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A B C ANALYSIS AND DESIGN OF BEAMS Roof beam Roof beam: RB 1 and RB2 BS 8110 Referenc e Calculations Output Table 3.3 Specification: Self weight of concrete = 24 KN/m 3 f cu = 30mm f y = 460 N/mm 2 Fire resistance = 1 hours Concrete exposure = mild Cover = 25mm Ceiling = 0.10KN/m 2 Roof = 0.75 KN/m 2 Tension φ = 12mm Link φ = 8mm f yv = 250 N/mm 2 Case 1 : Adverse + adverse 2m 4m
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ANALYSIS AND DESIGN OF BEAMS Roof beam Roof beam: RB 1 and RB2

BS 8110 Reference Specification:

Calculations

Output

Self weight of concrete = 24 KN/m3 f cu = 30mm f y = 460 N/mm2 Table 3.3 Fire resistance = 1 hours Concrete exposure = mild Cover = 25mm Ceiling = 0.10KN/m2 Roof = 0.75 KN/m2 Tension = 12mm Link = 8mm f yv = 250 N/mm2

A

B

C

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 Gk= 0.85 KN/m Qk= 0.5 KN/m

Design load = 1.4Gk+1.6Qk = 1.4( 1.44 + 0.85) + 1.6( 0.5) = 4.00KN/m 4 KN/m

A

A 2m

B

C

4m

i) Fixed end moment FEMAB = FEMCB = 0 FEMBA = FEMBC = =( ( )( )

= 2.00KN. m = - 8.00KN.m

=-

)( )

ii) Stiffness factor for each member (K) Kab = Kba = = = 1.5EI Kbc = Kcb = = = 0.75EI

iii) Joint stiffness factor ( K) Kb = Kba + Kbc = 1.5EI + 0.75EI = 2.25EI iv) Distribution factor, DF = = o.67 = 0.33

DFab = 1 DFba = DFbc = DFcb = 1 v)

Carry over 0.5 0 0 0.5

A

B

2m

4m

Calculations vi) Joint a Span DF FEM Balance Distribution Carry over Final moment Moment distribution table COF ab 1 0 0 0 0 0 ba 0.67 2.00 6.00 4.02 0 6.02 b bc 0.33 -8.00 6.00 1.98 0 -6.02 COF c cb 1 0 0 0 0 0

For span RB1(A-B) 4 KN/m 6.02 KN

B (L)8kN

4 KN/m 6.02 KN

A 2m

B 4m

C

Taking moment at BL + MB = 0 -2Vab + 8.00 6.02 = 0 VA = 0.99KN VB (L) = 7.01 KN Shear diagram 0.99 2-x

VA = 0.99KN VB (L) = 7.01 KN

7.01

= 1.98 0.99

= 7.01 = 0.2475

For left hand side Max. moment when V = 0 + Mmax. - 0.99 ( 0.2475 ) + Mmax. = 0.1225KN.m For right hand side 7.01KN( )

=0

Mmax. = 0.1225KN.m

1.7525 m

+ Mmax + 6.02 + 7.01( Mmax = 0.1225KN.m

) 1.7525 ( 7.01 ) = 0

For span RB2 ( B-C) 6.02KN.m 4 KN/m

VB (R)

4m 16 KN

Vc

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

9.505 4-x x 6.495 = 38.02 9.505

= 6.495 = 2.376m

6.02 KN.

4 KN/m Mmax 2.376 m

9.505

9.504 KN

+ Mmax + 6.02 + 9.504 ( Mmax = 5.273 KN.m

) 9.505 (2.376) = 0

Mmax = 5.273 KN.m

4 KN/m Mmax 1.624 m 6.496 KN

6.495 KN

+ Mmax + 6.496 ( Mmax = 5.273 KN.m

) + 6.495 (1.624) = 0

Case 2 : For adverse + Beneficial w = 4.00KN/m Adverse 2m A B w = 2.29 KN/m Beneficial

4m C

h = 300m b = 200m Self weight of the concrete = 24 0.3 0.2 = 1.44 KN/m Dead 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

Gk = 0.85 KN/m Qk =0.5 KN/m

4.00 KN/m

2.29 KN/m

A 2m

B 4m

C

i) Fixed end moment FEMAB = FEMCB = 0 FEMBA = FEMBC = =( ( )( )

= 2.00KN. m = - 4.58KN. m

=-

)( )

ii) Stiffness factor for each member (K) Kab = Kba = = = 1.5EI Kbc = Kcb = = = 0.75EI

iii) Joint stiffness factor ( K) Kb = Kba + Kbc = 1.5EI + 0.75EI = 2.25EI iv) Distribution factor, DF =

DFab = 1 DFba = DFbc = DFcb = 1 v) Carry over 0.5 0 0 0.5 = o.67

= 0.33

2m

4m

vi)

Moment distribution table

Joint a Span DF FEM Balance Distribution Carry over Final moment

COF ab 1 0 0 0 0 0 ba 0.67 2.00 2.58 1.729 0 3.729

b

COF c bc cb 0.33 1 -4.58 0 2.58 0 0.851 0 0 0 -3.729 0

For span RB1 (A B)

4 KN/m Mmax 2m 3.729 KN BL VB1(L)

VA

8.00 KN

Taking moment at B1L + Mmax = 0 -VA(2) + 8.00 3.729 = 0 VA = 2.136 KN VB(L) = 5.864 KN

VA = 2.136 KN VB(L) = 5.864 KN

2.136 2 x = 1.4762m x

5.864

=4.272 2.316 = 5.864 = 0.534m

For left hand side

3.729 KN.m

Mmax 1.466 m 5.864 KN Max. moment + Mmax 2.136 ( 0.534 ) + Mmax = 0.57 KN.m + Mmax + 5.864 ( Mmax = 0.57 KN.m For span RB2(BR C) 3.729 KN.m 2.29 KN/m( )

5.864 KN

=0

) 5.864 (1.466) + 3.729 = 0

Mmax = 0.57 KN.m

BR 4m VB (R) 9.16 KN Vc

C

Taking moment at BR + MB (R) = 0 3.729 9.16 (2) +Vc1 = 3.65 KN VB (R) = 5.51 KN 5.51 4x x

VB (R) = 5.51 KN

3.65

= 22.04 5.51

= 3.65 = 2.406m

Right hand side

2.29 KN/m Mmax 1.594 m 3.65 KN 3.65 KN

+ Mmax 3.65( Mmax = 2.90 KN.m Left hand side 3.729KNm

) + 3.65 ( 1.594 ) = 0

Mmax = 2.90 KN.m

2.29 KN/m Mmax 2.406 m

5.51 KN

5.51 KN ) (5.51)(2.406) = 0

+ Mmax + 3.729 + (5.51)( Mmax = 2.90 KN.m

2m A h = 300m b = 200m B

4m C

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 Live load = 0.5KN/m/m width = 0.5 1 m = 0.5 KN/m Design 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

Gk=0.85 KN/m Qk=0.5 KN/m

2.29 KN/m

4.00 KN/m

A 2m

B 4m

C

i) Fixed end moment FEMAB = FEMCB = 0 FEMBA = FEMBC = =( ( )( )

= 1.145 KN. m = - 8.00 KN. m

=-

)( )

ii) Stiffness factor for each member (K) Kab = Kba = = = 1.5EI Kbc = Kcb = = = 0.75EI

iii) Joint stiffness factor ( K) Kb = Kba + Kbc = 1.5EI + 0.75EI = 2.25EI iv) Distribution factor, DF =

DFab = 1 DFba = DFbc = DFcb = 1 vii) Carry over 0.5 0 0 0.5 = o.67

= 0.33

2mA B

4mC

v) Moment distribution table Joint a COF b COF c Span ab ba bc cb DF 1 0.67 0.33 1 FEM 0 1.145 -8.00 0 Balance 0 6.855 6.855 0 Distribution 0 4.593 2.262 0 Carry over 0 0 0 0 Final 0 5.738 -5.738 0 moment

For span RB1(A- BL) 5.738 KN 2.29 KN/m M=0 2m 4.58 KN BL VB1(L)

VA

Taking moment at B1L = 0 + MBL = 0 4.58(1) 2VA1 5.738 = 0 VA = -0.579 KN VBL = 5.159 KN

VA = -0.579 KN VBL = 5.159 KN

5.159

x 2-x 0.579

= 1.158 0.579 = 5.159 = 0.2018 Left Hand Side

2.29 KN/m Mmax 0.2018 m 0.462 KN

0.579 KN

+ Mmax + 0.579(0.2018) + 0.462( Mmax = - 0.163 KN.m Right Hand Side

)=0

Mmax = - 0.163 KN.m

2.29 KN/m Mmax 1.7982 m 4.118 KN + Mmax + 4.118( Mmax = - 0.163KN.m For span RB2(BR C) 5.738 KN.m

5.738

5.159 KN

) + 5.738 5.159(1.7982) = 0

4.0 KN/m

BR 4m VB(R) 16 KN Vc

C

Taking moment at C + MC = 0 16(2) + 5.738 4VB1(R) = 0 VB (R) = 9.435 KN VC = 6.565 KN 9.435 4x x

6.565 6.565

= 37.74 9.435 = 2.359 m Left Hand Side

= 6.565

5.738 KN.m 4.0 KN/m Mmax 2.359 m 9.436 KN VB1(L)

9.435

+ Mmax + 9.436( Mmax = 5.39 KN.m Right Hand Side

) 9.435( 2.35 ) = 0

4 KN/m Mmax 1.641 m 6.564 KN 6.565 KN

Mmax + 6.564 ( ) 6.565 ( 1.641 ) = 0 Mmax = 5.39 KN.m 9.505 2.136 9.435 5.51 0.99 0.579 5.519 5.864 7.01 Shear force envelope diagram 6.02 5.738 3.7292.90

3.65 6.495 6.565

0.1225 0.57

5.273 5.39

Moment force envelope diagram

Maximum shear diagram

Maximum moment diagram

Reinforced design For sagging at B Check K= =()( )( )

= 0.0147< 0.156 ok Clause 3.4.4.4 z = 261[ 0.5 + ]

= 256.66 > 0.95d = (247.95) use 0.95d

=(

)(

)(

)

= 55.58 = 226 For span A B Hogging Check K= =()( )( )

2T12, Asprov = 226mm2

K < K( = 0.00139 ) ok So the compression reinforcement is not required

= 0.00139 Find z Clause 3.4.4.4 z = d [ 0.5 + = 261 [ 0.5 + ] ] >0.95d, so use 0.95d

= 260.60mm > ( 0.95d = 247.95mm ) Find AS M = 0.95fyAsz =( )( )

= 5.26 For span B C (Hogging) Check K= =()( )( )

AS prov = 226mm2 2T12

K < K( = 0.0129 ) ok So the compression reinforcement is not required

= 0.0132

Find z Clause 3.4.4.4 z = d [ 0.5 + = 261 [ 0.5 + ] ] >0.95d, so use 0.95d

= 257.11mm > ( 0.95d = 247.95mm ) Find AS M = 0.95fyAsz =( )( )

As prov = 226mm2 2T12

= 48.66 Check max. & min. As ( Min. As = = Max. As = = As min = 78 mm2 As max = 2400 mm2

Table 3.25

)(

) = 78mm2

(200)(300) = 2400mm2

2T12

A1

B1

C1

Check deflection For span A B ( Hogging ) ( )basic = 26 ( For continues beam ) d = 261mm =( fs = = Table 3.10( ( )( ) ) )( )

= 0.04184

= 7.137 The modification factor for tension steel = 0.55 +( ( )

MFT > 2, so use 2

= 0.55 + = 4.707 > 2

)

( ) ( )

= 26 = 52 =( ) = 7.663

2

( ) ok

>( )

Table 3.10

For span B C ( )basic = 26 ( For continues beam ) d = 261mm =( fs = =( ( )( ) ) )( )

= 0.3956

= 65.95 N/mm2 The modification factor for tension steel = 0.55 +( ( )

More than 2, use 2

= 0.55 +

)

= 3.1939 > 2 ( ) = 26 2 = 52 ( ) =( ) = 15.33 Check shear At support A As prov = 226mm2 ( ) = ( )( ) = 0.433 < 3 ( )

( ) ok

>( )

Table 3.8

ok

=( ) = 1.1126 > 1

ok

= = 1.06

Vc = = = 0.5639 N/mm2

Table 3.7

V= =( )(

0.5Vc >V, minimum shear is provided)

= 0.0409 N/mm2 0.5Vc = 0.282N/mm2 At support B Take 9.05 KN to calculate the shear V = =( )( )

= 0.182 0.5Vc > V At support C Take 6.565 KN to calculate the shear V = =( )( )

= 0.126 0.5Vc > 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 = Let link = 8 mm Area = 101 mm ( for 2 leg ) ( ) = ( ) Sv = 299.84 mm Max. spacing = 0.75d = 195.75 mm Hence use [email protected] c/c

A

B

C

Crack checking [ only consider tension reinforcement] ( ) ( ) ( ) Clear spacing = = 110 mm Clear spacing 300 = 7126.6 300 Actual spacing ( = 110mm ) max. Spacing

So, we use max. 300 mm

Use type deformation type 2 (460) Use for round cross section with rip Ka= 40( Tension) Length of anchorge L= Ka x bar size L= 40 x 12 = 480mm

8mm 4T12 110mm 25mm

Roof Beam: RB3 and RB 4 REFERENCES: Figure 9 (Appendix) BS 8110 Reference Specification: Self weight of concrete = 24 KN/m3 f cu = 30mm Table 3.15 f y = 460 N/mm2 Fire resistance = 1 hours Concrete exposure = mild Cover = 25mm Ceiling = 0.10KN/m2 Roof = 0.75 KN/m2 Tension = 12mm Calculations Output

Link = 8mm f yv = 250 N/mm2 Estimation thickness of slab, h Table 3.9 With refer to beam RB3 ( ) = 20 ( For simply support) d min = = = 250mm (Assume, MF = 0.80)

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

6.965KN

6.965KN

6.965KN

8.064KN

1m VA1

1m

1m

1m VA2

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

6.965KN

6.965KN

6.965KN

a A1 1m 14.48KN 1m

b

8.064KN

c

D A2

1m

1m 14.48KN

+ F y = 0; 14.48 2.016 - V aL = 0 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:

14.48 12.464 5.499 3.483

-3.482

-5.498 -14.48

Bending moment diagram

17.963

13.472

13.473

0

Hence maximum moment = 17.963 KNm Check K= =()( )( )

K ( = 0.156 ) -ok0.5 -okSo, the compression reinforcement is not required. K

= 0.0439 d = h 25 8 6 = 300 25 8 6 = 261 mm

Find z z = d [ 0.5 + ] 0.95d, Use z = 247.58 mm ] 0.95d = 247.95 mm A s prov = 226 mm2 2T12

= 261 [ 0.5 + = 247.58 mm Find As : M = 0,95 f y A s Z A s req = =

(

)(

)

= 166.03 mm2

Check max and min: Min As =

bh 200 300

As min = 78.0 mm2 As max = 2400 mm2 -ok-

= 78 mm2 Min As =

bh 200 300

= 2400 mm2

Check deflection:

( ) basic = 20 ( for beam simply supported)( ) actual = = = 15.326 d = 261 mm ( ) =( )( )

= 1.318 f s= =( )( ( ) )

= 225.29 N/mm2 The modification factor for tension steel

= 0.55 +

(

)

= 0.55 +

(

)

= 1.4957 < 2

( ) all-ok-

( ) actual

( ) allowable = 20 1.4957= 29.914

( ) actual =Check Shear: As = 226mm2

= 15.326

-ok=( ( ) )

= 0.433 < 3

(

)=(= 1.1126

)1

-okOk for not greater than 40 N/mm2

= = 1.06

vc=

=

= 0.5639 N/mm2 v= V = 15.49 KN

=

(

)(

)

= 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 . = Assume link = 8mm, Area = 101 mm2 ( for 2 legs ) =( ( ) )

0.5Vc

V

S v = 299.84 mm Hence, max spacing = 0.75 d = 195.75mm

Apply the max R8 @175mm c/c R8 @175mm c/c

Crack checking ( only consider tension reinforcement ) ( ) ( ) ( ) Clear spacing =

= 110 mm Clear spacing ( ) 300 300 -ok-

actual spacing max. spacing 110mm = 208.62mm

25 mm

8 mm

2T12

110 mm

Use type deformation type 2 (460) Use for round cross section with rip Ka= 40( Tension) Length of anchorge L= Ka x bar size L= 40 x 12 = 480mm