Design of Beam-USD

8/10/2019 Design of Beam-USD

1/94

COMPUT TION FOR BE M USING UNIVERS L STRENGTH DESIGN

DESIGN OF FLOOR BEAM

(Considering Critical Beam Only along Row E)

DESIGN CRITERIA:f'c = 20.7 MPa

fy = 275 Mpa

DESIGN LOADS:

Dead Loadings;( assumed section 20cm x 30 cm)

Self weight = 0.20mx0.30mx 2.3x9.81 = 1.412 kN/m

a. Dead Loads

1. Weight of Concrete = 23.60 kN/m3

2. Utilities = 135 Pa

3. Wt. of Ceiling = 120 Pa

4. Hard Wooden flooring frames= 190 Pa

5. Floor finish = 158 kPa

6. Wood studs partition = 960 Pa

Total deadload = 1563 Pa

Total distributed load = 1563 (3.00 ) = 4689 N/m = 4.69 kN/m

Live Loads:

Floor Live LoadBedroom = 750 Pa

Hallway = 750 Pa

Stairs = 750 Pa

a) Factored load, Wu = 1.4(DL) + 1.7(LL)

Wu = 1.4(4.69) + 1.7(.75) = 7.841 kN/m

1) Max Positive bending moment

Max Mu = -0.10wL2kN.m.+Pab

2/L

2

b) Compute Ultimate moment, Mu

Mu = 0.10(7.841) (4.25)2+ 3.55(2.6) (1.4)

2/4

= 16.26 kN-/m


2/94


c) Compute maximum steel ratio, max = 0.75 b

b = 0.85 (20.7) (600) (0.85) = 0.037275(600+275)

max = 0.75(0.037) = 0.028Assume = (max) = max (0.028) = 0.014

min = 1.4 = 0.005 < min OK275

d) Compute strength ratio, w

w = (FY) = 0.011(275) = 0.186f'c 20.7

Compute for Ultimate reaction, Ru

Ru = f'c w (1- 0.59w)= 20.7 (0.18) (10.59(0.1860)) = 3.317 Mpa

e) Compute depth of beam (d)

Mu = Ru b d2 assume b = d

16.26 x 106

= 0.90 (3.3173) d (d)2

16.26 x 106

= 1.7712 d3

d3

= 10892392.41 mm

d = 221.67 say; 250 mm.B = 2/3 (275) = 183.330 mm say 200.

H = 250 + 30 +20 = 300 mm.

g) Compute new resisting force based bd.

MuT = Ru b d2

16.26 x 106= 0.90(200) (250)

2Ru

Ru = 1.445 Mpa

h) Compute for new steel ratio/ actual steel ratio, = 0.85fc [1 - 1 2 Ru]

FY 0.85f'c

= 0.85(20.7) [1 - 1 2 (1.445)] = 0.0053

275 0.85(20.7)

min < < max ok (singly Reinforced design)


3/94


I) compute steel area requirement; use 16 mm. Dia. Rebars

As = bd

= 0.0053 (200) (250) = 267.07 mm2

No. of Bars = As / Ab= 267.07 = 1.32 pcs.

(16)2

4

Say; 3 pcs16mm. Dia. RSB

Actual As = As

A16 =201.06 (3) = 603.16 mm2

> 267.07 mm2

2) Check for Ultimate moment Capacity, Mu

Mu = Asfy [da / 2]

Mu = 0.90 (603.16) (275) [250120/2]

Mu = 28.36 kN.m > MuT = 16.26 kN-mOk. (SAFE)

Use 2 pcs16 mm dia cont. main bot. bars and 1 -16 mm extr a bars and

216mm dia cont. top bars near the suppor t w/ 1 -16 mm extr a top bars

(To be applied to all beams of the structu re)

k) Check for web Reinforcement

Vu = (WuL) + Pab2/L3 (3a + b) - Wud

2

Vu = 7.841(4.25) + (3.55) (1.8) (2.43)2/ (4.25)3

(3(1.8) + 2.43 - 7.841(0.250)

2

Vu = 14.76 kN.

VC = 1/6 f'c (b) (d) = 1/6 20.7 (200) (250) = 37.92 KN

VC = 0.85 (37.92) = 16.116 KN > Vu (no need web reinforcements)

2 2

But Provide close stirrups w/ the ff spacing:

10 mm 1 @ 50cm; 5@ 150 cm, rest @ 200 cm

TO BE APPLIED/USED FOR ALL FLOOR BEAMS


4/94


2) Max. Positive moment ;

Max Mu = -0.1071wL2+2Pa

2b

2/L

3

b) Compute Ultimate moment, Mu

Mu = 0.1071(1.9768) 42+2(97.88) (2.6)

2(1.4)

2/ (4)

3

= 43.907 kN-m


max = 0.75 (20.7) (600) (0.85) = 0.0189414(600+414)

max = 0.75(0.01891) = 0.0141

Assume = (max) = max (0.0141) = 0.0071

min = 1.4 = 0.0034 < max OK414


w = (FY) = 0.0071(414) = 0.1415f'c 20.7

Compute for Ultimate reaction, Ru

Ru = f'c w (1- 0.59w)

= 20.7 (0.1415) (10.59(0.1415) = 2.6846 Mpa



43.907 x 106

= 0.90 (2.6846) d (d)2

43.907 x 106

= 1.208 d3

d3

= 36346854.3 mm

d = 331.25 say; 420 mm.


5/94


B = (420) = 210.

H = 420 + 40 +20 = 480 mm.

g) Compute new resisting force based bd.

MuT = Ru bd2

43.907 x 106= 0.90(220) (420)

2Ru

Ru = 1.257 Mpa

h) Compute for new steel ratio/ actual steel ratio,

= 0.85fc [1 - 1 2 Ru]

FY 0.85f'c

= 0.85(20.7) [1 - 1 2 (1.257)] = 0.0032

414 0.85(20.7)min < < max ok (singly Reinforced design)


As = bd

= 0.0032 (220) (420) = 295.68 mm2

No. of Bars = As / Ab

= 295.68 = .9412 pcs.

(20)2

4Say; 3pcs20mm. Dia. RSB

Actual As = AsA16 =314.16(3) = 942.48 mm

2> 295.68 mm

j) Investigate the beam

From StressStrain Diagram

1) Check if steel yeilds; T = C

Asfy = 0.85 f'c a b

a = 942.48 mm2(414) = 100.8 mm Say; 101 mm.

0.85(20.7) (220)

a = c ; c = 101 = 118.82 mm. Say 120.00 mm0.85

By Ratio and Proportion from strain diagram

s = 0.003(dc)C


6/94


s = 0.003(420120) = 0.0075120

y = 414 = 0.00207 < s Ok, (Steel Yield)200,000


Mu = Asfy[da / 2]

Mu = 0.90 (942.48) (414) [420120/2]

Mu = 136.42 kN.m > MuT = 127.86 kN-m Ok. (SAFE)

k) Check for web Reinforcement

Vu = (WuL) + Pab2/L3 (3a + b) - Wud

2

Vu = 1.9768(4) + (97.88) (2.6) (1.4)2/ (4)3

(3(2.6) + 2.4 - 1.9768(0.420)

2

Vu = 82.50 kN.

VC = 1/6 f'c (b) (d) = 1/6 20.7 (220) (420)

= 80.07 KN

VC = 0.85 (80.07) = 34.030 KN < Vu (need web reinforcements)

2 2

Vs = Vu - VC = 82.5 - 80.07

0.85

= 16.98 KN

Compute Spacing of web reinforcements / stirrups; use 10 mm. Dia. Bars

S = Av FY d Av = (10)2 = 78.54 mm2

Vs 4

S = 78.54 (276) (420) = 536.18 mm.

16.98(1000)

Check Max Spacing.1/3 f'c (b) (d) = 1/3 20.7 (220) (420)

= 140.13 KN > Vs

Therefore Max Spacing = d/2

= 420 = 210 mm.2

S = 10 mm. Dia. RSB


7/94


Provide close stirrups:

10 mm 1 @ 50cm; 5@ 150 cm, rest @ 200 cm

DESIGN OF FLOOR BEAMS / GIRDER

DESIGN CRITERIA

f'c = 20.7MPaFY = 275 Mpa

Dead Loadings;

Window, glass frames and sash = 0.38 kPa

Self weight assumed section (0.220mx.350mx2.3) = 0.177kPa

FOR BEAM, FB1

Ws-6 DL = 2.9542(2.5) [3(0.625)2] = 3.215 kN/m3 2

Ws-6 LL = 4.8 (2.5) [3(0.625)2] = 5.218kN/m3 2

a) Factored load, Wu = 1.4(DL) + 1.7(LL)

Wu = 1.4(3.215) + 1.7(5.218) = 13.373 kN/m

Design the beam

1) Negative moment @ support;

Max Mu = -0.10wL2kN.m.

b) Compute Ultimate moment, Mu = -0.10wL2

Mu = 0.10(13.373.11) 42

= 21.3967 kN/m



8/94


b = 0.85 ((20.7)) (600) (0.85) = 0.0214414(600+414)

max = 0.75(0.0214) = 0.0160

Assume = (max) = max (0.016) = 0.008

min = 1.4 = 0.0034 < min OK414


w = (FY) = 0.008(414) = 0.1603f'c 20.7

Compute for Ultimate reaction, RuRu = f'c w (1- 0.59w)

= 20.7 (0.1603) (10.59(0.1603)) = 3.0043 Mpa



21.3967 x 106

= 0.90 (3.0043) d (d)2

21.3967 x 106

= 1.3520 d3

d

3

= 15825961.23 mmd = 251.06 say; 290 mm.

B = (290) = 145 mm say 200.

H = 290 + 40 +20 = 350 mm.

h) Compute for new steel ratio/ actual steel ratio,

= 0.85fc [1 - 1 2 Ru]

FY 0.85f'c

= 0.85(20.7) [1 - 1 2 (3.043)] = 0.0081

414 0.85(20.7)



As = bd

= 0.0081 (200) (290) = 469.8 mm2


9/94



= 469.8 = 1.49 pcs.(20)2

4

Say; 2pcs20 mm. Dia. RSB

Actual As = As

A20 =314.16 (2) = 628.32 mm2

> 402.24 mm2



1) Check if steel yields; T = C

Asfy = 0.85 f'c a b

a = 628.32 mm2(414) 73.92 mm Say; 74 mm.

0.85(20.7) (200)

a = c ; c = 74 = 87.06 mm. Say 88 mm0.85


s = 0.003(dc)C

s = 0.003(29088) = 0.0069

88y = 414 = 0.00207 < s Ok, (Steel Yeild)

200,000


Mu = Asfy[da / 2]

Mu = 0.90 (628.32) (414) [29074/2]

Mu = 59.23kN.m > 21.3967 MuT Ok. (SAFE)

k) Check for web ReinforcementVu = WuL - Wud

2

Vu = 13.373 (4) - 13.373(0.29)

2

Vu = 22.867 kN.

VC = 1/6 f'c (b) (d) = 1/6 28 (200) (290)


10/94


= 51.15 KN

VC = 0.85 (51.15) = 21.74 KN < Vu (need web reinforcements)

2 2

Vs = Vu - VC ; 51.15 - 43.478 0.85

= 16.70Kn


S = Av FY d Av = (10)2 = 78.54 mm2

Vs 4

S = 78.54 (276) (290) = 376.43 mm. say 350mm

16.70(1000)

Check Max Spacing.

1/3 f'c (b) (d) = 1/3 20.7 (200) (290)

= 87.96 KN > Vs


= 350 = 175 mm.

2

Provide close stirrups:

10 mm 1 @ 50cm; 5@ 150 cm, rest @ 200 cm

FOR BEAM, FB2

a) Slab loading

Ws-7 DL = 3.9542(2) [3(0.5)2] (2) = 6.73 kN/m3 2

Ws-6 LL = 4.8 (2) [3(0.5)2] (2) = 8.16 kN/m3 2

b) Factored load, Wu = 1.4(DL) + 1.7(LL)

Wu = 1.4 (6.73) + 1.7 (8.16) = 23.294 kN/m

c) Compute beam weight, Consider 1 meter strip

Use d = 300 mm.

B = 150 mm.

H = 300 + 40 + 10 + 10 = 360 mm.

Wb = (0.36) (0.15) (23.544) = 1.27 kN/m


11/94


Factored Load = 1.4(1.27) = 1.80 kN/m

WuT = 1.8 + 23.294 = 25.10 kN/m

MuT =Wu L2

= 0.90(25.1) (32) = 25.4 kN.m

8 8

d) Solve for Ru (Ultimate Reaction)

25.4 x 106 =

0.90 (150) (300)2

Ru

Ru = 2.091 MPa

e) Solvefor actual steel ratio,

= 0.85fc [1 - 1 2 Ru]FY 0.85f'c

= 0.85(28) [1 - 1 2(2.091)] = 0.0053414 0.85(28)


f) Compute steel area requirement; use 16 mm. Dia. Rebars

As = bd

= 0.0053 (150) (300) = 238.5 mm2


= 238.5 = 1.20 pcs.

(16)2

4Say; 2 pcs16 mm. Dia. RSB

Actual As = As = 2 [(16)2]= 402.12 mm2

4

g) Investigate the beam


1) Check if steel yields; T = C

Asfy = 0.85 f'c a b

a = 402.12 (414) = 46.63 mm Say; 47 mm.0.85(28) (150)

a = c ; c = 47 = 55.3 mm. Say 56 mm0.85


s = 0.003(dc)C


12/94


s = 0.003(30056) = 0.013156

y = 414 = 0.00207 < s Ok, (Steel Yeild)200,000


Mu = Asfy[da / 2]

Mu = 0.90 (402.12) (414) [30047/2]

Mu = 41.43 kN.m > MuT Ok. (SAFE)

h) Check for web Reinforcement

Vu = WuL - Wud

2

Vu = 25.1 (3) - 25.1(0.30)2

Vu = 30.12 kN.

Vc = 1/6 f'c (b)(d) = 1/6 28 (150)(300)

= 39.69 KN

Vc = 0.85 (39.69) = 16.87 kN < Vu (need web reinforcements)2 2

Vs = Vu - Vc ; 30.12 - 39.69

0.85= - 4.25 KN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4


= 300 = 150 mm.

2

S = 10 mm. Dia. RSB @ 150 mm. o.c. Ok...

FOR BEAM , B3 (FB2)

a) Slab loading

Ws-4 DL = 3.9542(4) [3(1)2] = 5.27 kN/m3 2

Ws-3 DL = 3.9542(3) [3(0.75)2] = 4.82 kN/m


13/94


3 2 10.09 kN/m

Ws-4 LL = 4.8 (4) [3(1)2](2) = 6.4 kN/m3 2

Ws-3 LL = 4.8 (3) [3(0.75)2] (2) = 5.85 kN/m

3 2 12.25 kN/m


Wu = 1.4 (10.09) + 1.7 (12.25) = 35.66 kN/m


Use d = 340 mm.

B = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

DL = 150 mm. CHB exterior wall both face plastered = 2.4 kPaDL = 2.4 (1) = 2.4 kN/m

Wb = (0.20)(.40)(23.544) = 1.88 kN/m

4.28 kN/m

Factored Load = 1.4 (4.28) = 6.0 kN/m

Total Factored Load;

= 6.0 + 35.66 = 41.66 kN/m

f) Design the beam

1) Negative moment @ support;

Max Mu = -0.10wL2kN.m.

= -0.10(41.66)(4)

= 66.656

a) Compute for Ru (Ultimate Reaction)

66.656X 106

= 0.90 (200) (340)2

Ru

Ru = 3.203 MPa

b) Solve for actual steel ratio,

= 0.85fc [1 - 1 2 Ru]fy 0.85f'c

= 0.85(21) [ 1 - 1 2(3.203) ] = 0.0086414 0.85(21)

min= < < max ok ( singly Reinforced design )


14/94


c) compute steel area requirement ; use 16 mm. Dia. Rebars

As = bd

= 0.0086 (200)(340) = 584.8 mm2

No. of Bars = As / Ab= 584.8 = 1.86 pcs. Say 2 pcs

(20)2

4

A16 = 201.1 (1) = 201.1 mm2

A20 = 314.16 (2) = 628.32 mm2

829.42 mm2

> 584.8 mm2

say ; 2 pcs20 mm Dia.RSB cont top bars and 1pc16 mm Dia. RSB extra top bar

Actual As = A's 829.42 mm2

1) Positive moment @ Midspan, Max

Max Mu = +0.08wL2kN.m.

= 0.08(41.66) (4)2

= 53.325 kN-m

a) Compute for Ru (Ultimate Reaction)

53.325 x 106

= 0.90 (200)(340)2

Ru

Ru = 2.563 MPa

d) Solve for actual steel ratio,

= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(21) [ 1 - 1 2(2.563) ] = 0.0067414 0.85(21)

min < < max ok ( singly Reinforced design )

c) Compute steel area requirement; use 20 mm. Dia. RebarsAs = bd

= 0.0067 (200) (340) = 456.46 mm2


= 456.46 = 1.45 pcs. (20)2

4


15/94


say ; 2 pcs20 mm Dia.RSB con. Bot bar 1 pc-16mm dia extra bar

Actual As

A16 = 201.1 (1) = 201.1 mm2

A20 = 314.16 (2) = 628.32 mm2

829.42 mm2 > 456.46 mm2


from StressStrain Diagram

1) check if steel yeilds ; T = C

Asfy = 0.85 f'c a b

a = 829.42 (414) = 96.18 mm Say; 96 mm.

0.85(21) (200)

a = c ; c = 96 = 113.16 mm. Say 114.00mm0.85


s = 0.003 ( dc )c

s = 0.003 ( 340114 ) = 0.00591114

y = 414 = 0.00207 < s Ok, (Steel Yield)200,000


Mu = Asfy [ d a / 2 ]

Mu = 0.90 (829.42) (414) [ 34096/2 ]

Mu = 90.24 kN.m > MuT Ok. ( SAFE )

f) Check for web Reinforcement

Vu = VmaxWu (d)

Vmax = 0.60Wl = 0.60(35.66)(4) = 85.584 kn

Vu = 85.584 - 35.66(0.34)

Vu = 73.46 kN

Vc = 1/6 f'c (b)(d) = 1/6 21 (200)(340)

= 199.16 kN

Vc = 0.85 ( 199.1) = 84.64 kN > Vu ( no need web reinforcements )2 2


16/94


From the code provide min web reinforcements/stirrups

Compute Spacing of web reinforcements / stirrups ; use 10 mm. Dia. Bars

Av = (8)2 = 50.2656 mm2

4

Provide Spacing = d/2= 340 = 170 mm.

2


FOR BEAM , B3 (FB3)

a) Slab loading

Ws-7 DL = 3.9542(7) [ 3(1 )2

] 2 = 18.453 kN/m3 2

Ws-4DL = 3.9542(4) [ 3(1 )2 ] 2 = 10.54 kN/m3 2 28.993 kN/m

Ws-4 LL = 4.8 (7) [ 3(1)2](2) = 22.4 kN/m3 2

Ws-4 LL = 4.8 (4) [ 3(1)2](2) = 12.8 kN/m3 2 35.2 kN/m


Wu = 1.4 (28.993) + 1.7 (35.2) = 100.43 kN/m


Use d = 340 mm.

B = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(.40)(23.544) = 1.88 kN/m


Total Factored Load ;

= 2.632 + 100.43 = 103.062 kN/mf) Design the beam

1) Negative moment @ support ;

Max Mu = -0.10wL2kN.m. = -0.10(41.66)(4) = 66.656

a) Compute for Ru ( Ultimate Reaction )

66.656X 106

= 0.90 (200)(340)2

Ru


17/94


Ru = 3.203 MPa

d) solve for actual steel ratio ,

= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(21) [ 1 - 1 2(3.203) ] = 0.0086414 0.85(21)

min= < < max ok ( singly Reinforced design )


As = bd

= 0.0086 (200)(340) = 584.8 mm2


= 584.8 = 1.86 pcs. Say 2 pcs (20)2

4A16 = 201.1 (1) = 201.1 mm

2

A20 = 314.16 (2) = 628.32 mm2

829.42 mm2

> 584.8 mm2

say ; 2 pcs20 mm Dia.RSB cont top bars and 1pc16 mm Dia. RSB extra top bar


1) Positive moment @ Midspan, Max

Max Mu = +0.08wL2kN.m. = 0.08(41.66)(4)

2 = 53.325 kN-m


53.325 x 106

= 0.90 (200)(340)2

Ru

Ru = 2.563 MPa

b) solve for actual steel ratio ,

= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(21) [ 1 - 1 2(2.563) ] = 0.0067414 0.85(21)




18/94


As = bd

= 0.0067 (200)(340) = 456.46 mm2


= 456.46 = 1.45 pcs. (20)2

4

say ; 2 pcs20 mm Dia.RSB con. Bot bar 1 pc-16mm dia extra bar

Actual As

A16 = 201.1 (1) = 201.1 mm2

A20 = 314.16 (2) = 628.32 mm2

829.42 mm2

> 456.46 mm2


from StressStrain Diagram

1) check if steel yeilds ; T = C

Asfy = 0.85 f'c a b

a = 829.42 (414) = 96.18 mm Say; 96 mm.

0.85(21) (200)

a = c ; c = 96 = 113.16 mm. Say 114.00mm0.85


s = 0.003 ( dc )c

s = 0.003 ( 340114 ) = 0.00591114

y = 414 = 0.00207 < s Ok, ( Steel Yeild )200,000


Mu = Asfy [ d a / 2 ]

Mu = 0.90 (829.42) (414) [ 34096/2 ]

Mu = 90.24 kN.m > MuT Ok. ( SAFE )


Vu = VmaxWu (d)

Vmax = 0.60wL = 0.60(35.66)(4) = 85.584 kn

Vu = 85.584 - 35.66(0.34) = 73.46 kN


19/94


Vc = 1/6 f'c (b)(d) = 1/6 21 (200)(340) = 199.16 kN

Vc = 0.85 ( 199.1) = 84.64 kN > Vu ( no need web reinforcements )2 2

From the code provide min web reinforcements/stirrups


Av = (8)2 = 50.2656 mm2

4

Provide Spacing = d/2

= 340 = 170 mm.

2


FOR BEAM , B

4

a) slab loading

Ws-1 DL = 4.47(1.5) [ 3(0.3)2 ] = 3.25 kN/m3 2

Ws- LL = 4.8 (1.5) [ 3(0.3)2] = 3.492 kN/m3 2

b) factored load, Wu = 1.4(DL) + 1.7(LL)

Wu = 1.4 (3.25) + 1.7 (3.492) = 10.49 kN/m


Use d = 300 mm.

B = 150 mm.

H = 300 + 40 + 10 + 10 = 360 mm.

Wb = (0.150)(0.36)(23.544) = 1.27 kN/m

dead loads = steel studs = 0.38 kN/m1.65 kN/m



= 2.31 + 10.49 = 12.8 kN/m

d) Compute for moment and Reaction ( see Table 4 )


20/94


e) Shear and moment diagram ( see figure 4 )

f) Design the beam

1) Negative moment @ support ; Max Mu = 29.44 kN.m.

a) Compute for Ru ( Ultimate Reaction )29.44 x 10

6= 0.90 (150)(300)

2Ru

Ru = 2.423 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.423) ] = 0.0062414 0.85(28)



As = bd

= 0.0062 (150)(300) = 279 mm2


= 279 = 1.39 pcs.

say ; 2 pcs16 mm Dia.RSB

Actual As = A's = (2) 279 = 402.12 (16)2

4

1)Positive moment @ Midspan, Max Mu = 26.65 kN.m.


Ru = 26.65 x 106

0.90 (150)(300)2

Ru = 2.423 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.423) ] = 0.0062414 0.85(28)


21/94




As = bd

= 0.0062 (150)(300) = 279 mm2


= 279 = 1.39 pcs.

(16)2

4


Actual As = A's = (2) (16)2 = 402.124

d) Check for web ReinforcementVu = VmaxWu (d)37.9 - 12.8(0.30) = 34.06 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (150)(300)

= 39.69 kN

Vc = 0.85 ( 39.69) = 16.87 kN < Vu ( need web reinforcements )2 2

Vs = Vu - Vc ; 34.06 - 39.69

0.85= 0.38 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

0.38(1000)

= 34,209.5 mm.Therefore Max Spacing = d/2

= 300 = 150 mm.2


FOR BEAM , B5


22/94


a) slab loading

Ws-1 DL = 4.47(1.5) [ 3(0.3)2 ] = 3.25 kN/m3 2

Ws-1 LL = 4.8 (1.5) [ 3(0.3)2] = 3.492 kN/m3 2

b) factored load, Wu = 1.4(DL) + 1.7(LL)

Wu = 1.4 (3.25) + 1.7 (3.492) = 10.49 kN/m


Use d = 300 mm.

B = 150 mm.

H = 300 + 40 + 10 + 10 = 360 mm.Wb = (0.150)(0.36)(23.544) = 1.27 kN/m

dead loads = steel studs = 0.38 kN/m

1.65 kN/m



= 2.31 + 10.49 = 12.8 kN/m


e) Shear and moment diagram (see figure 5 )

f) Design the beam



Ru = 34.2 x 106

0.90 (150)(300)2

Ru = 2.815 MPab) solve for actual steel ratio ,

= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.815) ] = 0.0073414 0.85(28)



23/94


c) Compute steel area requirement; use 16 mm. Dia. Rebars

As = bd

= 0.0073(150)(300) = 328.5 mm2


= 328.5 = 1.63 pcs. (16)2

4


Actual As = A's = (2) 279 = 402.12

(16)2

4

1) Positive moment @ Midspan, Max Mu = 24.81 kN.m.


Ru = 24.81 x 106

0.90 (150)(300)2

Ru = 2.042 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.042) ] = 0.0052414 0.85(28)

min < < max ok


As = bd

= 0.0052 (150)(300) = 234 mm2


= 234 = 1.2 pcs.say ; 2 pcs16 mm Dia.RSB

Actual As = A's = (2) 279 = 402.12 (16)2

4

d) Check for web Reinforcement

Vu = VmaxWu (d)


24/94


38.8 - 12.8(0.30) = 34.96 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (150)(300)

= 39.69 kN


Vs = Vu - Vc ; 34.96 - 39.69

0.85

= 1.44 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

1.44(1000)

= 9,027.5 mm.


= 300 = 150 mm.

2


FOR BEAM , B6A

a) Slab loading

Ws-a DL = 3.9542(1) [ 3(0.5)2 ] = 1.81 kN/m3 2

Ws- LL = 4.8 (1) [ 3(0.5)2] = 2.2 kN/m3 2


Wu = 1.4 (1.81) + 1.7 (2.2) = 6.274 kN/m


Use d = 200 mm.

B = 150 mm.

H = 200 + 40 + 10 + 10 = 260 mm.


25/94


Wb = (0.150)(0.26)(23.544) = 0.92 kN/m

dead loads = 100mm CHB both = 1.76 kN/m

face pastered 2.68 kN/m

Factored Load = 1.4 (2.68) = 3.8 kN/mTotal Factored Load ;

= 3.8 + 6.274 = 10.10 kN/m

d) Compute for moment and Reaction ( see Table 6a )

e) Shear and moment diagram ( see figure 6a )

f) Design the beam



Ru = 6.33 x 106

0.90 (150)(200)2

Ru = 1.172 MPa

b) Solve for actual steel ratio ,

= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(1.172) ] = 0.0029414 0.85(28)



As = bd

= 0.0057(2000)(300) = 342 mm2


= 342 = 1.7 pcs.


Actual As = A's = (16)2 = 201.06mm24


Vu = WuL - Wud

2


26/94


Vu = 10.10 (4) - 10.10( 0.30)

2

Vu = 9.30 kN.

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(150)

= 26.46 kN

Vc = 0.85 ( 26.46) = 11.25 kN > Vu ( no web reinforcements need )2 2

Check Min. area of web reinforcement.

Min. Av = b S / 3 fy

As = 78.54

Av = 2As = 2 (78.54) = 157.10 mm2

S = 157.10(3)(276) = 867.192 mm150


FOR BEAM , B6B

a) Slab loading

Ws-11 DL = 4.47(1) [ 3(0.375)2 ] = 3.20 kN/m

3 2Ws-11 LL = 4.8 (1) [ 3(0.375)2] = 3.43 kN/m

3 2


Wu = 1.4 (3.20) + 1.7 (3.43) = 10.311 kN/m


Use b = 200 mm.

d = 3000 mm.

H = 300 + 40 + 10 + 10 = 360 mm.

Wb = (0.20)(0.36)(23.544) = 1.7 kN/m

dead loads = 100mm CHB both = 1.76 kN/m

face pastered 3.46 kN/m




27/94


= 4.84 + 10.31 = 15.16 kN/m

d) Compute for moment and Reaction ( see Table 6b)

e) Shear and moment diagram ( see figure 6b )

f) Design the beam1) Negative moment @ support ; Max Mu = 36.24 kN.m.


Ru = 36.24 x 106

0.90 (200)(300)2

Ru = 2.237 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.237) ] = 0.0057414 0.85(28)

min > use minc) compute steel area requirement ; use 16 mm. Dia. Rebars

As = bd

= 0.0034(150)(200) = 102 mm2


= 102 = 0.51 pcs. (16)2

4


Actual As = A's = (16)2 = 201.1 mm2

4


Vu = WuL - Wud

2

Vu = 18 (4) - 18( 0.30)

2

Vu = 30.6 kN.

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(300)

= 52.91 kN

Vc = 0.85 ( 52.91) = 22.49 kN < Vu ( need web reinforcements )


28/94


2 2

Vs = Vu - Vc ; 30.6 - 52.91

0.85

= - 16.91 kN


Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 179.91 kN > Vs


= 300 = 150 mm.

2


FOR BEAM , B7

a) Slab loading

Bleacher slab DL = 6.85(4) [ 3(0.8)2 ] = 10.78kN/m3 2

Bleacher slab LL = 4.8(3) [ 3(0.8 )2 ] = 7.552 kN/m3 2


Wu = 1.4 (10.78) + 1.7 (7.552) = 27.93 kN/m


Use d = 340 mm.

B = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

DL = steel studs = 2.4 kN/m

Wb = (0.20)(.40)(23.544) = 1.88 kN/m



= 3.164 + 27.93 = Ru = 36.24 x 106

0.90 (200)(300)2

31.1 kN/m


29/94



e) Shear and moment diagram (see figure 7)

f) Design the beam1) Negative moment @ support ; Max Mu = - 83.56kN.m.


Ru = 83.56 x 106

0.90 (200)(340)2

Ru = 4.016 MPa

d) Solve for actual steel ratio ,

= 0.85 f'c [ 1 - 1 2 Ru ]

fy 0.85f'c = 0.85(28) [ 1 - 1 2(4.016) ] = 0.0107

414 0.85(28)


c) Compute steel area requirement ; use 20 mm. Dia. Rebars

As = bd

= 0.0107 (200)(340) = 727.6 mm2


= 727.6 = 2.31 pcs. (20)2

4A16 = 201.1 (1) = 201.1 mm

2

A20 = 314.16 (2) = 628.32 mm2

829.42 mm2

> 727.6

say ; 2 pcs20 mm Dia.RSB and 1pcs16 mm Dia. RSB

Actual As = A's 829.42 mm

2



Ru = 67.88 x 106

0.90 (200)(340)2

Ru = 3.262 MPa


30/94


d) Solve for actual steel ratio ,

= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(3.262) ] = 0.0085414 0.85(28)

min < < max ok


As = bd

= 0.0085 (200)(340) = 578 mm2


= 578 = 2.87 pcs.

(20)2

4


Actual As = A's = (3) (16)2 = 603.2 mm2

4

g) Check for web Reinforcement

Vu = VmaxWu (d)

Vu = 126.99 - 37.73(0.34)

Vu = 114.2 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)

= 59.97 kN


Vs = Vu - Vc ; 114.2 - 59.97

0.85

= 74.38 kN


S = Av fy d

Vs Av = (10)2 = 157 mm2

4

S = 157(276)(340)

74.38(1000)

= 198.1 mm.


31/94


Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


= 340 = 170 mm.

2S = 10 mm. Dia. RSB @ 170 mm. o.c. Ok...

FOR BEAM , B8

a) Slab loading

Ws-10 DL = 3.9542(2.5) [ 3(0.5)2 ] = 4.531 kN/m3 2

Ws-10 LL = 4.8 (2.5) [ 3(0.5)2] = 5.50 kN/m3 2

Ws-11 DL = 4.47(1.5) 2 = 2.235 kN/m

3

Ws-11 LL = 4.8 (1.5) 2 = 2.4 kN/m

3


Wu1= 1.4 (4.531) + 1.7 (5.50) = 15.70 kN/m

Wu2 = 1.4 (2.235) + 1.7 (2.4) = 7.21 kN/m


Use d = 340 mm.

b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(0.40)(23.544) = 1.88 kN/mDead load = 100mm CHB interior wall = 1.76 kN/m

3.64 kN/m



Wu1 = 15.7 kN/m + 5.1kN/m = 20.8 kN/m

Wu2 = 7.21 kN/m + 5.1 kN/m = 28.0 kN/m


32/94




f) Design the beam

1) Negative moment @ support ; Max Mu = 99.03 kN.m.a) Compute for Ru ( Ultimate Reaction )

Ru = 99.03 x 106

0.90 (200)(340)2

Ru = 4.760 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(4.760) ] = 0.0130414 0.85(28)



As = bd

= 0.0130 (200)(340) = 884 mm2


= 884 = 2.81 pcs. (20)2

4


Actual As = A's = (3) (20)2 = 942.48 mm2

4



Ru = 41.44 x 10

6

0.90 (200)(340)2

Ru = 1.992 MPa


= 0.85 f'c [ 1 - 12 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(1.992) ] = 0.005


33/94


414 0.85(28)

min < < max ok


As = bd

= 0.005 (200)(340) = 340 mm2


= 340 = 1.7 pcs.

(16)2

4


Actual As = A's = (2) (16) = 402.124


Vu = VmaxWu (d)

114.15 - 38 (0.34) = 101.23 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)

= 59.97 kN


2 2Vs = Vu - Vc ; 84.03 - 59.97

0.85

= 59.12 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

59.12(1000)

= 249.2 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


34/94



= 340 = 170 mm.2


FOR BEAM , B9

a) Slab loading

Ws-9,10 DL = 3.9542(4) [ 3(0.5)2 ](2) = 9.06 kN/m3 2

Ws-9,10 LL = 4.8 (4) [ 3(0.5)2](2) = 11.0 kN/m3 2


Wu1= 1.4 (9.06) + 1.7 (11.0) = 31.384 kN/m


Use d = 340 mm.

b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(0.40)(23.544) = 1.88 kN/m

Dead load = steel studs = 0.38 kN/m

2.26 kN/m



Wu1 = 31.384 kN/m + 3.164 kN/m = 41.62kN/m



f) Design the beam

1) Negative moment @ support ; Max Mu = 108 kN.m.



35/94


Ru = 108 x 106

0.90 (200)(340)2

Ru = 5.190 MPa

b) Solve for actual steel ratio , = 0.85 f'c [ 1 - 1 2 Ru ]

fy 0.85f'c

= 0.85(28) [ 1 - 1 2(5.190) ] = 0.0143414 0.85(28)



As = bd

= 0.0143 (200)(340) = 972.4 mm2


= (2) (20)2 = 628.32 mm2

4

= (2) (16)2 = 402.12 mm2

4 1030.44 mm2 > 972.4

say ; 2 pcs20 mm and 2-16mm. Dia.RSB




Ru = 48.37 x 106

0.90 (200)(340)2

Ru = 0.006 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.691) ] = 0.006414 0.85(28)

min < < max ok



36/94


As = bd

= 0.006 (200)(340) = 408 mm2


= 469.2 = 2.1 pcs. (16)2

4


Actual As = A's = (2) (16)2 = 402.124

\d) Check for web Reinforcement

Vu = VmaxWu (d)

95.37 - 34.55 (0.34) = 83.62 kNVc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)

= 59.97 kN


Vs = Vu - Vc ; 83.62 - 59.97

0.85

= 38.41 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

38.89(1000)

= 378.33 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs



37/94


= 340 = 170 mm.


FOR BEAM , B10

a) Slab loading

Ws-10 DL = 3.9542( 2.5) [ 3(0.5)2 ] = 4.53 kN/m3 2

Ws-10 LL = 4.8 (2.5) [ 3(0.5)2] = 5.5 kN/m3 2

factored load, Wu = 1.4(DL) + 1.7(LL)

Wu1= 1.4 (4.53) + 1.7 (5.5) = 15.7 kN/m

Ws-3 DL = 3.9542( 3) [ 3(0.75)2 ] = 4.82 kN/m3 2

Ws-3 LL = 4.8 (3) [ 3(0.75)2] = 5.85 kN/m3 2


Wu1= 1.4 (4.82) + 1.7 (5.85) = 16.7 kN/mWs-7 DL = 3.9542( 2) = 2.64 kN/m

3

Ws-7 LL = 4.8 (2) = 3.2 kN/m

3


Wu1= 1.4 (2.64) + 1.7 (3.2) = 9.14 kN/m

b) Compute beam weight, Consider 1 meter strip

Use d = 340 mm.b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(0.40)(23.544) = 1.88 kN/m

Dead load = 150mm exterior wall = 2.4 kN/m= window glass frame and sash = 0.38 kN.m

4.66 kN/m


38/94




Wu1 = 15.7 kN/m + 6.524 kN/m = 22.22 kN/m

Wu2 = 16.7 kN/m + 6.524 kN/m = 23.22 kN/m

Wu3 = 9.14 kN/m + 6.524 kN/m = 15.66 kN/m


d) Shear and moment diagram ( see figure 10 )

e) Design the beam



Ru = 62.3 x 106

0.90 (200)(340)2

Ru = 2.994 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.994) ] = 0.0078414 0.85(28)



As = bd

= 0.0078 (200)(340) = 530.4 mm2


= 530.4 = 1.67 pcs.

(20)2

4

say ; 2 pcs20 mm and 2-16mm. Dia.RSB

Actual As = A's = (2) (20)2 = 628.32 mm2

4




39/94


Ru = 49.28 x 106

0.90 (200)(340)2

Ru = 2.368 MPa

b) Solve for actual steel ratio , = 0.85 f'c [ 1 - 1 2 Ru ]

fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.368) ] = 0.006414 0.85(28)

min < < max ok


As = bd

= 0.006 (200)(340) = 408 mm2


= 469.2 = 1.3 pcs.

(20)2

4

say ; 2pcs20mm Dia.RSB

Actual As = A's = (2) (16)2 = 402.12 mm2

4


Vu = VmaxWu (d)

68.39 - 22.22 (0.34) = 60.84 kNVc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)

= 59.97 kN


Vs = Vu - Vc ; 60.84 - 59.97

0.85

= 11.61 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4


40/94


S = 157(276)(300)

38.89(1000)

= 378.33 mm.

Check Max Spacing.1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


= 340 = 170 mm.

2


FOR BEAM , B11

a) Slab loading

Ws-3 DL = 3.9542( 4) [ 3(1)2 ] = 10.54 kN/m3 2

Ws-3 LL = 4.8 (4) [ 3(1)2] = 12.8 kN/m3 2

Factored load, Wu = 1.4(DL) + 1.7(LL)

Wu1= 1.4 (10.54) + 1.7 (12.8) = 36.52 kN/m

Ws-11 DL = 3.6484( 1.5) = 1.82 kN/m

3

Ws-4 DL = 3.9542 (4) = 5.27 kN/m

3

Ws-11 LL = 4.8( 1.5) = 2.4 kN/m

3

Ws-4 LL = 4.8 (4) = 6.4 kN/m

3


Wu1= 1.4 (7.09) + 1.7 (8.8) = 24.9 kN/m


Use d = 340 mm.


41/94


b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(0.40)(23.544) = 1.88 kN/m

Dead load = 100mm exterior wall = 2.4 kN/m3.64 kN/m



Wu1 = 36.52 kN/m + 5.10 kN/m = 41.62 kN/m

Wu2 = 24.9 kN/m + 5.10 kN/m = 28.10 kN/m



e) Design the beam

1) Negative moment @ support ; Max Mu = - 82.0 kN.m.


Ru = 82.0x 106

0.90 (200)(340)2

Ru = 3.941 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(3.941) ] = 0.0105414 0.85(28)



As = bd

= 0.0105 (200)(340) = 714.0 mm2


= 714.0 = 2.27 pcs. (20)2

4

A16 = 201.1 (1) = 201.1 mm2

A20 = 314.16 (2) = 628.32 mm2

829.42 mm2

> 714.0 mm2


42/94


say use; 2 pcs20 mm and 1-16mm. Dia.RSB

Actual As = A's = (2) (20)2 = 829.42 mm2

4



Ru = 63.82 x 106

0.90 (200)(340)2

Ru = 3.067 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]

fy 0.85f'c = 0.85(28) [ 1 - 1 2(3.067) ] = 0.008

414 0.85(28)

min < < max ok


As = bd

= 0.008 (200)(340) = 544 mm2


= 469.2 = 2.71 pcs. (16)2

4


Actual As = A's = (2) (16)2 = 402.12 mm2

4


Vu = VmaxWu (d)

100.79 - 41.62 (0.34) = 86.64 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)

= 59.97 kN


Vs = Vu - Vc ; 686.64 - 59.97


43/94


0.85

= 41.96 kN


S = Av fy d Av = (10)2

= 157 mm2

Vs 4

S = 157(276)(300)

41.96(1000)

= 378.33 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


= 340 = 170 mm.

2


FOR BEAM , B12A

a) Slab loading

Ws-4,3 DL = 3.9542(4) [ 3(1)2 ](2) = 10.545 kN/m3 2

Ws-4,3 LL = 4.8 (4) [ 3(1)2](2) = 12.8 kN/m3 2

Factored load, Wu

Wu1= 1.4 (10.545) + 1.7 (12.8) = 36.523 kN/m

Ws-11 DL = 3.6484(4) (2) = 3.6484 kN/m3

Ws-11 LL = 4.8 (4) (2) = 4.8 kN/m

3

Factored load, Wu

Wu2= 1.4 (3.6484) + 1.7 (4.8) = 13.27 kN/m



44/94


Use d = 340 mm.

b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(0.40)(23.544) = 1.88 kN/mDead load = 100 mm. CHB = 1.76 kN/m

3.64 kN/m



Wu1 = 36.523 kN/m + 5.10 kN/m = 41.62 kN/m

Wu2 = 13.27 kN/m + 5.10 kN/m = 18.37kN/m

c) Compute for moment and Reaction ( see Table 12A )d) Shear and moment diagram ( see figure 12A )

e) Design the beam



Ru = 94.0 x 106

0.90 (200)(340)2

Ru = 4.517 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(4.517) ] = 0.0122414 0.85(28)



As = bd= 0.0122 (200)(340) = 829.6 mm

2


= 829.6 = 2.64 pcs. (20)2

4



45/94


Actual As = A's = (3) (20)2 = 942.48 mm2

4



Ru = 56.0 x 106

0.90 (200)(340)2

Ru = 2.691 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]

fy 0.85f'c = 0.85(28) [ 1 - 1 2(2.691) ] = 0.0069

414 0.85(28)

min < < max ok


As = bd

= 0.0069 (200)(340) = 469.2 mm2


= 469.2 = 1.49 pcs. (20)2

4


Actual As = A's = (2) (16)2 = 402.12 mm2

4


Vu = VmaxWu (d)

98.18 - 41.62 (0.34) = 84.03 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)

= 59.97 kN



46/94


Vs = Vu - Vc ; 84.03 - 59.97

0.85

= 38.89 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

38.89(1000)

= 378.33 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


= 340 = 170 mm.

2


FOR BEAM , B12B (FB3)

a) Slab loading

Ws-4,3 DL = 3.9542(4) [ 3(1)2 ](2) = 10.545 kN/m3 2

Ws-4,3 LL = 4.8 (4) [ 3(1)2](2) = 12.8 kN/m3 2

factored load, Wu

Wu1= 1.4 (10.545) + 1.7 (12.8) = 36.523 kN/m

Ws-11 DL = 3.6484(4) (2) = 3.6484 kN/m3

Ws-11 LL = 4.8 (4) (2) = 4.8 kN/m

3

factored load, Wu

Wu1= 1.4 (3.6484) + 1.7 (4.8) = 13.27 kN/m



47/94


Use d = 340 mm.

b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(0.40)(23.544) = 1.88 kN/mFactored Load = 1.4 (1.88) = 2.632 kN/m


Wu1 = 36.523 kN/m + 2.632 kN/m = 39.15 kN/m

Wu2 = 13.27 kN/m + 2.632 kN/m = 15.90 kN/m

55.05 kN-m

c) Max Ultimate Moment , Mu = 0.10wL2

= 0.1(55.05)(4)(4) = 88.08 kN-m

e) Design the beam



Ru = 88.08 x 106

0.90 (200)(340)2

Ru = 4.233 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(4.233) ] = 0.0113414 0.85(28)



As = bd

= 0.0113 (200)(340) = 771.38 mm2


= 771.38 = 2.455 pcs say 3.

(20)2

4

A20 = 314.16 (3) = 942.48 mm2

> 771.38 mm2

say use; 3 pcs20 mm. Dia.RSB


48/94


Actual As = A's = 942.48 mm2

1) Positive moment @ Midspan, Mu = 0.08wL2 = 0.08(55.05)(4)4

Max Mu = 70.464 kN.m.


Ru = 70.464 x 106

0.90 (200)(340)2

Ru = 3.386 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(3.386) ] = 0.00886414 0.85(28)

min < < max ok


As = bd

= 0.00886 (200)(340) = 602.48

mm

2


= 602.48 = 1.918 pcs. (20)2

4


f) Check for web Reinforcement Vmax = 0.60wL = 0.6(55.05)(4) = 132.12

Vu = VmaxWu (d)

132.12 - 55.05 (0.34) = 113.403 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)

= 59.97 kN


Vs = Vu - Vc ; 113.403 - 59.97


49/94


0.85

= 73.445 kN

Compute Spacing of web reinforcements / stirrups ; use 8mm. Dia. Bars

S = Av fy d Av = (8)2

= 50.265 mm2

Vs 4

S = 50.265(276)(340)

73.445(1000)

= 64.223 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


= 340 = 170 mm.2


FOR BEAM , B13

a) Slab loading

Ws-2 DL = 3.9542(4) [ 3(1)2 ] = 5.27 kN/m3 2

Ws-2 LL = 4.8 (4) [ 3(1)2] = 6.4 kN/m3 2

factored load, Wu

Wu1= 1.4 (5.27) + 1.7 (6.4) = 18.258 kN/m

Ws-3 DL = 3.9542 (4) [ 3(1)2] = 5.27 kN/m3 2

Ws-6 DL = 3.9542 (4) [ 3(1)2] = 3.62 kN/m3 2 8.89 kN/m

Ws-3 LL = 4.8 (4) [ 3(1)2] = 6.4 kN/m3 2


50/94


Ws-6 LL = 4.8 (4) [ 3(0.5)2] = 4.4 kN/m3 2 10.8 kN/m

Factored load, Wu

Wu2= 1.4 (8.89) + 1.7 (10.8) = 30.81 kN/m

Ws-4 DL = 3.9542 (4) [ 3(1)2] = 5.27 kN/m3 2

Ws-6 DL = 3.9542 (4) (2) = 5.27 kN/m

3 10.54 kN/m

Ws-4 LL = 4.8 (4) [ 3(1)2] = 6.4 kN/m3 2

Ws-6 LL = 4.8 (2) = 6.4 kN/m

3 12.8 kN/m

Factored load, WuWu2= 1.4 (10.54) + 1.7 (12.8) = 36.51 kN/m

Ws-5 DL = 3.9542 (4) [ 3(0.8)2] = 6.22 kN/m3 2

Ws-1 DL = 4.47 (1.5) [ 3(0.3)2] = 3.25 kN/m3 2 9.47 kN/m

Ws-5 LL = 4.8 (4) [ 3(0.8)2] = 7.55 kN/m3 2

Ws-1 LL = 4.8 (1.5) [ 3(0.3)2] = 3.492 kN/m3 2 11.0 kN/m

factored load, Wu

Wu2= 1.4 (9.47) + 1.7 (11.0) = 31.96 kN/m


Use d = 340 mm.

b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.Wb = (0.20)(0.40)(23.544) = 1.88 kN/m

DL = 150mm CHB = 2.40 kN/m

4.28 kN/m



Wu1 = 18.258 kN/m + 6.0 kN/m = 24.26 kN/m


51/94


Wu2 = 30.81 kN/m + 6.0 kN/m = 36.81 kN/m

Wu3 = 36.51 kN/m = 36.51 kN/m

Wu4 = 31.96 kN/m = 31.96 kN/m

c) Compute for moment and Reaction ( see Table 13 )d) Shear and moment diagram ( see figure 13 )

e) Design the beam



Ru = 96.8 x 106

0.90 (200)(340)2

Ru = 4.652 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(4.652) ] = 0.0126414 0.85(28)



As = bd= 0.0126 (200)(340) = 856.8 mm

2


= 856.8 = 2.73 pcs. (20)2

4

say use; 3pcs20 mm. Dia.RSB

Actual As = A's = (3) (20)2 = 942.48 mm2

4



Ru = 57.55 x 106

0.90 (200)(340)2

Ru = 2.766 MPa



52/94


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.766) ] = 0.0071414 0.85(28)

min < < max ok


As = bd

= 0.0071 (200)(340) = 482.8 mm2


= 482.8 = 2.4 pcs.

(16)2

4



Vu = VmaxWu (d)

111.49 - 36.51 (0.34) = 99.10 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)= 59.97 kN


2 2Vs = Vu - Vc ; 99.10 - 59.97

0.85

= 56.62 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

56.62(1000)= 260.61 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs



53/94


= 340 = 170 mm.


FOR BEAM , B14

a) Slab loading

Ws-3 DL = 3.9542(4) [ 3(1)2 ](2) = 10.545 kN/m3 2

Ws-3 LL = 4.8 (4) [ 3(1)2](2) = 12.8 kN/m3 2

factored load, Wu

Wu1= 1.4 (10.545) + 1.7 (12.8) = 36.523 kN/m

Ws-3 DL = 3.9542 (4) [ 3(1)2] = 5.27 kN/m3 2

Ws-6 DL = 3.9542 (2) = 2.64 kN/m

3 7.91 kN/m

Ws-3 LL = 4.8 (4) [ 3(1)2] = 6.4 kN/m3 2

Ws-6 LL = 4.8 (2) = 3.2 kN/m3 9.6 kN/m

factored load, Wu

Wu2= 1.4 (7.91) + 1.7 (9.6) = 25.63 kN/m

Ws-5 DL = 3.9542 (4) [ 3(0.8)2](2) = 12.44 kN/m3 2

Ws-5 LL = 4.8 (4) [ 3(0.8)2](2) = 15.104 kN/m3 2

factored load, Wu

Wu3= 1.4 (12.44) + 1.7 (15.104) = 43.10 kN/m

Ws-5 DL = 3.9542 (4) [ 3(0.8)2] = 6.22 kN/m3 2

Ws-1 DL = 4.47 (1.5) [ 3(0.3)2] = 3.25 kN/m3 2 9.47 kN/m

Ws-5 LL = 4.8 (4) [ 3(0.8)2] = 7.522 kN/m3 2


54/94


Ws-1 LL = 4.8 (1.5) [ 3(0.3)2] = 3.492 kN/m3 2 11.094 kN/m

Factored load, Wu

Wu4= 1.4 (9.47) + 1.7 (11.094) = 32.03 kN/m


Use d = 340 mm.

b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(0.40)(23.544) = 1.88 kN/m

DL = steel studs, gypsum board each side = 0.38kN/m

2.26 kN/m


Wu1 = 36.523 kN/m + 3.164 kN/m = 39.7 kN/m

Wu2 = 25.63 kN/m + 3.164 kN/m = 28.8 kN/m

Wu3 = 43.10 kN/m + 3.164 kN/m = 46.3 kN/m

Wu4 = 32.03 kN/m + 3.164 kN/m = 35.2 kN/m

c) Compute for moment and Reaction ( see Table 14 )


e) Design the beam



Ru = 88.6 x 106

0.90 (200)(340)2

Ru = 4.258 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(4.258) ] = 0.01142414 0.85(28)



55/94



As = bd

= 0.01142 (200)(340) = 776.56 mm2


(20)2

4

say use; 3pcs20 mm. Dia.RSB

Actual As = A's = (3) (20)2 = 942.48 mm2

4



Ru = 85.74 x 106

0.90 (200)(340)2

Ru = 4.121 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(4.121) ] = 0.0110414 0.85(28)

min < < max ok


As = bd

= 0.0110 (200)(340) = 748 mm2


= 748 = 2.4 pcs. (20)2

4


Actual As = A's = (3) (20)2 = 942.48 mm2

4


Vu = VmaxWu (d)

116.31 - 46.3. (0.34) = 100.57 kN


56/94


Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)= 59.97 kN


Vs = Vu - Vc ; 100.57 - 59.97 0.85

= 58.35 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

58.35(1000)

= 252.51 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


= 340 = 170 mm.

2


FOR BEAM , B15

a) Slab loading

Ws-2 DL = 3.9542(4) [ 3(1)2 ] = 5.27 kN/m3 2

Ws-2 LL = 4.8 (4) [ 3(1)2] = 6.4 kN/m

3 2

factored load, Wu

Wu1= 1.4 (5.27) + 1.7 (6.4) = 18.258 kN/m

Ws-5 DL = 3.9542 (4) [ 3(1)2] = 5.27 kN/m3 2

Ws-1 DL = 4.47 (1.5) [ 3(0.3)2] = 3.252 kN/m3 2 8.522 kN/m


57/94


Ws-5 LL = 4.8 (4) [ 3(1)2] = 6.4 kN/m3 2

Ws-1 LL = 4.8 (1.5) [ 3(0.3)2] = 3.492 kN/m3 2 9.892 kN/m

Factored load, Wu

Wu2= 1.4 (8.522) + 1.7 (9.892) = 28.75 kN/m


Use d = 340 mm.

b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(0.40)(23.544) = 1.88 kN/m

DL = 150 mm CHB = 2.4 kN/m4.28 kN/m



Wu1 = 18.258 kN/m + 5.992 kN/m = 24.25 kN/m

Wu2 = 28.75 kN/m + 1.4(1.88) kN/m = 31.38 kN/m



e) Design the beam



Ru = 69.8 x 106

0.90 (200)(340)2

Ru = 3.354 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(3.354) ] = 0.0088414 0.85(28)



As = bd


58/94


= 0.0088 (200)(340) = 598.4 mm2


= 598.4 = 1.90 pcs. (20)2

4


Actual As = A's = (2) (20)2 = 628.32 mm2

4



Ru = 39.15 x 106

0.90 (200)(340)2

Ru = 1.881 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(1.881) ] = 0.0047414 0.85(28)

min < < max ok


As = bd

= 0.0047(200)(340) = 319.6 mm2


= 319.6 = 1.6 pcs. (16)2

4


Actual As = A's = (2) (16)2 = 402.12 mm2

4


Vu = VmaxWu (d)

82.6 - 31.38. (0.34) = 71.93 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)


59/94


= 59.97 kN


Vs = Vu - Vc ; 71.93 - 59.97

0.85

= 24.65 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

24.65(1000)

= 597.7 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


= 340 = 170 mm.2


FOR BEAM , B16

a) Slab loading

Ws-1 DL = 4.47(1.5) [ 3(0.3)2 ] = 3.252 kN/m3 2

Ws-1 LL = 4.8 (1.5) [ 3(0.3)2] = 3.492 kN/m3 2

Factored load, Wu

Wu1= 1.4 (3.252) + 1.7 (3.492) = 10.49 kN/m


Use d = 300 mm.

b = 150 mm.

H = 300 + 40 + 10 + 10 = 360 mm.


60/94


Wb = (0.150)(0.360)(23.544) = 1.27 kN/m

DL = 150 mm CHB = 2.4 kN/m

3.67 kN/m



Wu1 = 10.49 kN/m + 5.138 kN/m = 15.63 kN/m



e) Design the beam



Ru = 41.31 x 106

0.90 (200)(340)2

Ru = 3.40 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(3.40) ] = 0.0089414 0.85(28)



As = bd

= 0.0089 (150)(300) = 400.5 mm2


= 400.5 = 1.99 pcs. (16)2

4


Actual As = A's = (2) (16)2 = 402.12 mm2

4



Ru = 30.38 x 106


61/94


0.90 (200)(340)2

Ru = 2.50 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.50) ] = 0.0064414 0.85(28)

min < < max ok


As = bd

= 0.0064(150)(300) = 288 mm2


= 288m = 1.4 pcs.

(16)2

4


Actual As = A's = (2) (16)2 = 402.12 mm2

4


Vu = VmaxWu (d)

47.31 - 15.63. (0.34) = 42.62 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)

= 59.97 kN


Vs = Vu - Vc ; 42.62 - 59.97

0.85

= 10.45 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

10.45(1000)


62/94


= 1243.98 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (150)(300)

= 79.4 kN > VsTherefore Max Spacing = d/2

= 300 = 150 mm.


FOR GIRDER/ BEAM along A and M

a) slab loading

Ws-3,2 DL = 3.9542(4) [ 3(1)2 ] = 5.27 kN/m3 2

Ws-3,2 LL = 4.8 (4) [ 3(1)2] = 6.4 kN/m3 2

factored load, Wu

Wu1= 1.4 (5.27) + 1.7 (6.4) = 18.258 kN/m

b) Compute beam weight, Consider 1 meter stripUse d = 340 mm.

b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(0.40)(23.544) = 1.88 kN/m

DL = 150 mm CHB = 2.4 kN/m

Windows glass frames and sash = 0.38 kN/m

4.66 kN/m


Wu1 = 18.258 kN/m + 6.524 kN/m = 24.8 kN/m

c) Compute for moment and Reaction ( see Table A and M )

d) Shear and moment diagram ( see figure A and M )

e) Design the beam


63/94




Ru = 42.0 x 106

0.90 (200)(340)2

Ru = 2.018 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.018) ] = 0.0051414 0.85(28)


c) Compute steel area requirement ; use16 mm. Dia. Rebars

As = bd

= 0.0051 (200)(340) = 346.8 mm2


= 346.8 = 1.25 pcs.

(16)2

4


Actual As = A's = (2) (16)2 = 402.12 mm2

4



Ru = 39.15 x 106

0.90 (200)(340)2

Ru = 1.485 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(1.485) ] = 0.0037414 0.85(28)

min < < max ok


64/94



As = bd

= 0.0037(200)(340) = 251.6 mm2


(16)2

4


Actual As = A's = (2) (16)2 = 402.12 mm2

4


Vu = VmaxWu (d)

60.1 - 24.8 (0.34) = 51.67 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)

= 59.97 kN


Vs = Vu - Vc ; 51.67 - 59.97

0.85

= 0.82 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

0.82(1000)

= 17,966.9 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


= 340 = 170 mm.

2



65/94


FOR GIRDER/ BEAM along B and L

a) Slab loading

Ws-3,2 DL = 3.9542(4) [ 3(1)2 ] = 5.27 kN/m3 2

Ws-3,2 LL = 4.8 (4) [ 3(1)2] = 6.4 kN/m3 2

factored load, Wu

Wu1= 1.4 (5.27) + 1.7 (6.4) = 18.258 kN/m

Ws-4DL = 3.9542 (4) [ 3(1)2 ](2) = 10.54 kN/m3 2

Ws-4 LL = 4.8 (4) [ 3(1)2](2) = 12.8 kN/m3 2

factored load, Wu

Wu1= 1.4 (10.54) + 1.7 (12.8) = 36.516 kN/m


Use d = 340 mm.

b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.Wb = (0.20)(0.40)(23.544) = 1.88 kN/m

DL = 150 mm CHB = 2.4 kN/mWindows glass frames and sash = 0.38 kN/m

4.66 kN/m



Wu1 = 18.258 kN/m + 6.524 kN/m = 24.78 kN/m

Wu2 = 36.516 kN/m + 6.524 kN/m = 43.04 kN/mWu3 = 36.516 kN/m + 1.4(1.88) = 39.15 kN/m

Wu3 = 1.4 (1.88) = 6.52 kN/m

c) Compute for moment and Reaction ( see Table B and L)

d) Shear and moment diagram ( see figure Band L )


66/94


e) Design the beam

1) Negative moment @ support ; Max Mu = -76.73 kN.m.


Ru = 76.73 x 106

0.90 (200)(340)2

Ru = 3.668 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(3.668) ] = 0.0097414 0.85(28)


c) Compute steel area requirement ; use16 mm. Dia. Rebars

As = bd

= 0.0097 (200)(340) = 659.6 mm2


= 238.5 = 2.14 pcs.

(20)2

4

A16 = 201.1 (2) = 402.12 mm

2

A20 = 314.16 (1) = 314.16 mm2

716.28 mm2

> 659.6 mm2

say ; 1 pcs20 mm Dia.RSB and 2 pcs16 mm Dia. RSB




Ru = 75.57 x 106

0.90 (200)(340)2

Ru = 3.632 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(3.632) ] = 0.0096414 0.85(28)


67/94


min < < max ok


As = bd

= 0.0096(200)(340) = 652.8 mm2


= 652.8 = 2.1 pcs.

(20)2

4A16 = 201.1 (2) = 402.12 mm

2

A20 = 314.16 (1) = 314.16 mm2

716.28 mm2

> 659.8 mm2

say ; 1 pcs20 mm Dia.RSB and 2 pcs16 mm Dia. RSB



Vu = VmaxWu (d)

118.99 - 57.715 (0.34) = 99.37 kN

Vc = 1/6 f'c (b)(d) = 1/6 28(200)(340)

= 59.97 kN


Vs = Vu - Vc ; 99.37 - 59.97

0.85

= 56.94 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

56.94(1000)

= 258.74 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


68/94



= 340 = 170 mm.


FOR GIRDER/ BEAM along C and K

a) Slab loading

Ws-7 DL = 3.9542(2) [ 3(0.67)2 ] = 3.363 kN/m3 2

Ws-8 DL = 3.9542 (3) = 3.9542 kN/m3 7.32 kN/m

Ws-7 LL = 4.8(2) [ 3(0.67)2 ] = 4.082 kN/m3 2

Ws-8 LL = 4.8 (3) = 4.8 kN/m

3 8.88 kN/m

factored load, Wu

Wu1= 1.4 (7.32) + 1.7 (8.88) = 23.34 kN/m

Ws-4DL = 3.9542 (4) [ 3(1)2

](2) = 10.54 kN/m3 2

Ws-4 LL = 4.8 (4) [ 3(1)2](2) = 12.8 kN/m3 2

Factored load, Wu

Wu2= 1.4 (10.54) + 1.7 (12.8) = 36.52 kN/m

Ws-4DL = 3.9542 (4) [ 3(1)2 ] = 5.27 kN/m3 2

Ws-11DL = 3.6484 (1.5) [ 3(0.375)2 ] = 2.61 kN/m

3 2 7.88 kN/m

Ws-4 LL = 4.8 (4) [ 3(1)2] = 6.4 kN/m3 2

Ws-11 LL = 4.8 (4) [ 3(0.375)2] = 3.43 kN/m3 2 9.83 kN/m

Factored load, Wu

Wu3= 1.4 (7.88) + 1.7 (9.83) = 27.74 kN/m


69/94


Ws-6DL = 3.9542 (2) [ 3(0.5)2 ] = 3.625 kN/m3 2

Ws-3DL = 4.8 (4) [ 3(1)2] = 5.27 kN/m

3 2 8.90 kN/mWs-6LL = 4.8 (2) [ 3(0.5)2 ] = 4.40 kN/m

3 2

Ws-3LL = 4.8 (4) [ 3(1)2] = 6.40 kN/m3 2 10.8 kN/m

factored load, Wu

Wu4= 1.4 (8.90) + 1.7 (10.8) = 30.82 kN/m


Use d = 340 mm.b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(0.40)(23.544) = 1.88 kN/m

DL = 100 mm CHB = 1.76 kN/m


4.02 kN/m


Total Factored Load ;Wu1 = 23.34 kN/m + 1.4 (1.88) = 26.0 kN/m

Wu2 = 36.52 kN/m + 1.4 (1.88) = 39.2 kN/m

Wu3 = 27.74 kN/m + 5.63kN/m = 33.37 kN/m

Wu3 = 3.82 kN/m + 5.63 kN/m = 36.45 kN/m

c) Compute for moment and Reaction ( see Table C and K)

d) Shear and moment diagram ( see figure C and K )

e) Design the beam



Ru = 63.5 x 106

0.90 (200)(340)2

Ru = 3.052 MPa



70/94


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(3.052) ] = 0.0080414 0.85(28)


c) Compute steel area requirement ; use 20mm. Dia. Rebars

As = bd

= 0.0080 (200)(340) = 544 mm2


= 238.5 = 1.73 pcs.

(20)2

4

say ; 2 pcs20 mm Dia. RSB.

Actual As = A's = (2) (20)2 = 628.32 mm2

4



Ru = 62.52 x 106

0.90 (200)(340)2

Ru = 3.005 MPab) Solve for actual steel ratio ,

= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(3.005) ] = 0.0078414 0.85(28)

min < < max ok


As = bd

= 0.0078 (200)(340) = 530.4 mm2


= 652.8 = 1.7 pcs. (20)2

4



71/94


Actual As = A's = (2) (20)2 = 628.32 mm2

4

f) Check for web ReinforcementVu = VmaxWu (d)

96.38 - 48.08 (0.34) = 80.05 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)

= 59.97 kN


Vs = Vu - Vc ; 80.05 - 59.97

0.85= 34.21 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

34.21(1000)

= 430.66 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


= 340 = 170 mm.

2


FOR GIRDER/ BEAM along D and J

a) Slab loading

Ws-1 DL = 4.47(1.5) = 2.235 kN/m

3


72/94


Ws-1 LL = 4.8 (1.5) = 2.4 kN/m

3

Factored load, Wu

Wu1= 1.4 (2.235) + 1.7 (2.4) = 7.21 kN/m

Ws-3 DL = 3.9542(4) [ 3(1)2 ] = 5.27 kN/m3 2

Ws-5 DL = 3.9542(4) = 5.27 kN/m

3 10.54 kN/m

Ws-3 LL = 4.8(4) [ 3(1)2 ] = 6.4 kN/m3 2

Ws-5 LL = 4.8 (4) = 6.4 kN/m

3 12.80 kN/m

Factored load, WuWu2= 1.4 (10.54) + 1.7 (12.80) = 36.52 kN/m

Ws-5DL = 3.9542 (4) = 5.27 kN/m

3

Ws-5 LL = 4.8 (4) = 6.4 kN/m

3

Factored load, Wu

Wu3= 1.4 (5.27) + 1.7 (6.4) = 18.258 kN/m

b) Compute beam weight, Consider 1 meter stripUse d = 340 mm.

b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(0.40)(23.544) = 1.88 kN/m

DL = 100 mm CHB = 1.76 kN/m


4.02 kN/m


Wu1 = 7.21 kN/m + 1.4 (1.88 + 1.76) = 12.31 kN/m

Wu2 = 36.52 kN/m + 1.4 (1.88 + 1.76) = 41.62 kN/m

Wu3 = 27.74 kN/m + 1.4 (1.88 + 0.38) = 21.42 kN/m

Wu3 = 3.82 kN/m + 1.4 (1.88) = 17.1 kN/m


73/94


c) Compute for moment and Reaction ( see Table D and J)

d) Shear and moment diagram ( see figure D and J )

e) Design the beam

1) Negative moment @ support ; Max Mu = -118.24 kN.m.a) Compute for Ru ( Ultimate Reaction )

Ru = 118.24 x 106

0.90 (200)(340)2

Ru = 5.682 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(5.682) ] = 0.01593414 0.85(28)


c) Compute steel area requirement ; use 20mm. Dia. Rebars

As = bd

= 0.01593 (200)(340) = 1088 mm2


= 1088 = 3.46 pcs. (20)2

4

A16 = 201.1 (1) = 201.1 mm2

A20 = 314.16 (3) = 942.48 mm2

1143.58 mm2

> 1088 mm2

say ; 3 pcs20 mm Dia.RSB and 1 pc.16 mm Dia. RSB


1) Positive moment @ Midspan, Max Mu = 61.52 kN.m.a) Compute for Ru ( Ultimate Reaction )

Ru = 61.52 x 106

0.90 (200)(340)2

Ru = 2.957 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]


74/94


fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.957) ] = 0.0077414 0.85(28)

min < < max ok


As = bd

= 0.0077 (200)(340) = 523.6 mm2


= 523.6 = 1.67 pcs.

(20)2

4


Actual As = A's = (2) (20)2 = 628.32 mm2

4


Vu = VmaxWu (d)

91.65 - 61.1 (0.34) = 70.88 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)

= 59.97 kN


Vs = Vu - Vc ; 70.88 - 59.97

0.85

= 23.42 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)23.42(1000)

= 629.10 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


75/94



= 340 = 170 mm.


FOR GIRDER/ BEAM along E and I

a) Slab loading

Ws-1 DL = 4.47(1.5) (2) = 4.47 kN/m3

Ws-1 LL = 4.8 (1.5) (2) = 4.8 kN/m

3

Factored load, Wu

Wu1= 1.4 (4.47) + 1.7 (4.8) = 14.418 kN/m

Ws-5 DL = 3.9542(4) (2) = 10.54 kN/m

3

Ws-5 LL = 4.8 (4) (2) = 12.8 kN/m

3

Factored load, Wu

Wu2= 1.4 (10.54) + 1.7 (12.80) = 36.52 kN/m

Ws-5DL = 3.9542 (4) = 5.27 kN/m

3

Ws-1DL = 4.47 (1.5) = 2.235 kN/m

3 7.505 kN/m

Ws-5LL = 4.8 (4) = 6.4 kN/m

3Ws-1LL = 4.8 (1.5) = 2.4 kN/m

3 8.8

Factored load, Wu

Wu3= 1.4 (7.505) + 1.7 (8.8) = 25.47 kN/m


Use d = 340 mm.


76/94


b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.

Wb = (0.20)(0.40)(23.544) = 1.88 kN/m

DL = 100 mm CHB = 1.76 kN/mWindows glass frames and sash = 0.38 kN/m

4.02 kN/m



Wu1 = 14.418 kN/m + 1.4 (1.88 + 1.76) = 19.51 kN/m

Wu2 = 36.52 kN/m + 1.4 (1.88 + 1.76) = 41.62 kN/mWu3 = 25.47 kN/m + 1.4 (1.88 + 0.38) = 28.63 kN/m

Wu4 = 14.418 kN/m + 1.4 (1.88) = 17.1 kN/m

c) Compute for moment and Reaction ( see Table E and I)

d) Shear and moment diagram ( see figure E and I )

e) Design the beam



Ru = 61.1 x 106

0.90 (200)(340)2

Ru = 2.936 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(2.936) ] = 0.0076414 0.85(28)



As = bd

= 0.0076 (200)(340) = 516.8 mm2



77/94


= 516.8 = 1.65 pcs. (20)2

4


Actual As = A's = (2) (20)2 = 628.32 mm24



Ru = 78.89 x 106

0.90 (200)(340)2

Ru = 3.791 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(3.791) ] = 0.010414 0.85(28)

min < < max ok


As = bd

= 0.010 (200)(340) = 680 mm2

No. of Bars = As / Ab= 680 = 2.16 pcs.

(20)2

4

A16 = 201.1 (1) = 201.1 mm2

A20 = 314.16 (2) = 628.32 mm2

829.24 mm2

> 680 mm2

say ; 2 pcs20 mm Dia.RSB and 1 pc.16 mm Dia. RSB



Vu = VmaxWu (d)

92.54 - 41.62 (0.34) = 78.39 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)

= 59.97 kN



78/94


2 2

Vs = Vu - Vc ; 78.39 - 59.97

0.85

= 32.25 kN


S = Av fy d Av = (10)2 = 157 mm2

Vs 4

S = 157(276)(300)

32.25(1000)

= 456.83 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)= 119.94 kN > Vs


= 340 = 170 mm.


FOR GIRDER/ BEAM along F, G and H

a) Slab loading

Ws-1 DL = 4.47(1.5) (2) = 4.47 kN/m

3

Ws-1 LL = 4.8 (1.5) (2) = 4.8 kN/m

3

Factored load, Wu

Wu1= 1.4 (4.47) + 1.7 (4.8) = 14.418 kN/mWs-5 DL = 3.9542(4) (2) = 10.54 kN/m

3

Ws-5 LL = 4.8 (4) (2) = 12.8 kN/m

3

Factored load, Wu

Wu2= 1.4 (10.54) + 1.7 (12.80) = 36.52 kN/m


79/94



Use d = 340 mm.

b = 200 mm.

H = 340 + 40 + 10 + 10 = 400 mm.Wb = (0.20)(0.40)(23.544) = 1.88 kN/m

DL = 100 mm CHB interior wall = 1.76 kN/m

3.64 kN/m



Wu1 = 14.418 kN/m + 5.10 kN/m = 19.52 kN/m

Wu2 = 36.52 kN/m + 5.10 kN/m = 41.62 kN/mWu3 = 14.418 kN/m + 1.4 (1.88) = 17.10 kN/m

c) Compute for moment and Reaction ( see Table F,G and H)

d) Shear and moment diagram ( see figure F,G and H)

e) Design the beam



Ru = 141.41 x 106

0.90 (200)(340)2

Ru = 6.796 MPa


= 0.85 f'c [ 1 - 1 2 Ru ]fy 0.85f'c

= 0.85(28) [ 1 - 1 2(6.796) ] = 0.0198414 0.85(28)



As = bd

= 0.0198 (200)(340) = 1346.4 mm2



80/94


= 1346.4 = 4.28 pcs. (20)2

4

A16 = 201.1 (4) = 804.25 mm2

A20 = 314.16 (2) = 628.32 mm2

1432.57 mm2

> 1346.4 mm2

say ; 2 pcs20 mm Dia.RSB and 4 pcs.16 mm Dia. RSB



Vu = VmaxWu (d)

108.96 - 72.63 (0.34) = 84.27 kN

Vc = 1/6 f'c (b)(d) = 1/6 28 (200)(340)= 59.97 kN


Vs = Vu - Vc ; 84.27 - 59.97

0.85

= 39.17 kN


S = Av fy d Av = (10)

2

= 157 mm

2

Vs 4

S = 157(276)(300)

39.17(1000)

= 376.12 mm.

Check Max Spacing.

1/3 f'c (b)(d) = 1/3 28 (200)(340)

= 119.94 kN > Vs


= 340 = 170 mm.



81/94


FOR GIRDER/ BEAM B-G2

a) Slab loading

Ws-9,10 DL = 3.9542(2.5) (2) = 6.59 kN/m3

Ws-9,10 LL = 4.8 (2.5) (2) = 8.0 kN/m

3

factored load, Wu

Wu1= 1.4 (6.59) + 1.7 (8.0) = 22.826 kN/m


Use d = 400 mm.

b = 200 mm.

H = 400 + 40 + 10 + 10 = 460 mm.

Wb = (0.20)(0.40)(23.544) = 2.17 kN/m


Total Factored Load

WuT = 22.826 + 3.125 = 25.95 kN/m

c) Compute for moment and Reaction ( see Table G2)

d) Shear and moment diagram ( see figure G2)e) Design the beam


a) Compute for Ru ( Ulti

Documents

Design of Beam-USD