6.00 0
Distribution 0 4.02 1.98 0Carry over 0 0 0 0
Final moment
0 6.02 -6.02 0
For span RB1(A-B)
4 KN/m 6.02 KN
B (L)
4 KN/m
A B C
2m 4m
6.02 KN 8kN
B
Taking moment at BL
+ MB = 0-2Vab + 8.00 – 6.02 = 0 VA = 0.99KN VB (L) = 7.01 KN
VA = 0.99KNVB (L) = 7.01 KN
Shear diagram
0.99x
= 7.012−x
1.98 – 0.99 x = 7.01x x = 0.2475
For left hand sideMax. moment when V = 0
+ Mmax. - 0.99 ( 0.2475 ) + 4 (0.2475)2
2 = 0
Mmax. = 0.1225KN.m
Mmax. = 0.1225KN.m
For right hand side
+ Mmax + 6.02 + 7.01( 1.7525
2 ) – 1.7525 ( 7.01 ) = 0
Mmax = 0.1225KN.m
0.99
2 - x
7.01
7.01KN
1.7525 m
For span RB2 ( B-C)
Taking moment at BR
+ MB (R) = 0-6.02 + 16(2) – 4Vc = 0 Vc = 6.495 KN VB (R) = 9.505 KN
Vc = 6.495 KN
VB (R) = 9.505 KN
9.505x
= 6.4954−x
38.02 – 9.505x = 6.495x x = 2.376m
4 m
16 KN
6.02KN.m
VB (R) Vc
4 KN/m
9.505
x
6.495
4 - x
+ Mmax + 6.02 + 9.504 (2.376
2) – 9.505 (2.376) = 0
Mmax = 5.273 KN.mMmax = 5.273 KN.m
+ Mmax + 6.496 (1.624
2) + 6.495 (1.624) = 0
Mmax = 5.273 KN.m
9.504 KN
2.376 m
6.02 KN.
Mmax
9.505
4 KN/m
6.496 KN
1.624 m
6.495 KN
4 KN/m
Mmax
Case 2 : For adverse + Beneficial
h = 300mb = 200m
Self weight of the concrete = 24 × 0.3 × 0.2 = 1.44 KN/m Gk = 0.85 KN/m
Qk =0.5 KN/mDead load of root + ceiling = (0.75 + 0.1) 1m width = 0.85 KN/m Live load = 0.5KN/m/m width = 0.5 ×1 m = 0.5 KN/m
Design load for span RB1( A – B) ( adverse )Design load = 1.4Gk+1.6Qk
= 1.4( 1.44 + 0.85) + 1.6( 0.5) = 4.00KN/m
Design load for span RB2(B – C) ( Beneficial )Design load = 1.0 Gk + 0 Qk
= 1.0 Gk
= 1.0 ( 1.44 + 0.85 ) = 2.29 KN/m
2m
w = 2.29 KN/mBeneficial
w = 4.00KN/m
4m CBA
Adverse
i) Fixed end momentFEMAB = FEMCB = 0
FEMBA = wL2
8 =
(4.00 )(2)2
8 = 2.00KN. m
FEMBC = -wL2
8 = -
(2.29 )(4 )2
8 = - 4.58KN. m
ii) Stiffness factor for each member (K)
Kab = Kba = 3EIL
= 3EI
2 = 1.5EI
Kbc = Kcb = 3EIL
= 3EI
4 = 0.75EI
iii) Joint stiffness factor (∑K)∑ Kb = Kba + Kbc
= 1.5EI + 0.75EI = 2.25EI
iv) Distribution factor, DF = K∑K
DFab = 1
DFba = 1.5EI
2.25EI = o.67
DFbc = 0.75 EI2.25 EI
= 0.33
DFcb = 1
v) Carry over
0.5 0 0 0.5
4m2m
A B C
4.00 KN/m 2.29 KN/m
vi) Moment distribution table
Joint a COF b COF cSpan ab ba bc cbDF 1 0.67 0.33 1FEM 0 2.00 -4.58 0Balance 0 2.58 2.58 0Distribution 0 1.729 0.851 0Carry over 0 0 0 0Final moment
0 3.729 -3.729 0
For span RB1 (A – B)
Taking moment at B1L
4m2m
8.00 KN
2 m
Mmax
VA
4 KN/m
BL
VB1(L)
3.729 KN
+ Mmax = 0-VA(2) + 8.00 – 3.729 = 0VA = 2.136 KNVB(L) = 5.864 KN
VA = 2.136 KNVB(L) = 5.864 KN
2.136x = 5.864
2−x
4.272 – 2.316x = 5.864x x = 0.534m
For left hand side
Max. moment
2.136
x
5.864
2 – x = 1.4762m
5.864 KN
1.466 m
5.864 KN
Mmax
3.729 KN.m
+ Mmax – 2.136 ( 0.534 ) + 4.0(0.534)2
2 = 0
Mmax = 0.57 KN.m
+ Mmax + 5.864 ( 1.466
2) – 5.864 (1.466) + 3.729 = 0
Mmax = 0.57 KN.m
Mmax = 0.57 KN.m
For span RB2(BR – C)
Taking moment at BR
+ MB (R) = 03.729 – 9.16 (2) +Vc1 = 3.65 KNVB (R) = 5.51 KN
VB (R) = 5.51 KN
5.51x
= 3.654−x
22.04 – 5.51x= 3.65x x = 2.406m
Right hand side
4 m
9.16 KN
3.729 KN.m
VB (R) Vc
2.29 KN/m
CBR
5.51
x
3.65
4 – x
+ Mmax – 3.65( 1.594
2 ) + 3.65 ( 1.594 ) = 0
Mmax = 2.90 KN.m
Mmax = 2.90 KN.m
Left hand side
+ Mmax + 3.729 + (5.51)(2.406
2) – (5.51)(2.406) = 0
Mmax = 2.90 KN.m
Case 3:For Beneficial + adverse
h = 300mb = 200m
Self weight of the concrete = 24 × 0.3 × 0.2 = 1.44 KN/m
Dead load of root + ceiling = 0.75 + 0.1 = 0.85KN/m/m width = 0.85× 1m = 0.85 KN/m
3.65 KN
1.594 m
3.65 KN
2.29 KN/mMmax
5.51 KN
2.406 m
Mmax
5.51 KN
2.29 KN/m3.729KNm
2m
Beneficial
4m CBA
adverse
Live load = 0.5KN/m/m width = 0.5 ×1 m = 0.5 KN/m Gk=0.85 KN/m
Qk=0.5 KN/mDesign load for span AB (Beneficial) Design load = 1.0 Gk + 0 Qk
= 1.0 Gk
= 1.0 ( 1.44 + 0.85 ) = 2.29 KN/m
Design load for span BC (Adverse)Design load = 1.4Gk+1.6Qk
= 1.4( 1.44 + 0.85) + 1.6( 0.5) = 4.00KN/m
i) Fixed end momentFEMAB = FEMCB = 0
FEMBA = wL2
8 =
(2.29 )(2)2
8 = 1.145 KN. m
FEMBC = -wL2
8 = -
(4.00 )(4)2
8 = - 8.00 KN. m
ii) Stiffness factor for each member (K)
Kab = Kba = 3EIL
= 3EI
2 = 1.5EI
Kbc = Kcb = 3EIL
= 3EI
4 = 0.75EI
iii) Joint stiffness factor (∑K)∑ Kb = Kba + Kbc
= 1.5EI + 0.75EI = 2.25EI
4m2m
CBA
4.00 KN/m2.29 KN/m
iv) Distribution factor, DF = K∑K
DFab = 1
DFba = 1.5EI
2.25EI = o.67
DFbc = 0.75 EI2.25 EI
= 0.33
DFcb = 1
vii) Carry over
0.5 0 0 0.5
v) Moment distribution tableJoint a COF b COF cSpan ab ba bc cbDF 1 0.67 0.33 1FEM 0 1.145 -8.00 0Balance 0 6.855 6.855 0Distribution 0 4.593 2.262 0Carry over 0 0 0 0Final moment
0 5.738 -5.738 0
For span RB1(A- BL)
4m2m
A B C
5.738 KN
Taking moment at B1L = 0+ MBL = 04.58(1) – 2VA1 – 5.738 = 0VA = -0.579 KNVBL = 5.159 KN
VA = -0.579 KNVBL = 5.159 KN
0.579x
= −5.159
2−x1.158 – 0.579x = 5.159xx = 0.2018Left Hand Side
VA 4.58 KN
2.29 KN/m
VB1(L)
2 m BL
5.159
2 - x
x
0.579
M = 0
0.462 KN
0.2018 m
Mmax
0.579 KN
2.29 KN/m
+ Mmax + 0.579(0.2018) + 0.462( 0.2018
2) = 0
Mmax = - 0.163 KN.mMmax = - 0.163 KN.m
Right Hand Side
+ Mmax + 4.118(1.7982
2) + 5.738 – 5.159(1.7982) = 0
Mmax = - 0.163KN.m
For span RB2(BR – C)
Taking moment at C + MC = 016(2) + 5.738 – 4VB1(R) = 0VB (R) = 9.435 KNVC = 6.565 KN
4.118 KN
1.7982 m
5.159 KN
2.29 KN/m
Mmax
5.738
4 m
16 KN
5.738 KN.m
VB(R) Vc
4.0 KN/m
CBR
9.435x
= 6.5654−x
37.74 – 9.435x = 6.565xx = 2.359 m
Left Hand Side
+ Mmax + 9.436( 2.359
2 ) – 9.435( 2.35 ) = 0
Mmax = 5.39 KN.m
Right Hand Side
Mmax + 6.564 ( 1.641
2 ) – 6.565 ( 1.641 ) = 0
Mmax = 5.39 KN.m
9.435
x
6.565
4 – x
6.565
9.436 KN
2.359 m
Mmax
9.435
5.738 KN.m
VB1(L)
4.0 KN/m
4 KN/m
1.641 m
6.565 KN6.564 KN
Mmax
2.90
5.273
Shear force envelope diagram
Moment force envelope diagram
Maximum shear diagram
Maximum moment diagram
Reinforced designFor sagging at B
3.729
9.505
7.01
0.99
6.565
0.579
5.8646.495
6.02
5.738
5.5193.65
0.57
5.51
0.1225
9.4352.136
5.39
Check xd
K = Mf cu bd2
= 6.02×106
(30 )(200)(261)2
= 0.0147< 0.156 ok
Clause 3.4.4.4
z = 261[ 0.5 + ¿ ] = 256.66 > 0.95d = (247.95) use 0.95d
ASreq = 6.02×106
(0.95 ) (460 )(247.95) = 55.58 mm2
Aprov = 226 mm2 2T12, Asprov = 226mm2
For span A – B Hogging
Check xd
K = Mf cu bd2
= 0.57×106
(30 )(200)(261)2
= 0.00139
K < K’( = 0.00139 ) okSo the compression reinforcement is not required
Find zClause 3.4.4.4 z = d [ 0.5 + √0.25− K
0.9 ]
= 261 [ 0.5 + √0.25−0.001390.9
]
= 260.60mm > ( 0.95d = 247.95mm )
>0.95d, so use 0.95d
Find AS
M = 0.95fyAsz
ASreq = 0.57×106
0.95 (460 )(247.95) = 5.26mm2
AS prov = 226mm2
2T12
For span B – C (Hogging)
Check xd
K = Mf cu bd2
K < K’( = 0.0129 ) okSo the compression reinforcement is not required
= 5.39×106
(30 )(200)(261)2
= 0.0132
Find zClause 3.4.4.4 z = d [ 0.5 + √0.25− K
0.9 ]
= 261 [ 0.5 + √0.25−0.01320.9
]
= 257.11mm > ( 0.95d = 247.95mm )
>0.95d, so use 0.95d
Find AS
M = 0.95fyAsz
ASreq = 5.273×106
0.95 (460 )(247.95) = 48.66mm2
As prov = 226mm2
2T12
Check max. & min. As
Table 3.25Min. As =
0.13bh100
= 0.13100
(200 )(300) = 78mm2
Max. As = 4
100bh =
4100
(200)(300) = 2400mm2
As min = 78 mm2
As max = 2400 mm2
Check deflectionFor span A – B ( Hogging )
( Ld
)basic = 26 ( For continues beam )
d = 261mmM
bd2 = 0.57×106
(200 )(261)2
= 0.04184
fs = 2 f y A s req
3 A s prov
= 2(460)(5.26)
3(226) = 7.137
Table 3.10 The modification factor for tension steelMFT > 2, so use 2
2T12
C1B1A1
= 0.55 +
477−f s
120(0.9+ Mbd2 )
= 0.55 + 477−7.137
120(0.9+0.04184 )= 4.707 > 2
( Ld
)allowable
= 26 × 2
= 52
( Ld
)actual
= ( 2000261
)
= 7.663
( Ld
)allowable
> (Ld
)actual
ok
For span B – CTable 3.10
( Ld
)basic = 26 ( For continues beam )
d = 261mmM
bd2 = 5.39×106
(200 )(261)2
= 0.3956
fs = 2 f y A s req
3 A s prov
= 2(460)(48.60)
3 (226) = 65.95 N/mm2
The modification factor for tension steel
= 0.55 +
477−f s
120(0.9+ Mbd2 )
= 0.55 + 477−65.9
120(0.9+0.3956)= 3.1939 > 2
More than 2, use 2
( Ld
)allowable
= 26 × 2
= 52
( Ld
)actual
= ( 4000261
)
= 15.33
( Ld
)allowable
> (Ld
)actual
ok
Check shearAt support AAs prov = 226mm2
Table 3.8 100 A s
bd =
100 (226)(200 )(261)
= 0.433 < 3
( 400d
)14 = ( 400
261)
14
= 1.1126 > 1
3√ f cu25 = 3√ 30
25 = 1.06
Vc = 0.79× 3√ 100 As
bd× 4√ 400
4×
3√ f cu25Υ m
= 0.79× 3√0.433×1.1126×1.06
1.25= 0.5639 N/mm2
ok
ok
Table 3.7V =
vbd
= 2.136×103
(200 )(261)= 0.0409 N/mm2
0.5Vc = 0.282N/mm2
0.5Vc >V, minimum shear is provided
At support BTake 9.05 KN to calculate the shear
V = vbd
= 9.505×103
(200 )(261) = 0.1820.5Vc > V
At support CTake 6.565 KN to calculate the shear
V = vbd
= 6.565×103
(200 )(261) = 0.1260.5Vc > V
Hence, for the beam RB1 and RB2. Minimum link should be provided in all beams of structural
importance, it will be satisfactory to omit them in members of minor structural importance such as lintels or where the maximum design shear stress is less than half Vc.
Table 3.7 A s
Sv =
0.4 bv0.9 f yv
Let link = 8 mmArea = 101 mm ( for 2 leg )101Sv
= 0.4(200)
0.95(250)
Sv = 299.84 mmMax. spacing = 0.75d = 195.75 mm
Hence use R8@175mm c/c
Crack checking [ only consider tension reinforcement]
Clear spacing = 200−2 (25 )−2 (8 )−2(12)
1 = 110 mm
Clear spacing ≤ 47000f s
≤ 300
4700065.95
= 7126.6 ≤ 300
Actual spacing ( = 110mm ) ≤ max. Spacing
Use type deformation type 2 (460) Use for round cross section with rip
Ka= 40( Tension)
Length of anchorgeL= Ka x bar sizeL= 40 x 12 = 480mm
So, we use max. 300 mm
CBA
Roof Beam: RB3 and RB 4
REFERENCES: Figure 9 (Appendix)
BS 8110 Reference
Calculations Output
Table 3.15
Specification:
Self weight of concrete = 24 KN/m3
f cu = 30mm
f y = 460 N/mm2
Fire resistance = 1 hours
Concrete exposure = mild
Cover = 25mm
Ceiling = 0.10KN/m2
Roof = 0.75 KN/m2
Tension φ = 12mm
Link φ = 8mm
f yv = 250 N/mm2
Table 3.9Estimation thickness of slab, h
With refer to beam RB3
( Ld
) = 20 ( For simply support)
d min = L
20×MF (Assume, MF = 0.80)
= 4000
20×0.80
= 250mm
25mm
8mm
110mm
4T12
Take d = 250mm,
Assume tension steel = 12mm
Link = 8mm
Cover = 25mm
h = 289mm
So, h = 300mm
b = 200mm
Loading
Self-weight of the concrete = 24 × 0.3 × 0.2 × 1.4
= 2.016KN/m
Dead load of roof + ceiling = 0.75 + 0.1
= 0.85KN/m/m width × 1m
= 0.85 KN/m width
Live load = 0.5KN/m/m width
Design load = 1.4(0.85) + 1.6(0.5)
= 1.99KN/m/m width
Point load of roofing + ceiling = 1.99 × 3.5 ×1
= 6.965KN
Taking moment at Vab = 0
MA1 = 0,
-6.965(1) - 6.965(2) – 6.965(3) + 4 VA2 – 2(8.064) =0
V A2 = 14.48KN
V A1 = 14.48KN
+ F y = 0; 14.48 – 2.016 - V aL = 0
VA1 VA2
1m 1m 1m 1m
6.965KN 6.965KN 6.965KN
8.064KN
+
14.48KN
1m 1m 1m 1m
8.064KNA2
6.965KN
ca
A1
b
6.965KN 6.965KN
D
V aL = 12.464KN
+ F y = 0; 14.48 – 2.016 – 6.695 – V aR =0
V aR = 5.499KN
+ F y = 0; 14.48 – 4.032 – 6.695 – V DL =0
V DL = 3.483KN
+ F y = 0; 14.48 – 4.032 – 6.965 – 6.965 - V DR = 0
V DR = -3.482KN
+ F y = 0; 14.48 – 6.048 – 6.695 – 6.695 - V CL =0
V CL = -5.498KN
+ F y = 0; 14.48 – 6.048 – 6.965(3) - V CR = 0
V CR = -12.463KN
+ F y = 0; 14.48 – 8.064 – 3(6.965) – V A2L =0
V A2L = -14.48KN
Shear Diagram:
Bending moment diagram
14.48
12.464
5.4993.483
-3.482
-5.498
-14.48
Hence maximum moment = 17.963 KNm
Check xd
K = M
f cub d2
= 17.963×106
(30 )(200)(261)2
= 0.0439
d = h – 25 – 8 – 6
= 300 – 25 – 8 – 6
= 261 mm
K ¿ K’ ( = 0.156 ) -ok-xd≤ 0.5 -ok-
So, the compression reinforcement is not required.
Find z
z = d [ 0.5 + √0.25− K0.9
]
= 261 [ 0.5 +√0.25−0.04390.9
]
= 247.58 mm ¿ 0.95d = 247.95 mm
¿ 0.95d, Use z = 247.58 mm
Find As :
M = 0,95 f y A s Z
A s req = M
0.95 fy Z
A s prov = 226 mm2
2T12
13.472
17.963
13.473
0
= 17.963×10 6
0.95(460)(247.58)
= 166.03 mm2
Check max and min:
Min As = 0.13100
× b × h
¿0.13100
× 200 × 300
= 78 mm2
Min As = 4
100 × b × h
¿4
100 × 200 × 300
= 2400 mm2
Check deflection:
( Ld ) basic = 20 ( for beam simply supported)
( Ld
) actual = Ld
= 4000261
= 15.326 d = 261 mm
( Mbd2
) = 17.963×10 6(200 ) (261 ) 2
= 1.318
f s = 2 f y A sreq3 As prov
= 2(460)(166.03)
3(226)
As min = 78.0 mm2
As max = 2400 mm2
-ok-
= 225.29 N/mm2
The modification factor for tension steel
= 0.55 +
477−f s
120(0.9+ Mbd 2 )
= 0.55 + 477−225.29
120 (0.9+1.318 )
= 1.4957 < 2
( Ld ) allowable = 20× 1.4957
= 29.914
( Ld ) actual =
4000261 = 15.326
Check Shear:
As = 226mm2
100 A sbd
= 100(226)200(261)
= 0.433 < 3
(400d ) ¼ = (
400261 ) ¼
= 1.1126 ¿ 1
3√ f y25 = 3√ 30
25 = 1.06
v c = 0.79×
3√ 100 Asbd
×4√ 400d
×3√ f cu25
γm
( Ld ) all ¿ (
Ld ) actual
-ok-
-ok-
-ok-Ok for not greater than 40 N/mm2
= 0.79×√0.433×1.1126×1.06
1.25 = 0.5639 N/mm2
v = Vbd
V = 15.49 KN
= 14.48×103(200 )(261)
= 0.2774 N/mm2
0.5Vc = 0.282 N/mm2
Note 1 : Minimum links should be provided in all beams of structural importance, it will be satisfactory to omit them in members of minor structural importance such as lintels or where the maximum design shear stress is less than half Vc . A sSv
= 0.4bv
0.95 fyv
Assume link = 8mm,
Area = 101 mm2 ( for 2 legs )
101Sv
= 0.4(200)
0.95(250)
S v = 299.84 mm
Hence, max spacing = 0.75 d
= 195.75mm
Apply the max R8 @175mm c/c
0.5Vc ¿ V
R8 @175mm c/c
Crack checking ( only consider tension reinforcement )
Clear spacing = 200−2 (25 )−2 (8 )−2(12)
1 = 110 mm
Clear spacing ≤ 47000f s
≤ 300
47000225.29
(¿208.62 )≤ 300
actual spacing ≤ max. spacing 110mm = 208.62mm
-ok-
2T12
Use type deformation type 2 (460) Use for round cross section with rip
Ka= 40( Tension)
Length of anchorgeL= Ka x bar sizeL= 40 x 12 = 480mm
8 mm25 mm
110 mm