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7/27/2019 Beam & Colum Design - CE 161
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa1 = 0.85
= 0.90
u = 0.003
Cover= 70.00 mm
Assume Section : l= 1.90 m
b = 230 mm
d= 340
h = 410 mm
slab thcikness 160 mmService Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 25.00 ---- 2.36Weight of Slab (Ws) 25.00 ---- 7.6
Parapet (Normal weight concrete,Wp1) 21.20 ---- 1.91
Parapet (Decorative Tile Finished, Wp2) ---- 0.77 0.77
Live Loads: Kpa KN/m
2.4 4.56
Ultimate Load: KN/m
Wu(beam)= 1.2DL= 2.83
Wu(slab)= 1.2DL + 1.6LL = 16.42
Wu(Parapet)= 1.2DL= 3.21
Wu= 22.46
Beam RB-1:
Line of symmetry
Note: Choose biggest value of Shear & Moment.
Shear Diagram:
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam RB-1 - Roof Deck Framing Plan Rating:
Wb=
Wu
Ws=
L = 7.60 m L = 7.60 m
Wu = +
Wp2=
Wp1=(
l
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Design For Flexure:
Design Moment:
Mu= 26 KN-m
Designer's Preferred Steel Ratio:
b = 0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 28.89 Kn-m
bd2
= 5018688.488 mm3
b = 250 mm Fexural Resistance factor
d = 141.69 mm
h' = 141.69 mm R= 5.75626h (assume) > h', ok!
Final:
b = 250 mm
d = 330 mm
h = 400 mm
Area of Steel:
Roots:
a = 46793235022 1 = 0.14046
b = -6757789500 2 = 0.00396
c = 26000000 new = 0.00396 choose the smaller
rho new < rho min, use rho min
As= 418.78 mm
2Final: = 16 mm
Bar Diam.()= 16 mm n= 3 bars
no. of bars (n)= 2.08 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00731
min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 33.31345 mm Strain of steel Yeilds:
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=Mu/R Mn = Mu/
a^2+b+c=0
a = (As
n =
As/(((2)
= - ^ -
Mu = Mn= fybd^2 (1-0.59 fy/(f^ c))
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Design Moment:
(+)Mu= 15 KN-m
Cross Section of Beam:
b = 250 mm
d = 330 mm
h = 400 mm
Area of Steel:
Roots:
a = 46793235022 1 = 0.14216
b = -6757789500 2 = 0.00225
c = 15000000 new = 0.00225 choose the smaller
rho new < rho min, use rho min
As= 418.78 mm
2Final: = 16 mm
Bar Diam.()= 16 mm n= 3 bars
no. of bars (n)= 2.08 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00731
min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are needed:
Ultimate Shear (Vu)= 40.00 KN/m
Shear Strenght by Concrete (Vc) = 67.99 KN/m
Vc= 50.99 KN/m
Vu < 0.5Vc, Stirrups not needed!
Vu < Vc, Provide minimum shear reinforcement!
Design for Shear Reinforcement:
= 8 mm
Av min.= 100.5312 mm2
Maximum Spacing (smax):
a.) smax1= 165 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 165 mm
Positive Moment
a^2+b+c=0= - ^ -
Mu = Mn= fybd^2 (1-0.59 fy/(f^ c))
Av(min) =
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Bar cut-off and Development Length:
Moment Diagram:
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1102
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 603.19 mm2
positive moment
As= 603.19 mm2
negative moment
ln= 3.40 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off Positive 1 201.06 66.67 0.260 0.88
---- ---- ---- ---- ---- ---- ----
Negative 2 201.06 66.67 0.102 0.35
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 23 mm
s < 2db, use other cases equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s= 1
= 1
Developement Length:
ld= 0.44 m
12db= 0.19 m
d= 0.33 m
ld=12fyte/(25(fc)) db
As%= (As -As1)/As x 100
Cut-Off= factor x 100
s =(b-2(cover+stirrup)-
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Total 7288*Verify actual to minimize cost
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa1 = 0.85
= 0.90
u = 0.003
Cover= 60.00 mm
Assume Section : l1= 1.90 m
b = 250 mm l2= 1.90 m
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80Weight of Slab (Ws) 24.00 ---- 11.4
Live Loads: Kpa KN/m
2.4 9.12
Ultimate Load: KN/mWu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DL + 1.6LL = 28.27
Wu= 30.43
Beam RB-2:
Line of symmetry
Note: Choose biggest value of Shear & Moment.
Shear Diagram:
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam RB-2 - Roof Deck Framing Plan Rating:
Wb=
l1
Wu
Ws=
L = 7.60 m L = 7.60 m
Ws + Wp
Wu=Wb +
1
)
l2
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Design For Flexure:
Design Moment:
Mu= 165 KN-m
Designer's Preferred Steel Ratio:
b = 0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 183.33 Kn-m
bd2
= 31849369.25 mm3
b = 250 mm Fexural Resistance factor
d = 356.93 mm
h' = 356.93 mm R= 5.75626h' > h(assume), Revise Section!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 49672157654 1 = 0.11571
b = -7173558000 2 = 0.02871
c = 165000000 new = 0.02871 choose the smaller
rho new > rho min, use rho new
As= 2440.16 mm
2Final: = 20 mm
()= 20 mm n= 8 bars
(n)= 7.77 bars As= 2513.280 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.02957
(governs) min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 138.80603 mm Strain of steel Yeilds:
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=/
desired =0.6
Mn = /
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
a = (
n =
= 0.851
(^)/
(600/(600+))
=
(10.59/(
^))
=((^24))/2
Mu = = ^2(10.59/(^))
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Design Moment:
(+)Mu= 84 KN-m
Cross Section of Beam:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 49672157654 1 = 0.13156
b = -7173558000 2 = 0.01285
c = 84000000 new = 0.01285 choose the smaller
rho new > rho min, use rho new
As= 1092.56 mm
2Final: = 20 mm
()= 20 mm n= 4 bars
(n)= 3.48 bars As= 1256.640 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.01478
(governs) min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are needed:
Ultimate Shear (Vu)= 123.10 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu > Vc, calculate Shear Strenght of steel!
Positive Moment
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
=0.17(^ ) bwd
=((^24))/2
Mu = = ^2(10.59/(^))
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Design For Shear: Shear Diagram:
At Right Support:
= 8 mm
= 0.75 ACI CODE 2010
Factored Shear Force:
Vu= 123.10 KN
Shear Strength in Concrete:
Vc= 70.05 KN
Vc= 52.54 KN
0.5Vc= 26.27 KN
Spacing (s) Required:
Av= 100.5312 mm2
S= 100.20 mm
say: 100 mm
Shear Strenght in Steel:
Vs= 70.70 KN
Point of Minimum Stirrups starts:
By ratio and proportion:
Xu= 4.04 m
X1 = 2.32 m
Point where Stirrups are not needed:By ratio and proportion:
Xu= 4.04 m
X2 = 3.18 m
Design For Shear Reinforcement:
Vs
70.70 271954.84 ok, proceed to smax
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Check for maximun spacing:
Vs
Vs
70.70 135.98
smax will not be halved!
Smax = 170 mm
/=/(1)
/=(0.5)/(2)
0.66(^)bwd
Vs = ()/
0.33(^)bwd
s = (
)/()
, smax /224"
0.33(^)bwd
, smax /424"
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Design For Shear: Shear Diagram:
At Left Support:
Right = 8 mm
8mm = 0.75 ACI CODE 2010
Factored Shear Force:Vu= 108.20 KN
SAME
Shear Strength in Concrete:
Vc= 70.05 KN
Vc= 52.54 KN
0.5Vc= 26.27 KN
Spacing (s) Required:
Av= 100.5312 mm2
S= 127.02 mmsay: 125 mm
Shear Strenght in Steel:
Vs= 56.56 KN
Point of Minimum Stirrups starts:
By ratio and proportion:
Xu= 3.56 m
X1 = 1.83 m
Point where Stirrups are not needed:
By ratio and proportion:
Xu= 3.56 m
X2 = 2.70 m
Design For Shear Reinforcement:
Vs
56.56 271954.84 ok, proceed to smax
Maximum Spacing (smax):a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Check for maximun spacing:
Vs
Vs
56.56 135.98
smax will not be halved!
/=/(1)
/=(0.5)/(2)
0.66(^)bwd
Vs = ()/
0.33(^)bwd
s = (
)/()
, smax /224"
0.33(^)bwd
, smax /424"
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15 @ 0.125 m
Rest @ 0.170 m
Bar cut-off and Development Length:
Moment Diagram:
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1102
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 1256.64 mm2
positive moment
As= 2513.28 mm2
negative moment
ln= 7.20 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off
Positive 1 418.88 66.67 0.260 1.87
---- ---- ---- ---- ---- ---- ----
Negative 2 837.76 66.67 0.102 0.73
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 34 mm
s < 2db, use other cases equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s= 1
s =(b-2(cover+stirrup)-
As%= (As -As1)/As x 100
Cut-Off= factor x 100
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Beam Details:
Beam Sections Stirrups
Left Midspan Right Left
8mm
*Closed Stirrups were used for the ease of construction*
*Safe & Aesthetic design acquires if beams are uniformly tied (Structral Integrity)
Section:
Estimate for Beam:
Total Volume of Concrete: Class A Mixture: Factor
Vol.conc.= 0.720 m2
Cement (40kg) 9.00
sand 0.50
Use Class A Mixture (1:2:4) Gravel 1.00
No. of Bags of Cement:
No.bags= 6.48 bags
say: 7
Volume of Sand:
Vsand= 0.36 m
3
say: 1
Volume of Gravel:
Vgravel= 0.72 m3
say: 1
Reinforcing Bars:
Top bars: 7 bars (20mm)
Bot. bars: 5 bars (20mm) Direct Counting Method
Stirrups: 10 bars (8mm)
Summary: 2 RB-2
Item Quantity Unit Cost Total
1 @ 0.065 m15 @ 0.125 m
Rest @ 0.170 mO.C
1 @23 @
RestO.C
Vol =L x W x H
No. bags =Volume x Factor
Vol=Vsand x Factor
Vol=Vgravel x Factor
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa1 = 0.85
= 0.90 l
u = 0.003
Cover= 60.00 mm
Assume Section : l= 1.90 m
b = 250 mm
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80Weight of Slab (Ws) 24.00 ---- 5.7
Parapet (Normal weight concrete,Wp1) 21.20 ---- 1.91
Parapet (Decorative Tile Finished, Wp2) ---- 0.77 0.77
Live Loads: Kpa KN/m
2.4 4.56
Ultimate Load: KN/m
Wu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DL + 1.6LL = 14.14
Wu(Parapet)= 1.2DL= 3.21
Wu= 19.51
Beam RB-3:
Shear Diagram:
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam RB-3 - Roof Deck Framing Plan Rating:
(1+2)
s
Wb=
Wu
Ws=
L = 3.80 m
Wu = +
Wp2=
Wp1=(
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Design For Flexure:
Design Moment:
Mu= 56 KN-m
Designer's Preferred Steel Ratio:
b = 0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 62.22 Kn-m
bd2
= 10809482.9 mm3
b = 250 mm Fexural Resistance factor
d = 207.94 mm
h' = 207.94 mm R= 5.75626h (assume) > h', ok!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 49672157654 1 = 0.13614
b = -7173558000 2 = 0.00828
c = 56000000 new = 0.00828 choose the smaller
rho new > rho min, use rho new
As= 703.91 mm
2Final: = 16 mm
()= 16 mm n= 4 bars
(n)= 3.50 bars As= 804.250 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00946
(governs) min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 44.41793 mm Strain of steel Yeilds:
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=/
desired =0.6
Mn = /
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
a = (
n =
= 0.851
(^)/
(600/(600+))
=
(10.59/(
^))
=((^24))/2
Mu = = ^2(10.59/(^))
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Design Moment:
(+)Mu= 8 KN-m
Cross Section of Beam:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 49672157654 1 = 0.14329
b = -7173558000 2 = 0.00112
c = 8000000 new = 0.00112 choose the smaller
rho new < rho min, use rho min
As= 431.47 mm
2Final: = 16 mm
()= 16 mm n= 4 bars
(n)= 2.15 bars As= 804.250 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00946
(governs) min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are needed:
Ultimate Shear (Vu)= 49.80 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu < Vc, Provide minimum shear reinforcement!
Design for Shear Reinforcement:
= 8 mm
Av min.= 100.5312 mm2
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Positive Moment
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
=0.17(^ ) bwd
=((^24))/2
Mu = = ^2(10.59/(^))
Av(min) = (
2)/4
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Bar cut-off and Development Length:
Moment Diagram:
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1102
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 804.25 mm2
positive moment
As= 804.25 mm2
negative moment
ln= 3.40 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off Positive 1 268.08 66.67 0.260 0.88
---- ---- ---- ---- ---- ---- ----
Negative 2 268.08 66.67 0.102 0.35
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 82 mm
s > 2db, use case 1 or 2 equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s= 1
= 1
Developement Length:
ld= 0.44 m
12db= 0.19 m
d= 0.34 m
ld=12/(25()) db
As%= (1)/ x100
Cut-Off= factor x 100
s =(2(+)())/(1)
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Beam Details:
Beam Sections Stirrups
Left Midspan Right Left
8mm
*Closed Stirrups were used for the ease of construction*
*Safe & Aesthetic design acquires if beams are uniformly tied (Structral Integrity)
Section:
Estimate for Beam:
Total Volume of Concrete: Class A Mixture: Factor
Vol.conc.= 0.340 m2
Cement (40kg) 9.00
sand 0.50
Use Class A Mixture (1:2:4) Gravel 1.00
No. of Bags of Cement:
No.bags= 3.06 bags
say: 4
Volume of Sand:Vsand= 0.17 m
3
say: 1
Volume of Gravel:
Vgravel= 0.34 m3
say: 1
Reinforcing Bars:
Top bars: 3 bars (16mm)
Bot. bars: 3 bars (16mm) Direct Counting Method
Stirrups: 4 bars (8mm)
1 @ 0.085 m
Rest @ 0.170 m
O.C
Vol =L x W x H
No. bags =Volume x Factor
Vol=Vsand x Factor
Vol=Vgravel x Factor
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*Verify actual to minimize cost
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Right
8mm
0.075 m0.150 m
@ 0.170 m
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa1 = 0.85
= 0.90 l
u = 0.003
Cover= 60.00 mm
Assume Section : l= 1.90 m
b = 250 mm
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80Weight of Slab (Ws) 24.00 ---- 5.7
Parapet (Normal weight concrete,Wp1) 21.20 ---- 1.91
Parapet (Decorative Tile Finished, Wp2) ---- 0.77 0.77
Live Loads: Kpa KN/m
Office 2.4 4.56
Ultimate Load: KN/m
Wu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DL + 1.6LL= 14.14
Wu(Parapet)= 1.2DL= 3.21
Wu= 19.51
Beam RB-3A:
Shear Diagram:
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam RB-3A - Roof Deck Framing Plan Rating:
Wp2=
Wb=
Wu
Wp1=(
Ws=
L = 7.60 m
Ws + Wp
1
)
Wu = +
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Design For Flexure:
Design Moment:
Mu= 85 KN-m
Designer's Preferred Steel Ratio:
b = 0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 94.44 Kn-m
bd2
= 16407250.83 mm3
b = 250 mm Fexural Resistance factor
d = 256.18 mm
h' = 256.18 mm R= 5.75626h (assume) > h', ok!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 49672157654 1 = 0.13139
b = -7173558000 2 = 0.01302
c = 85000000 new = 0.01302 choose the smaller
rho new > rho min, use rho new
As= 1107.00 mm
2Final: = 16 mm
()= 16 mm n= 6 bars
(n)= 5.51 bars As= 1206.374 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.01419
(governs) min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 66.62690 mm Strain of steel Yeilds:
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=/
desired =0.6
Mn = /
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
a = (
n =
= 0.851
(^)/
(600/(600+))
=
(10.59/(
^))
=((^24))/2
Mu = = ^2(10.59/(^))
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Design Moment:
(+)Mu= 56 KN-m
Cross Section of Beam:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 49672157654 1 = 0.13614
b = -7173558000 2 = 0.00828
c = 56000000 new = 0.00828 choose the smaller
rho new > rho min, use rho new
As= 703.91 mm
2Final: = 16 mm
()= 16 mm n= 4 bars
(n)= 3.50 bars As= 804.250 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00946
(governs) min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are needed:
Ultimate Shear (Vu)= 74.10 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu > Vc, calculate Shear Strenght of steel!
Positive Moment
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
=0.17(^ ) bwd
=((^24))/2
Mu = = ^2(10.59/(^))
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Design For Shear: Shear Diagram:
At Right & Left Support:
= 8 mm
= 0.75 ACI CODE 2010
Factored Shear Force:
Vu= 74.10 KN
Shear Strength in Concrete:
Vc= 70.05 KN
Vc= 52.54 KN
0.5Vc= 26.27 KN
Spacing (s) Required:
Av= 100.5312 mm2
S= 327.88 mm
say: 325 mm
Shear Strenght in Steel:
Vs= 21.75 KN
Point of Minimum Stirrups starts:
By ratio and proportion:
Xu= 3.8 m
X1 = 1.11 m
Point where Stirrups are not needed:By ratio and proportion:
Xu= 3.8 m
X2 = 2.45 m
Design For Shear Reinforcement:
Vs
21.75 271954.84 ok, proceed to smax
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Check for maximun spacing:
Vs
Vs
21.75 135.98
smax will not be halved!
Smax = 170 mm
/=/(1)
/=(0.5)/(2)
0.66(^)bwd
Vs = ()/
0.33(^)bwd
s = (
)/()
, smax /224"
0.33(^)bwd
, smax /424"
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Bar cut-off and Development Length:
Moment Diagram:
Right
8mm
SAME
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1101
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 804.25 mm2
positive moment
As= 1206.37 mm2
negative moment
ln= 7.20 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off
Positive 1 268.08 66.67 0.310 2.23
---- ---- ---- ---- ---- ---- ----
Negative 2 402.12 66.67 0.095 0.68
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 33 mm
s > 2db, use case 1 or 2 equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s =(2(+)())/(1)
As%= (1)/ x100
Cut-Off= factor x 100
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Beam Details:
Beam Sections Stirrups
Left Midspan Right Left
8mm
*Closed Stirrups were used for the ease of construction*
*Safe & Aesthetic design acquires if beams are uniformly tied (Structral Integrity)
Section:
Estimate for Beam:
Total Volume of Concrete: Class A Mixture: Factor
Vol.conc.= 0.720 m2
Cement (40kg) 9.00
sand 0.50
Use Class A Mixture (1:2:4) Gravel 1.00
No. of Bags of Cement:
No.bags= 6.48 bags
say: 7
Volume of Sand:
Vsand= 0.36 m
3
say: 1
Volume of Gravel:
Vgravel= 0.72 m3
say: 1
Reinforcing Bars:
Top bars: 5 bars (16mm)
Bot. bars: 5 bars (16mm) Direct Counting Method
Stirrups: 8 bars (8mm)
Summary: 1 RB-3A
1 @ 0.085 m
Rest @ 0.170 m
O.C
Vol =L x W x H
No. bags =Volume x Factor
Vol=Vsand x Factor
Vol=Vgravel x Factor
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa1 = 0.85
= 0.90
u = 0.003
Cover= 60.00 mm
Assume Section : l= 1.90 m
b = 250 mm
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80Weight of Slab (Ws) 24.00 ---- 5.7
Parapet (Normal weight concrete,Wp1) 21.20 ---- 1.91
Parapet (Decorative Tile Finished, Wp2) ---- 0.77 0.77
Live Loads: Kpa KN/m
Office 2.4 4.56
Ultimate Load: KN/m
Wu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DL + 1.6LL = 14.14
Wu(Parapet)= 1.2DL= 3.21
Wu= 19.51
Beam RB-4:
Line of symmetry
Note: Choose biggest value of Shear & Moment.
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam RB-4 - Roof Deck Framing Plan Rating:
1
)
1
Wp2=
Wb=
Wu
l
Wp1=(
Ws=
L = 3.80 m L = 3.80 m
Ws + Wp Wu =Wb +
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Design For Flexure:
Design Moment:
Mu= 28 KN-m
Designer's Preferred Steel Ratio:
b = 0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 31.11 Kn-m
bd2
= 5404741.449 mm3
b = 250 mm Fexural Resistance factor
d = 147.03 mm
h' = 147.04 mm R= 5.75626h (assume) > h', ok!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 49672157654 1 = 0.14040
b = -7173558000 2 = 0.00401
c = 28000000 new = 0.00401 choose the smaller
rho new < rho min, use rho min
As= 431.47 mm
2Final: = 16 mm
()= 16 mm n= 3 bars
(n)= 2.15 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00710
(governs) min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 33.31345 mm Strain of steel Yeilds:
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=/
desired =0.6
Mn = /
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
a = (
n =
= 0.851
(^)/
(600/(600+))
=
(10.59/(
^))
=((^24))/2
Mu = = ^2(10.59/(^))
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Design Moment:
(+)Mu= 16 KN-m
Cross Section of Beam:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 49672157654 1 = 0.14215
b = -7173558000 2 = 0.00227
c = 16000000 new = 0.00227 choose the smaller
rho new < rho min, use rho min
As= 431.47 mm
2Final: = 16 mm
()= 16 mm n= 3 bars
(n)= 2.15 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00710
(governs) min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are needed:
Ultimate Shear (Vu)= 40.00 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu < Vc, Provide minimum shear reinforcement!
Design for Shear Reinforcement:
= 8 mm
Av min.= 100.5312 mm2
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Positive Moment
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
=0.17(^ ) bwd
=((^24))/2
Mu = = ^2(10.59/(^))
Av(min) = (
2)/4
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Bar cut-off and Development Length:
Moment Diagram:
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1102
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 603.19 mm2
positive moment
As= 603.19 mm2
negative moment
ln= 3.40 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off Positive 1 201.06 66.67 0.260 0.88
---- ---- ---- ---- ---- ---- ----
Negative 2 201.06 66.67 0.102 0.35
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 33 mm
s > 2db, use case 1 or 2 equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s= 1
= 1
Developement Length:
ld= 0.44 m
12db= 0.19 m
d= 0.34 m
ld=12/(25()) db
As%= (1)/ x100
Cut-Off= factor x 100
s =(2(+)())/(1)
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Beam Details:
Beam Sections Stirrups
Left Midspan Right Left
8mm
*Closed Stirrups were used for the ease of construction*
*Safe & Aesthetic design acquires if beams are uniformly tied (Structral Integrity)
Section:
Estimate for Beam:
Total Volume of Concrete: Class A Mixture: Factor
Vol.conc.= 0.340 m2
Cement (40kg) 9.00
sand 0.50
Use Class A Mixture (1:2:4) Gravel 1.00
No. of Bags of Cement:
No.bags= 3.06 bags
say: 4Volume of Sand:
Vsand= 0.17 m3
say: 1
Volume of Gravel:
Vgravel= 0.34 m3
say: 1
Reinforcing Bars:
Top bars: 2 bars (16mm)
Bot. bars: 2 bars (16mm) Direct Counting Method
Stirrups: 4 bars (8mm)
1 @ 0.085 m
Rest @ 0.170 m
O.C
Vol =L x W x H
No. bags =Volume x Factor
Vol=Vsand x Factor
Vol=Vgravel x Factor
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Total 3644*Verify actual to minimize cost
Right
8mm
SAME
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa1 = 0.85
= 0.90
u = 0.003
Cover= 60.00 mm
Assume Section : l1= 1.90 m
b = 250 mm l2= 1.90 m
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80Weight of Slab (Ws) 24.00 ---- 11.4
Live Loads: Kpa KN/m
Office 2.4 9.12
Ultimate Load: KN/mWu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DL + 1.6LL = 28.27
Wu= 30.43
Beam RB-5:
Shear Diagram:
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam RB-5 - Roof Deck Framing Plan Rating:
1
)
Wb=
Wu
L = 7.60 m
Ws + Wp
l1
l2
Ws=
Wu =Wb +
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Design For Flexure:
Design Moment:
Mu= 126 KN-m
Designer's Preferred Steel Ratio:
b = 0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 140.00 Kn-m
bd2
= 24321336.52 mm3
b = 250 mm Fexural Resistance factor
d = 311.91 mm
h' = 311.91 mm R= 5.75626h' > h(assume), Revise Section!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 49672157654 1 = 0.12395
b = -7173558000 2 = 0.02046
c = 126000000 new = 0.02046 choose the smaller
rho new > rho min, use rho new
As= 1739.47 mm
2Final: = 20 mm
()= 20 mm n= 6 bars
(n)= 5.54 bars As= 1884.960 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.02218
(governs) min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 104.10452 mm Strain of steel Yeilds:
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=/
desired =0.6
Mn = /
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
a = (
n =
= 0.851
(^)/
(600/(600+))
=
(10.59/(
^))
=((^24))/2
Mu = = ^2(10.59/(^))
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Design Moment:
(+)Mu= 101 KN-m
Cross Section of Beam:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 49672157654 1 = 0.12861
b = -7173558000 2 = 0.01581
c = 101000000 new = 0.01581 choose the smaller
rho new > rho min, use rho new
As= 1343.88 mm
2Final: = 20 mm
()= 20 mm n= 5 bars
(n)= 4.28 bars As= 1570.800 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.01848
(governs) min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are needed:
Ultimate Shear (Vu)= 117.40 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu > Vc, calculate Shear Strenght of steel!
Positive Moment
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
=0.17(^ ) bwd
=((^24))/2
Mu = = ^2(10.59/(^))
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Design For Shear: Shear Diagram:
At Right Support:
= 8 mm
= 0.75 ACI CODE 2010
Factored Shear Force:
Vu= 113.90 KN
Shear Strength in Concrete:
Vc= 70.05 KN
Vc= 52.54 KN
0.5Vc= 26.27 KN
Spacing (s) Required:
Av= 100.5312 mm2
S= 115.22 mm
say: 115 mm
Shear Strenght in Steel:
Vs= 61.48 KN
Point of Minimum Stirrups starts:
By ratio and proportion:
Xu= 3.74 m
X1 = 2.01 m
Point where Stirrups are not needed:By ratio and proportion:
Xu= 3.74 m
X2 = 2.88 m
Design For Shear Reinforcement:
Vs
61.48 271954.84 ok, proceed to smax
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Check for maximun spacing:
Vs
Vs
61.48 135.98
smax will not be halved!
Smax = 170 mm
/=/(1)
/=(0.5)/(2)
0.66(^)bwd
Vs = ()/
0.33(^)bwd
s = (
)/()
, smax /224"
0.33(^)bwd
, smax /424"
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Design For Shear: Shear Diagram:
At Left Support:
Right = 8 mm
8mm = 0.75 ACI CODE 2010
Factored Shear Force:Vu= 117.40 KN
SAME
Shear Strength in Concrete:
Vc= 70.05 KN
Vc= 52.54 KN
0.5Vc= 26.27 KN
Spacing (s) Required:
Av= 100.5312 mm2
S= 109.00 mmsay: 110 mm
Shear Strenght in Steel:
Vs= 64.28 KN
Point of Minimum Stirrups starts:
By ratio and proportion:
Xu= 3.86 m
X1 = 2.13 m
Point where Stirrups are not needed:
By ratio and proportion:
Xu= 3.86 m
X2 = 3.00 m
Design For Shear Reinforcement:
Vs
64.28 271954.84 ok, proceed to smax
Maximum Spacing (smax):a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Check for maximun spacing:
Vs
Vs
64.28 135.98
smax will not be halved!
/=/(1)
/=(0.5)/(2)
0.66(^)bwd
Vs = ()/
0.33(^)bwd
s = (
)/()
, smax /224"
0.33(^)bwd
, smax /424"
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19 @ 0.110 m
Rest @ 0.170 m
Bar cut-off and Development Length:
Moment Diagram:
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1102
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 1570.80 mm2
positive moment
As= 1884.96 mm2
negative moment
ln= 7.20 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off
Positive 1 523.60 66.67 0.260 1.87
---- ---- ---- ---- ---- ---- ----
Negative 2 628.32 66.67 0.095 0.68
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 27 mm
s < 2db, use other cases equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s= 1
s =(b-2(cover+stirrup)-
As%= (As -As1)/As x 100
Cut-Off= factor x 100
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Beam Details:
Beam Sections Stirrups
Left Midspan Right Left
8mm
*Closed Stirrups were used for the ease of construction*
*Safe & Aesthetic design acquires if beams are uniformly tied (Structral Integrity)
Section:
Estimate for Beam:
Total Volume of Concrete: Class A Mixture: Factor
Vol.conc.= 0.720 m2
Cement (40kg) 9.00
sand 0.50
Use Class A Mixture (1:2:4) Gravel 1.00
No. of Bags of Cement:
No.bags= 6.48 bags
say: 7
Volume of Sand:
Vsand= 0.36 m
3
say: 1
Volume of Gravel:
Vgravel= 0.72 m3
say: 1
Reinforcing Bars:
Top bars: 5 bars (20mm)
Bot. bars: 6 bars (20mm) Direct Counting Method
Stirrups: 10 bars (8mm)
Summary: 1 RB-5
Item Quantity Unit Cost Total
1 @ 0.055 m19 @ 0.110 m
Rest @ 0.170 mO.C
1 @18 @
RestO.C
Vol =L x W x H
No. bags =Volume x Factor
Vol=Vsand x Factor
Vol=Vgravel x Factor
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa1 = 0.85
= 0.90
u = 0.003
Cover= 60.00 mm
Assume Section : l1= 1.90 m
b = 250 mm l2= 1.90 m
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80Weight of Slab (Ws) 24.00 ---- 11.4
Live Loads: Kpa KN/m
Office 2.4 9.12
Ultimate Load: KN/mWu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DL + 1.6LL= 28.27
Wu= 30.43
Beam RB-6:
Note: Choose biggest value of Shear & Moment.
Shear Diagram:
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam RB-6 - Roof Deck Framing Plan Rating:
(1+2)
Ws
Wb=
Wu
L = 3.80 m
l1
l2
Ws=
Wu =Wb +
L = 3.80 m
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Design For Flexure:
Design Moment:
Mu= 45 KN-m
Designer's Preferred Steel Ratio:
b = 0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 50.00 Kn-m
bd2
= 8686191.614 mm3
b = 250 mm Fexural Resistance factor
d = 186.40 mm
h' = 186.40 mm R= 5.75626h (assume) > h', ok!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 49672157654 1 = 0.13785
b = -7173558000 2 = 0.00657
c = 45000000 new = 0.00657 choose the smaller
rho new > rho min, use rho new
As= 558.63 mm
2Final: = 16 mm
()= 16 mm n= 3 bars
(n)= 2.78 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00710
(governs) min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 33.31345 mm Strain of steel Yeilds:
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=/
desired =0.6
Mn = /
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
a = (
n =
= 0.851
(^)/
(600/(600+))
=
(10.59/(
^))
=((^24))/2
Mu = = ^2(10.59/(^))
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Design Moment:
(+)Mu= 25 KN-m
Cross Section of Beam:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 49672157654 1 = 0.14084
b = -7173558000 2 = 0.00357
c = 25000000 new = 0.00357 choose the smaller
rho new < rho min, use rho min
As= 431.47 mm
2Final: = 16 mm
()= 16 mm n= 3 bars
(n)= 2.15 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00710
(governs) min = 0.00508 0.00439 0.00508
rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are needed:
At Right Support:
Ultimate Shear Moment (Vu)= 52.00 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu < Vc, Provide minimum shear reinforcement!
At Left Support:
Ultimate Shear Moment (Vu)= 64.10 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu > Vc, calculate Shear Strenght of steel!
Positive Moment
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
=((^24))/2
Mu = = ^2(10.59/(^))
=0.17(^ ) bwd
=0.17(^ ) bwd
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Design For Shear: Shear Diagram:
At Left Support:
= 8 mm
= 0.75 ACI CODE 2010
Factored Shear Force:
Vu= 64.00 KN
Shear Strength in Concrete:
Vc= 70.05 KN
Vc= 52.54 KN
0.5Vc= 26.27 KN
Spacing (s) Required:
Av= 100.5312 mm2
S= 616.78 mm
say: 420 mm
Shear Strenght in Steel:
Vs= 16.83 KN
Point of Minimum Stirrups starts:
By ratio and proportion:
Xu= 2.1 m
X1 = 0.38 m
Point where Stirrups are not needed:By ratio and proportion:
Xu= 2.1 m
X2 = 1.24 m
Design For Shear Reinforcement:
Vs
16.83 271954.84 ok, proceed to smax
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Check for maximun spacing:
Vs
Vs
16.83 135.98
smax will not be halved!
Smax = 170 mm
/=/(1)
/=(0.5)/(2)
0.66(^)bwd
Vs = ()/
0.33(^)bwd
s = (
)/()
, smax /224"
0.33(^)bwd
, smax /424"
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Design For Shear: Shear Diagram:
At Right Support:
= 8 mm
= 0.75 ACI CODE 2010Factored Shear Force:
Vu= 52.00 KN
Design for Shear Reinforcement:
= 8 mm
Av min.= 100.5312 mm2
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mmc.)smax3= 327 mm
Use smax= 170 mm
Actual Spacing (s):
s= 170 mm
First Stirrup:
s1= 85 mm
Stirrups:
1 @ 0.085 m
Rest @ 0.170 m
Av(min) = (
2)/4
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Bar cut-off and Development Length:
Moment Diagram:
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1102
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 603.19 mm2
positive moment
As= 603.19 mm2
negative moment
ln= 3.40 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off
Positive 1 201.06 66.67 0.260 0.88
---- ---- ---- ---- ---- ---- ----
Negative 2 201.06 66.67 0.102 0.35
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 33 mm
s > 2db, use case 1 or 2 equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s =(b-2(cover+stirrup)-
As%= (As -As1)/As x 100Cut-Off= factor x 100
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Beam Details:
Beam Sections Stirrups
Right Left Midspan Right Left
8mm 8mm
1 @ 0.085 m
Rest @ 0.170 m
O.C.
*Closed Stirrups were used for the ease of construction*
*Safe & Aesthetic design acquires if beams are uniformly tied (Structral Integrity)
Section:
Estimate for Beam:
Total Volume of Concrete: Class A Mixture: Factor
Vol.conc.= 0.340 m2
Cement (40kg) 9.00
sand 0.50
Use Class A Mixture (1:2:4) Gravel 1.00
No. of Bags of Cement:
No.bags= 3.06 bags
say: 4
Volume of Sand:
Vsand= 0.17 m
3
say: 1
Volume of Gravel:
Vgravel= 0.34 m3
say: 1
Reinforcing Bars:
Top bars: 2 bars (16mm)
Bot. bars: 2 bars (16mm) Direct Counting Method
Stirrups: 4 bars (8mm)
Summary: 2 RB-6
Item Quantity Unit Cost Total
0.055 m0.155 m
@ 0.170 m
Vol =L x W x H
No. bags =Volume x Factor
Vol=Vsand x Factor
Vol=Vgravel x Factor
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(1+2)
Ws
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(governs)
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(governs)
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Right
8mm
Same
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa
1 = 0.85
= 0.90
u = 0.003
Cover= 60.00 mm
Assume Section: l1= 0.50 m
b = 250 mm l2= 1.90 m
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80
Weight of Slab (Ws1 + Ws2 ) 24.00 ---- 7.20
Wall (Normal weight concrete, Ww1) 21.20 ---- 6.36
Wall (Foam/Plastered Finished, Ww2) ---- 0.48 0.024
Live Loads: Kpa KN/m
Office 2.4 5.76
Ultimate Load: KN/m
Wu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DL + 1.6LL = 17.86
Wu(wall)= 1.2DL= 7.66
Wu1= 27.68
Wu2= 13.54
Beam B-1:
Line of symmetry
Note: Choose biggest value of Shear & Moment.
Shear Diagram:
Two-Story office Building
A. Design of Beams
Beam B-1 - 2nd Floor Framing Plan
Date Prepared:
Checked By:
Rating:
Ww2= 1
Wb=
Ww1=()
Wu1=Wb+ Ws+ Ww
Wu2=Wb+Ws2+Ww
Wu1
Ws= (1+2)
L = 7.60 m L = 7.60 m
l1
l2
Wu2Wu1
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Design For Flexure:
Design Moment:
Mu= 25 KN-m
Designer's Preferred Steel Ratio:
b
=0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 27.78 Kn-m
bd2
= 4825662.01 mm3
b = 250 mm Fexural Resistance factor
d = 138.93 mm
h' = 138.94 mm R= 5.75626
h (assume) > h', ok!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 4.9672E+10 1 = 0.14084
b = -7173558000 2 = 0.00357
c = 25000000 new = 0.00357 choose the smaller
rho new < rho min, use rho min
As= 431.47 mm
2Final: = 16 mm
Diam.()= 16 mm n= 3 bars
f bars (n)= 2.15 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:max = 0.03163
actual = 0.00710
min = 0.00508 0.00439 0.00508 (governs)
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 33.31345 mm Strain of steel Yeilds: = 0 001379
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=/
desired =0.6
Mn = /
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
a = (
)/(0 85^ =/
n =
= 0.851 (^)/(600/(600+))
= (10.59/(^))
=((^24))/2
Mu = = ^2(10.59/(^))
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Design Moment:
(+)Mu= 15 KN-m
Cross Section of Beam:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 4.9672E+10 1 = 0.14230
b = -7173558000 2 = 0.00212
c = 15000000 new = 0.00212 choose the smaller
rho new < rho min, use rho min
As= 431.47 mm
2Final: = 16 mm
Diam.()= 16 mm n= 3 bars
f bars (n)= 2.15 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00710
min = 0.00508 0.00439 0.00508 (governs)
rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are needed:
Ultimate Shear Moment (Vu)= 46.90 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu < Vc, Provide minimum shear reinforcement!
Design for Shear Reinforcement:
= 8 mm
Av min.= 100.5312 mm2
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Positive Moment
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
=0.17(^ ) bwd
=((^24))/2
Mu = = ^2(10.59/(^))
Av(min) = (
2)/4
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Bar cut-off and Development Length:
Moment Diagram:
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1102
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 603.19 mm2
positive moment
As= 603.19 mm2
negative moment
ln= 3.40 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off
Positive 1 201.06 66.67 0.260 0.88
---- ---- ---- ---- ---- ---- ----
Negative 2 201.06 66.67 0.102 0.35
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 33 mm
s > 2db, use case 1 or 2 equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s= 1
= 1
Developement Length:
s =(b-2(cover+stirrup)-
As%= (As -As1)/As x 100
Cut-Off= factor x 100
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Beam Details:
Beam Sections Stirrups
Left Midspan Right Left Right
8mm 8mm
1 @ 0.085 m
SAME
Rest @ 0.170 mO.C.
*Closed Stirrups were used for the ease of construction*
*Safe & Aesthetic design acquires if beams are uniformly tied (Structral Integrity)
Section:
Estimate for Beam:
Total Volume of Concrete: Class A Mixture: Factor
Vol.conc.= 0.340 m2
Cement (40kg) 9.00
sand 0.50
Use Class A Mixture (1:2:4) Gravel 1.00
No. of Bags of Cement:
No.bags= 3.06 bags
say: 4
Volume of Sand:
Vsand= 0.17 m3
say: 1
Volume of Gravel:
Vgravel= 0.34 m3
say: 1
Reinforcing Bars:
Top bars: 2 bars (16mm)
Bot. bars: 2 bars (16mm) Direct Counting Method
Stirrups: 4 bars (8mm)
Vol =L x W x H
No. bags =Volume x Factor
Vol=Vsand x Factor
Vol=Vgravel x Factor
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa
1 = 0.85 l1
= 0.90
u = 0.003
Cover= 60.00 mm
Assume Section: l1= 1.90 m
b = 250 mm l2= 1.90 m
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80
Weight of Slab (Ws1+ Ws2) 24.00 ---- 11.40
Wall (Normal weight concrete, Ww1) 21.20 ---- 6.36
Wall (Foam/Plastered Finished, Ww2) ---- 0.48 0.024
Live Loads: Kpa KN/m
Office 2.4 9.12
Ultimate Load: KN/m
Wu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DL + 1.6LL= 28.27
Wu(wall)= 1.2DL = 7.66
Wu1= 30.43
Wu2= 23.96
Beam B-2:
Line of symmetry
Note: Choose biggest value of Shear & Moment.
Shear Diagram:
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam B-2 - 2nd Floor Framing Plan Rating:
Wb=
Wu1Wu1
Ws= (1+2)
Wu2
L = 7.60 m L = 7.60 m
l2
Ww2= 1
Ww1=()
Wu2=Wb+ Ws1+ Ww
Wu1=Wb+Ws
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Design For Flexure:
Design Moment:
Mu= 134 KN-m
Designer's Preferred Steel Ratio:
b
=0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 148.89 Kn-m
bd2
= 25865548.4 mm3
b = 250 mm Fexural Resistance factor
d = 321.66 mm
h' = 321.66 mm R= 5.75626
h' > h(assume),Revise Section!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 4.9672E+10 1 = 0.12237
b = -7173558000 2 = 0.02204
c = 134000000 new = 0.02204 choose the smaller
rho new > rho min, use rho new
As= 1873.80 mm
2Final: = 20 mm
= 20 mm n= 6 bars
n= 5.96 bars As= 1884.960 mm2
Checking for ACI requirements:
Steel ratio:max = 0.03163
actual = 0.02218
min = 0.00508 0.00439 0.00508 (governs)
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 104.10452 mm Strain of steel Yeilds: = 0 001379
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=/
desired =0.6
Mn = /
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
a = (
)/(0 85^ =/
n =
= 0.851 (^)/(600/(600+))
= (10.59/(^))
=((^24))/2
Mu = = ^2(10.59/(^))
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Design Moment:
(+)Mu= 73 KN-m
Cross Section of Beam:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 4.9672E+10 1 = 0.13340
b = -7173558000 2 = 0.01102
c = 73000000 new = 0.01102 choose the smaller
rho new > rho min, use rho new
As= 936.41 mm
2Final: = 20 mm
= 20 mm n= 3 bars
n= 2.98 bars As= 942.480 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.01109
min = 0.00508 0.00439 0.00508 (governs)
rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are needed:
Ultimate Shear Moment (Vu)= 107.40 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu > Vc, calculate Shear Strenght of steel!
Positive Moment
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
=0.17(^ ) bwd
=((^24))/2
Mu = = ^2(10.59/(^))
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Design For Shear: Shear Diagram:
At Left Support:
= 8 mm
= 0.75 ACI CODE 2010
Factored Shear Force:
Vu= 107.40 KN
Shear Strength in Concrete:Vc= 70.05 KN
Vc= 52.54 KN
0.5Vc= 26.27 KN
Spacing (s) Required:
Av= 100.5312 mm2
S= 128.87 mm
say: 125 mm
Shear Strenght in Steel:
Vs= 56.56 KN
Point of Minimum Stirrups starts:
By ratio and proportion:
Xu= 3.95 m
X1 = 2.02 m
Point where Stirrups are not needed:
By ratio and proportion:
Xu= 3.95 m
X2 = 2.98 m
Design For Shear Reinforcement:
Vs
56.56 271954.84 ok, proceed to smax
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Check for maximun spacing:
Vs
Vs
56.56 135.98
smax will not be halved!
Smax = 170 mm
Actual Spacing (s):
= -
-
Vs =
s =
, smax d/2
, smax d/4
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Bar cut-off and Development Length:
Moment Diagram:
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1102
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 942.48 mm2
positive moment
As= 1884.96 mm2
negative moment
ln= 7.20 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off
Positive 1 314.16 66.67 0.260 1.87
---- ---- ---- ---- ---- ---- ----
Negative 2 628.32 66.67 0.102 0.73
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 27 mm
s < 2db, use other cases equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s= 1
= 1
Developement Length:
ld= 0.82 m
s =(b-2(cover+stirrup)-
ld=18fyte/(25(fc)) db
As%= (As -As1)/As x 100
Cut-Off= factor x 100
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Beam Details:
Left Midspan Right Left Right
8mm 8mm
*Closed Stirrups were used for the ease of construction*
*Safe & Aesthetic design acquires if beams are uniformly tied (Structral Integrity)
Sections:
Estimate for Beam:Total Volume of Concrete: Class A Mixture: Factor
Vol.conc.= 0.720 m2
Cement (40kg) 9.00
sand 0.50
Use Class A Mixture (1:2:4) Gravel 1.00
No. of Bags of Cement:
No.bags= 6.48 bags
say: 7
Volume of Sand:
Vsand= 0.36 m3
say: 1Volume of Gravel:
Vgravel= 0.72 m3
say: 1
Reinforcing Bars:
Top bars: 5 bars (20mm)
Bot. bars: 4 bars (20mm) Direct Counting Method
Stirrups: 9 bars (8mm)
Summary: 2 B-2
Item Quantity Unit Cost Total
Concrete C 14 220 3080
1 @ 0.065 m
16 @ 0.125 m
Rest @ 0.170 m
1 @ 0.075 m
11 @ 0.150 m
Rest @ 0.170 m
Vol =L x W x H
No. bags =Volume x Factor
Vol=Vsand x Factor
Vol=Vgravel x Factor
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa
1 = 0.85
= 0.90 l
u = 0.003
Cover= 60.00 mm
Assume Section: l= 1.90 m
b = 250 mm
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80
Weight of Slab (Ws) 24.00 ---- 5.7
Wall (Normal weight concrete, Ww1) 21.20 ---- 6.36
Wall (Foam/Plastered Finished, Ww2) ---- 0.48 0.024
Live Loads: Kpa KN/m
Office 2.4 4.56
Ultimate Load: KN/m
Wu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DL + 1.6LL = 14.14
Wu(wall)= 1.2DL= 7.66
Wu= 23.96
Beam B-3:
Shear Diagram:
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam B-3 - 2nd Floor Framing Plan Rating:
Wb=
Wu
Ws=
L = 3.80 m
Ww2= 1
Ww1=()
Wu =Wb+Ws+Ww
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Design For Flexure:
Design Moment:
Mu= 25 KN-m
Designer's Preferred Steel Ratio:
b
=0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 27.78 Kn-m
bd2
= 4825662.01 mm3
b = 250 mm Fexural Resistance factor
d = 138.93 mm
h' = 138.94 mm R= 5.75626
h (assume) > h', ok!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 4.9672E+10 1 = 0.14084
b = -7173558000 2 = 0.00357
c = 25000000 new = 0.00357 choose the smaller
rho new < rho min, use rho min
As= 431.47 mm
2Final: = 16 mm
= 16 mm n= 3 bars
n= 2.15 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:max = 0.03163
actual = 0.00710
min = 0.00508 0.00439 0.00508 (governs)
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 33.31345 mm Strain of steel Yeilds: = 0 001379
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=/
desired =0.6
Mn = /
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
a = (
)/(0 85^ =/
n =
= 0.851 (^)/(600/(600+))
= (10.59/(^))
=((^24))/2
Mu = = ^2(10.59/(^))
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Design Moment:
(+)Mu= 22 KN-m
Cross Secti 250 mm
b = 340 mm
d = 400 mmh =
Area of Steel: Roots:
4.9672E+10 1 = 0.14128
a = -7173558000 2 = 0.00313
b = 22000000 new = 0.00313 choose the smaller
c = rho new < rho min, use rho min
431.47 mm2
Final: = 16 mm
As= 16 mm n= 3 bars
()= 2.15 bars As= 603.187 mm2
(n)=
Checking for ACI requirements:
Steel ratio: 0.03163
max = 0.00710
actual =
0.00508 0.00439 0.00508 (governs)
min = rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are neeUltimate Shear Moment (Vu)= 47.50 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu < Vc, Provide minimum shear reinforcement!
Design for Shear Reinforcement:
= 8 mm
Av min.= 100.5312 mm2
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Positive Moment
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
=0.17(^ ) bwd
=((^24))/2
Mu = = ^2(10.59/(^))
Av(min) = (
2)/4
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Bar cut-off and Development Length:
Moment Diagram:
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1102
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 603.19 mm2
positive moment
As= 603.19 mm2
negative moment
ln= 3.40 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off
Positive 1 201.06 66.67 0.260 0.88
---- ---- ---- ---- ---- ---- ----
Negative 2 201.06 66.67 0.102 0.35
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 33 mm
s > 2db, use case 1 or 2 equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s= 1
= 1
Developement Length:
s =(b-2(cover+stirrup)-
As%= (As -As1)/As x 100
Cut-Off= factor x 100
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Beam Details:
Beam Sections Stirrups
Left Midspan Right Left Right
8mm 8mm
1 @ 0.085 m
SAME
Rest @ 0.170 mO.C.
*Closed Stirrups were used for the ease of construction*
*Safe & Aesthetic design acquires if beams are uniformly tied (Structral Integrity)
Sections:
Estimate for Beam:
Total Volume of Concrete: Class A Mixture: Factor
Vol.conc.= 0.340 m2
Cement (40kg) 9.00
sand 0.50
Use Class A Mixture (1:2:4) Gravel 1.00
No. of Bags of Cement:
No.bags= 3.06 bags
say: 4
Volume of Sand:
Vsand= 0.17 m3
say: 1
Volume of Gravel:
Vgravel= 0.34 m3
say: 1
Reinforcing Bars:
Top bars: 2 bars (16mm)
Bot. bars: 2 bars (16mm) Direct Counting Method
Stirrups: 4 bars (8mm)
Vol =L x W x H
No. bags =Volume x Factor
Vol=Vsand x Factor
Vol=Vgravel x Factor
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa
1 = 0.85 l
= 0.90
u = 0.003
Cover= 60.00 mm
Assume Section: l= 1.90 m
b = 250 mm
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80
Weight of Slab (Ws) 24.00 ---- 5.7
Wall (Normal weight concrete, Ww1) 21.20 ---- 6.36
Wall (Foam/Plastered Finished, Ww2) ---- 0.48 0.024
Live Loads: Kpa KN/m
Office 2.4 4.56
Ultimate Load: KN/m
Wu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DL + 1.6LL = 14.14
Wu(wall)= 1.2DL= 7.66
Wu= 23.96
Beam B-4:
Line of symmetry
Note: Choose biggest value of Shear & Moment.
Shear Diagram:
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam B-4 - 2nd Floor Framing Plan Rating:
Wb=
Wu
Ws=
L = 3.80 m
Ww2= 1
Ww1=()
L = 3.80 m
Wu =Wb+Ws+Ww
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Design For Flexure:
Design Moment:
Mu= 32 KN-m
Designer's Preferred Steel Ratio:
b
=0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 35.56 Kn-m
bd2
= 6176847.37 mm3
b = 250 mm Fexural Resistance factor
d = 157.19 mm
h' = 157.19 mm R= 5.75626
h (assume) > h', ok!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 4.9672E+10 1 = 0.13981
b = -7173558000 2 = 0.00461
c = 32000000 new = 0.00461 choose the smaller
rho new < rho min, use rho min
As= 431.47 mm
2Final: = 16 mm
= 16 mm n= 3 bars
n= 2.15 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:max = 0.03163
actual = 0.00710
min = 0.00508 0.00439 0.00508 (governs)
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 33.31345 mm Strain of steel Yeilds: = 0 001379
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=/
desired =0.6
Mn = /
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
a = (
)/(0 85^ =/
n =
= 0.851 (^)/(600/(600+))
= (10.59/(^))
=((^24))/2
Mu = = ^2(10.59/(^))
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Design Moment:
(+)Mu= 17 KN-m
Cross Section of Beam:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 4.9672E+10 1 = 0.14201
b = -7173558000 2 = 0.00241
c = 17000000 new = 0.00241 choose the smaller
rho new < rho min, use rho min
As= 431.47 mm
2Final: = 16 mm
()= 16 mm n= 3 bars
(n)= 2.15 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00710
min = 0.00508 0.00439 0.00508 (governs)
rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are needed:
Ultimate Shear Moment (Vu)= 48.20 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu < Vc, Provide minimum shear reinforcement!
Design for Shear Reinforcement:
= 8 mm
Av min.= 100.5312 mm2
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Positive Moment
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
=0.17(^ ) bwd
=((^24))/2
Mu = = ^2(10.59/(^))
Av(min) = (
2)/4
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Bar cut-off and Development Length:
Moment Diagram:
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1102
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 603.19 mm2
positive moment
As= 603.19 mm2
negative moment
ln= 3.40 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off
Positive 1 201.06 66.67 0.260 0.88
---- ---- ---- ---- ---- ---- ----
Negative 2 201.06 66.67 0.102 0.35
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 33 mm
s > 2db, use case 1 or 2 equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s= 1
= 1
Developement Length:
s =(b-2(cover+stirrup)-
As%= (As -As1)/As x 100
Cut-Off= factor x 100
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Beam Details:
Beam Sections Stirrups
Left Midspan Right Left Right
8mm 8mm
1 @ 0.085 m
SAME
Rest @ 0.170 mO.C.
*Closed Stirrups were used for the ease of construction*
*Safe & Aesthetic design acquires if beams are uniformly tied (Structral Integrity)
Section:
Estimate for Beam:
Total Volume of Concrete: Class A Mixture: Factor
Vol.conc.= 0.340 m2
Cement (40kg) 9.00
sand 0.50
Use Class A Mixture (1:2:4) Gravel 1.00
No. of Bags of Cement:
No.bags= 3.06 bags
say: 4
Volume of Sand:
Vsand= 0.17 m3
say: 1
Volume of Gravel:
Vgravel= 0.34 m3
say: 1
Reinforcing Bars:
Top bars: 2 bars (16mm)
Bot. bars: 2 bars (16mm) Direct Counting Method
Stirrups: 4 bars (8mm)
Vol =L x W x H
No. bags =Volume x Factor
Vol=Vsand x Factor
Vol=Vgravel x Factor
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa
1 = 0.85
= 0.90
u = 0.003
Cover= 60.00 mm
Assume Section: l1= 0.85 m
b = 250 mm l2= 1.90 m
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80
Weight of Slab (Ws1+ Ws2) 24.00 ---- 8.25
Live Loads: Kpa KN/m
Office 2.4 6.6
Ultimate Load: KN/m
Wu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DL + 1.6LL = 20.46
Wu1= 22.62
Wu2= 12.04
Beam B-5:
Note: Choose biggest value of Shear & Moment.
Shear Diagram:
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam B-5 - 2nd Floor Framing Plan Rating:
Wb=
Wu1Wu2
Ws= (1 + l2)
l1
l2
Wu1= Ws+ Ww
Wu2=Ws2+Ww
L = 7.60 m
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Design For Flexure:
Design Moment:
Mu= 110 KN-m
Designer's Preferred Steel Ratio:
b
=0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 122.22 Kn-m
bd2
= 21232912.8 mm3
b = 250 mm Fexural Resistance factor
d = 291.43 mm
h' = 291.43 mm R= 5.75626
h (assume) > h', ok!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 4.9672E+10 1 = 0.12698
b = -7173558000 2 = 0.01744
c = 110000000 new = 0.01744 choose the smaller
rho new > rho min, use rho new
As= 1482.42 mm
2Final: = 20 mm
= 20 mm n= 5 bars
n= 4.72 bars As= 1570.800 mm2
Checking for ACI requirements:
Steel ratio:max = 0.03163
actual = 0.01848
min = 0.00508 0.00439 0.00508 (governs)
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 86.75377 mm Strain of steel Yeilds: = 0 001379
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=/
desired =0.6
Mn = /
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
a = (
)/(0 85^ =/
n =
= 0.851 (^)/(600/(600+))
= (10.59/(^))
=((^24))/2
Mu = = ^2(10.59/(^))
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Design Moment:
(+)Mu= 39 KN-m
Cross Section of Beam:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 4.9672E+10 1 = 0.13876
b = -7173558000 2 = 0.00566
c = 39000000 new = 0.00566 choose the smaller
rho new > rho min, use rho new
As= 480.96 mm
2Final: = 20 mm
= 20 mm n= 2 bars
n= 1.53 bars As= 628.320 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00739
min = 0.00508 0.00439 0.00508 (governs)
rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are needed:
At Right Support:
Ultimate Shear Moment (Vu)= 46.50 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu < Vc, Provide minimum shear reinforcement!
At Left Support:
Ultimate Shear Moment (Vu)= 69.30 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu > Vc, calculate Shear Strenght of steel!
Positive Moment
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
=0.17(^ ) bwd
=((^24))/2
Mu = = ^2(10.59/(^))
=0.17(^ ) bwd
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Design For Shear: Shear Diagram:
At Left Support:
= 8 mm
= 0.75 ACI CODE 2010
Factored Shear Force:
Vu= 69.30 KN
Shear Strength in Concrete:Vc= 70.05 KN
Vc= 52.54 KN
0.5Vc= 26.27 KN
Spacing (s) Required:
Av= 100.5312 mm2
S= 421.77 mm
say: 420 mm
Shear Strenght in Steel:
Vs= 16.83 KN
Point of Minimum Stirrups starts:
By ratio and proportion:
Xu= 3.8 m
X1 = 0.92 m
Point where Stirrups are not needed:
By ratio and proportion:
Xu= 3.8 m
X2 = 2.36 m
Design For Shear Reinforcement:
Vs
16.83 271954.84 ok, proceed to smax
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Check for maximun spacing:
Vs
Vs
16.83 135.98
smax will not be halved!
Smax = 170 mm
Actual Spacing (s):
= -
-
Vs =
s =
, smax d/2
, smax d/4
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Bar cut-off and Development Length:
Moment Diagram:
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1102
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 628.32 mm2
positive moment
As= 1570.80 mm2 negative moment
ln= 7.20 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off
Positive 1 209.44 66.67 0.260 1.87
---- ---- ---- ---- ---- ---- ----
Negative 2 523.60 66.67 0.102 0.73
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 27 mm
s < 2db, use other cases equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s= 1
= 1
Developement Length:
s =(b-2(cover+stirrup)-
ld 18f 25(f )) d
As%= (As -As1)/As x 100
Cut-Off= factor x 100
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Beam Details:
Left Midspan Right Left Right
8mm 8mm
1 @ 0.085 m
SAMERest @ 0.170 m
O.C.
*Closed Stirrups were used for the ease of construction*
*Safe & Aesthetic design acquires if beams are uniformly tied (Structral Integrity)
Sections:
Estimate for Beam:Total Volume of Concrete: Class A Mixture: Factor
Vol.conc.= 0.720 m2
Cement (40kg) 9.00
sand 0.50
Use Class A Mixture (1:2:4) Gravel 1.00
No. of Bags of Cement:
No.bags= 6.48 bags
say: 7
Volume of Sand:
Vsand= 0.36 m3
say: 1Volume of Gravel:
Vgravel= 0.72 m3
say: 1
Reinforcing Bars:
Top bars: 6 bars (20mm)
Bot. bars: 3 bars (20mm) Direct Counting Method
Stirrups: 8 bars (8mm)
Summary: 1 B-5
Item Quantity Unit Cost Total
Concrete C 7 220 1540
Vol =L x W x H
No. bags =Volume x Factor
Vol=Vsand x Factor
Vol=Vgravel x Factor
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa
1 = 0.85
= 0.90
u = 0.003
Cover= 60.00 mm
Assume Section: l1= 1.90 m
b = 250 mm l2= 1.90 m
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80
Weight of Slab (Ws1 + Ws2 ) 24.00 ---- 11.40
Wall (Normal weight concrete, Ww1) 21.20 ---- 6.36
Wall (Foam/Plastered Finished, Ww2) ---- 0.48 0.024
Live Loads: Kpa KN/m
office 2.4 4.56
Ultimate Load: KN/m
Wu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DL + 1.6LL = 20.98
Wu(wall)= 1.2DL= 7.66
Wu= 23.96
Beam B-6:
Shear Diagram:
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam B-6 - 2nd Floor Framing Plan Rating:
Wb=
Wu
l2
L = 3.80 m
l1
Ww2= 1
Ww1=()
Ws= (1+2)
Wu =Wb+ Ws+ Ww
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Design For Flexure:
Design Moment:
Mu= 24 KN-m
Designer's Preferred Steel Ratio:
b
=0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 26.67 Kn-m
bd2
= 4632635.53 mm3
b = 250 mm Fexural Resistance factor
d = 136.13 mm
h' = 136.13 mm R= 5.75626
h (assume) > h', ok!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 4.9672E+10 1 = 0.14099
b = -7173558000 2 = 0.00343
c = 24000000 new = 0.00343 choose the smaller
rho new < rho min, use rho min
As= 431.47 mm
2Final: = 16 mm
= 16 mm n= 3 bars
n= 2.15 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:max = 0.03163
actual = 0.00710
min = 0.00508 0.00439 0.00508 (governs)
rho actual > rho min, ok!
rho max> rho actual, ok!
Depth of Rectangular Stress Block (a): Check if Tension Steel Yields:
a = 33.31345 mm Strain of steel Yeilds: = 0 001379
Choose the biggest Mu for
beam dept(d)
Choose the biggest Mu for
beam depth (d)
bd2=/
desired =0.6
Mn = /
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
a = (
)/(0 85^ =/
n =
= 0.851 (^)/(600/(600+))
= (10.59/(^))
=((^24))/2
Mu = = ^2(10.59/(^))
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Design Moment:
(+)Mu= 19 KN-m
Cross Section of Beam:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 4.9672E+10 1 = 0.14172
b = -7173558000 2 = 0.00270
c = 19000000 new = 0.00270 choose the smaller
rho new < rho min, use rho min
As= 431.47 mm
2Final: = 16 mm
()= 16 mm n= 3 bars
(n)= 2.15 bars As= 603.187 mm2
Checking for ACI requirements:
Steel ratio:
max = 0.03163
actual = 0.00710
min = 0.00508 0.00439 0.00508 (governs)
rho actual > rho min, ok!
rho max > rho actual, ok!
Check if Stirrups are needed:
Ultimate Shear Moment (Vu)= 48.20 KN/m
Shear Strenght by Concrete (Vc) = 70.05 KN/m
Vc= 52.54 KN/m
Vu > 0.5Vc, Stirrups are needed!
Vu < Vc, Provide minimum shear reinforcement!
Design for Shear Reinforcement:
= 8 mm
Av min.= 100.5312 mm2
Maximum Spacing (smax):
a.) smax1= 170 mm
b.)smax2= 600 mm
c.)smax3= 327 mm
Use smax= 170 mm
Positive Moment
a^2+b+=0
max =0.75
= /
= (0.25(^ ))/
=0.17(^ ) bwd
=((^24))/2
Mu = = ^2(10.59/(^))
Av(min) = (
2)/4
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Bar cut-off and Development Length:
Moment Diagram:
Theoritical Bar Cut-off Graph: Wight & macGregor "Reinforced Concrete",6th edition, pp.1102
Bar Cut-off (1/3 and 2/3 of bars that satisfy the ACI code)
As= 603.19 mm2
positive moment
As= 603.19 mm2
negative moment
ln= 3.40 m
Moment x (1/3)As (2/3)As %As (cut-off) Factor Cut-off
Positive 1 201.06 66.67 0.260 0.88
---- ---- ---- ---- ---- ---- ----
Negative 2 201.06 66.67 0.102 0.35
* Refer to Theoritical Bar Cut-Off Graph
Bar Spacing (s): clear spacing between parallel bars
s= 33 mm
s > 2db, use case 1 or 2 equation for ld
Factors (ACI CODE 2010):
t= 1
e= 1
s= 1
= 1
Developement Length:
s =(b-2(cover+stirrup)-
As%= (As -As1)/As x 100
Cut-Off= factor x 100
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Beam Details:
Beam Sections Stirrups
Left Midspan Right Left Right
8mm 8mm
1 @ 0.085 m
SAME
Rest @ 0.170 mO.C.
*Closed Stirrups were used for the ease of construction*
*Safe & Aesthetic design acquires if beams are uniformly tied (Structral Integrity)
Sections:
Estimate for Beam:
Total Volume of Concrete: Class A Mixture: Factor
Vol.conc.= 0.340 m2
Cement (40kg) 9.00
sand 0.50
Use Class A Mixture (1:2:4) Gravel 1.00
No. of Bags of Cement:
No.bags= 3.06 bags
say: 4
Volume of Sand:
Vsand= 0.17 m3
say: 1
Volume of Gravel:
Vgravel= 0.34 m3
say: 1
Reinforcing Bars:
Top bars: 2 bars (16mm)
Bot. bars: 2 bars (16mm) Direct Counting Method
Stirrups: 4 bars (8mm)
Vol =L x W x H
No. bags =Volume x Factor
Vol=Vsand x Factor
Vol=Vgravel x Factor
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Material Property: Model: Tributary Area (refer to analysis)
fy= 275.80 Mpa
f'c= 23.50 Mpa
Es = 200000.00 Mpa
1 = 0.85
= 0.90
u = 0.003
Cover= 60.00 mm
Assume Section: l= 1.25 m
b = 250 mm
d= 240
h = 300 mm
Service Loads:
Dead loads: KN/m3
KN/m2
KN/m
Weight of beam (Wb) 24.00 ---- 1.80
Weight of Slab (Ws) 24.00 ---- 3.75
Railings (3" ) 77.30 ---- 5.22
Live Loads: Kpa KN/m
Building corridor/ hall way 3.8 4.75
Ultimate Load: KN/m
Wu(beam)= 1.2DL= 2.16
Wu(slab)= 1.2DLT + 1.6LLT= 12.10
Wu(Railings)= 1.2DL= 6.26
Wu= 20.52
Beam B-7:
Shear Diagram:
Two-Story office Building Date Prepared:
A. Design of Beams Checked By:
Beam B-7- 2nd Floor Framing Plan Rating:
Wb=
Wu
Ws=
Wu =++ Wr
L = 7.60 m
l
Wr=( )
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Design For Flexure:
Design Moment:
Mu= 148 KN-m
Designer's Preferred Steel Ratio:
b
=0.0421753
(desired) = 0.0253052
Cross Section of Beam: Nominal Moment:
Mn = 164.44 Kn-m
bd2
= 28567919.1 mm3
b = 250 mm Fexural Resistance factor
d = 338.04 mm
h' = 338.04 mm R= 5.75626
h' > h(assume),Revise Section!
Final:
b = 250 mm
d = 340 mm
h = 400 mm
Area of Steel:
Roots:
a = 4.9672E+10 1 = 0.11948
b = -7173558000 2 = 0.02494
c = 148000000 new = 0.02494 choose the smaller
rho new > rho min, use rho new
As= 2119.68 mm
2Final: = 20 mm
= 20 mm n= 8 bars
n= 6.75 bars As= 2513.280 mm2
Checking for ACI requirements:
Steel ra
Recommended