WESTERN MINDANAO STATE UNIVERSITY

College of Engineering and Technology

Department of Civil Engineering

CE 158
REINFORCED CONCRETED DESIGN II

Design of Slab using DDM

Eng’r Muhammadnur Muhammad

Professor

Juano, John Lloyd S.

BS-CE 5A
 Slab-1
1. Calculate the Loads
¿=5.0 kPa

DL=γ c h f =23.54 x 0. 165=3.884 kPa

w u=1.2 ( 3.884 ) +1.6 ( 5.0 )=12.661 kPa

2. Calculate the Total Moment

w u l 2 ln 2
M o=
8
M o=12.661(7.350)¿ ¿
M o=689.679 kN . m

Where:

ln=clear span=8.0−0.3=7.7 m

7.2+7.5
l 2=average of two adjacent spans perpendicular ¿ l 2= =7.35 m
2

3. Distribute the Total Moment M o into positive and negative moment

At Discontinuous Edge, 0.16 M o= 110.328 kN.m
At Continuous Edge, 0.70 M o= 482.775 kN.m
Positive Moment at Midspan, 0.57 M o= 110.328 kN.m

A. Consider the continuous edge, the maximum negative moment

M u= 482.775 kN.m, at continuous edge
l 2 7.35
= =0.919
l1 8
0.25 l 2=0.25 x 7.35=1.838 m
0.25 l 1=0.25 x 8=2 m

αm
fy
l n (0.8+ )
1400
0.2<α ≤2
36+5 β (α m −0.2)
≥
125 mm
fy
l n (0.8+ )
1400
α m >2
36+9 β
≥
90 mm
 α 2=5.757

l2
α 2 = (5,757 ) 0.919=5.291>1.0
l1
l βt
α 2
l 1 Portion of interior negative moment Mu in Column Strip
Table 408.10.5.2
l2
l1
0.5 1 2
0 1
0 ≥ 2.5
0.75
0 1
≥ 1.0
≥ 2.5
0.9 0.75 0.45

By Interpolation:

X 1−0.75 0.45−0.75
=
1.01−1 2−1

X 1 =0.747

Description Percentage Moment Factored Moment Unit

Column Strip Moment 74.70% 482.775 360.632925 kN.m
Beam Moment 85% 360.6329 306.5379863 kN.m
Column Strip Slab Moment 15% 360.6329 54.09493875 kN.m
Middle Strip Slab Moment - 482.775 122.142075 kN.m

a) Compute Rebars for Column Strip.

b=1850 mm
d=165−25=140 mm
54.095
M u= =27.048 kN . m
2

1.7 ( 27.048 x 1 06 )
ω=0.85− 0.85 −
ω=0.035846
√ 2
(0.9)(21)(1850)¿ ¿
¿

f c'
ρ=ω
fy
21
ρ=0.035846 x
275
ρ=0.0027373

A s=ρ bd=0.0027373 ( 1850 ) ( 140 )

A s=708.961 mm2

A smin=0.002 bd=0.002 ( 1650 ) ( 140 )

 A smin=462m m 2(NSCP 2015, Section 408.6.1.1

A s ≥ A smin Ok

Ab
S=1850
As
π
( x 1 02 )
4
S=1850 =204.946<2 h=320 Ok
708.961

Use 10 mm ∅ bars @200 mm O .C .

b) Compute Rebars for Middle Strip.

b=3350 mm
d=165−25=140 mm
M u=122.142 kN . m

1.7 ( 122.142 x 1 06 )

ω=0.10490
√
ω=0.85− 0.85 −
2
(0.9)(21)(3350)¿ ¿
¿

f c'
ρ=ω
fy
21
ρ=0.10490 x
275
ρ=0.0080105

A s=ρ bd=0.0080105 ( 3350 ) ( 140 )

2
A s=3756.925 mm

A smin=0.002 bd=0.002 ( 3750 ) ( 140 )

A smin=1050 m m2(NSCP 2015, Section 408.6.1.1

A s ≥ A smin Ok

Ab
S=3350
As
π
( x 1 22)
4
S=3350 =100.8472<2 h=320 Ok
3756.925

Use 12 mm ∅ bars @100 mm O. C .

B. Consider positive moment at midspan

M u= 393.117 kN.m, at continuous edge
l 2 7.35
= =0.919
l1 8

α 2=5.757
 l2
α 2 = (5,757 ) 0.919=5.291>1.0
l1
l βt
α 2
l 1 Portion of positive moment at midspan Mu in Column Strip
Table 408.10.5.2
l2
l1
0.5 1 2
0 1
0 ≥ 2.5
0.75
0 1
≥ 1.0
≥ 2.5
0.9 0.75 0.45

By Interpolation:

X 1−0.75 0.45−0.75
=
1.01−1 2−1

X 1 =0.747

Description Percentage Moment Factored Moment Unit

Column Strip Moment 74.70% 393.117 293.658399 kN.m
Beam Moment 85% 293.6584 249.6096392 kN.m
Column Strip Slab Moment 15% 293.6584 44.04875985 kN.m
Middle Strip Slab Moment - 393.117 99.458601 kN.m

a) Compute Rebars for Column Strip.

b=1850 mm
d=165−25=140 mm
44.049
M u= =22.025 kN . m
2

1.7 ( 22.025 x 1 06 )
ω=0.85− 0.85 −
ω=0.03 2770
√ 2
(0.9)(21)(1850)¿ ¿
¿

f c'
ρ=ω
fy
21
ρ=0.032770
275
ρ=0.00 25024
A s=ρ bd=0.0025024 ( 1850 )( 140 )
A s=648.122 mm 2

A smin=0.002 bd=0.002 ( 1650 ) ( 140 )

A smin=462m m 2(NSCP 2015, Section 408.6.1.1

A s ≥ A smin Ok
 Ab
S=1850
As
π
( x 1 02 )
4
S=1850 =224.184<2 h=320Ok
648.122

Use 10 mm ∅ bars @200 mm O .C .

b) Compute Rebars for Middle Strip.

b=3350 mm
d=165−25=140 mm
M u=99.459 kN . m

1.7 ( 99.459 x 10 6 )
ω=0.85− 0.85 −
ω=0.084329
√ 2
(0.9)(21)(3350)¿ ¿
¿

f c'
ρ=ω
fy
21
ρ=0.084329 x
275
ρ=0.0064397

A s=ρ bd=0.0064397 ( 3350 ) ( 140 )

A s=3020.219 mm 2

A smin=0.002 bd=0.002 ( 3750 ) ( 140 )

A smin=1050 m m2(NSCP 2015, Section 408.6.1.1

A s ≥ A smin Ok

Ab
S=3350
As
π
( x 1 22)
4
S=3350 =125.447<2 h=320Ok
3020.219

Use 12 mm ∅ bars @100 mm O. C .

C. Consider discontinuous edge, the exterior negative moment.

M u= 110.348 kN.m,
1
I s= x 4150 x 16 53=1553.527 x 1 06 m m 4 , moment inertiaof slab
12

 Torsional Constant.

x x3 y
(
C=Σ 1−0.63 )
y 3
 3 3
300 (300 ) (700) 165 ( 165 ) (535)
(
C=Σ 1−0.63
700 3 ) + Σ 1−0.63
535 (3 )
C=5244.443 x 1 06 mm4

C 5244.443 x 1 06
βt = = =1.688
2 I s 2 x 1553.527 x 1 06
l βt
α 2
l 1 Portion of exterior negative moment Mu in Column Strip
Table 408.10.5.2
l2
l1
0.5 1 2
0 1
0
≥ 2.5 0.75
0 1
≥ 1.0
≥ 2.5 0.9 0.75 0.45
βt

l2
l1
1 1.01 2
0 1 1 1
1.688 X2 X4 X3
≥ 2.5 .75 X1 .45

By Interpolation:

X 1−0.75 0.45−0.75 X 3−1 0.45−1

= =
1.01−1 2−1 1.688−0 2.5−0

X 1 =0.747 X 3 =0.629

X 2−1 0.75−1 X 4−1 0.747−1

= =
1.688−0 2.5−0 1.688−0 2.5−0

X 2 =0.831 X 4=0.829
 Description Percentage Moment Factored Moment Unit
Column Strip Moment 82.92% 110.348 91.49725116 kN.m
Beam Moment 85% 91.49725 77.77266349 kN.m
Column Strip Slab Moment 15% 91.49725 13.72458767 kN.m
Middle Strip Slab Moment - 110.348 18.85074884 kN.m

a) Compute Rebars for Column Strip.

b=1850 mm
d=165−25=140 mm
13.725
M u= =6.863 kN . m
2

1.7 ( 6.863 x 1 06 )

ω=0.010074
√
ω=0.85− 0.85 −
2
(0.9)(21)(1850)¿ ¿
¿

f c'
ρ=ω
fy
21
ρ=0.010074
275
ρ=0.0076929
A s=ρ bd=0.0076929 ( 1850 ) ( 140 )
A s=1992.461 mm 2

A smin=0.002 bd=0.002 ( 1650 ) ( 140 )

A smin=462m m 2(NSCP 2015, Section 408.6.1.1

A s ≥ A smin Ok

Ab
S=1850
As
π
( x 1 2z )
4
S=1850 =105.011<2 h=320 Ok
1992.461

Use 12 mm ∅ bars @100 mm O. C .

b) Compute Rebars for Middle Strip.

b=3350 mm
d=165−25=140 mm
M u=18.851 kN . m

1.7 ( 18.851 x 1 06 )

ω=0.015329
√
ω=0.85− 0.85 −
2
(0.9)(21)(3350)¿ ¿
¿

f c'
ρ=ω
fy
 21
ρ=0.015329 x
275
ρ=0.0011706

A s=ρ bd=0.0011706 (3350 )( 140 )

A s=549.011mm2

A smin=0.002 bd=0.002 ( 3350 ) ( 140 )

A smin=1050 m m2(NSCP 2015, Section 408.6.1.1

A s ≤ A smin N . G. use A s=1050 m m2

Ab
S=3350
As
π
( x 1 02 )
4
S=3350 =250.579<2 h=320 Ok
1050

Use 10 mm ∅ bars @250 mm O .C .

 Slab-2
1. Calculate the Loads
¿=5.0 kPa

DL=γ c h f =23.54 x 0. 165=3.884 kPa

w u=1.2 ( 3.884 ) +1.6 ( 5.0 )=12.661 kPa

2. Calculate the Total Moment

w u l 2 ln 2
M o=
8
M o=12.661(7.350)¿ ¿
M o=553.814 kN . m

Where:

ln=clear span=7.2−0.3=6.9 m

7.2+7.5
l 2=average of two adjacent spans perpendicular ¿ l 2= =7.35 m
2

3. Distribute the Total Moment M o into positive and negative moment

At Continuous Edge, 0.65 M o= 359.979 kN.m
At Continuous Edge, 0.65 M o= 359.979 kN.m
Positive Moment at Midspan, 0.35 M o= 193.835 kN.m

A. Consider the continuous edge, the maximum negative moment

M u= 359.979 kN.m, at continuous edge
 l 2 7.35
= =1.021
l 1 7.2
0.25 l 2=0.25 x 7.35=1.838 m
0.25 l 1=0.25 x 7.2=1.8 m

MODE 3-2
i h2 y(d) freq. (Area)
x( )
12
1 40,833.333 350 210,000
2 2268.750 82.5 176550

α 2=5.757

l2
α 2 = (5,757 ) 0.919=5.291>1.0
l1
l βt
α 2
l1
Table 408.10.5.2 Portion of interionegative moment Mu in Column Strip
l2
l1
0.5 1 2
0 1
0 ≥ 2.5
0.75
0 1
≥ 1.0
≥ 2.5
0.9 0.75 0.45

By Interpolation:

X 1−0.75 0.45−0.75
=
1.01−1 2−1

X 1 =0.747

Description Percent Moment Factored Moment Unit

Column Strip Moment 74.70% 359.979 268.904313 kN.m
Beam Moment 85% 268.9043 228.5686661 kN.m
Column Strip Slab Moment 15% 268.9043 40.33564695 kN.m
Middle Strip Slab Moment - 359.979 91.074687 kN.m

a) Compute Rebars for Column Strip.

b=1650 mm
d=165−25=140 mm
 40.336
M u= =20.168 kN .m
2

1.7 ( 20.168 x 1 06 )
ω=0.85− 0.85 −
ω=0.033663
√ 2
(0.9)(21)(1650)¿ ¿
¿

f c'
ρ=ω
fy
21
ρ=0.033663 x
275
ρ=0.0025706

A s=ρ bd=0.0025706 ( 1650 ) ( 140 )

A s=593.809 mm 2

A smin=0.002 bd=0.002 ( 1650 ) ( 140 )

A smin=462m m 2(NSCP 2015, Section 408.6.1.1

A s ≥ A smin Ok

Ab
S=1650
As
π
( x 1 02 )
4
S=1650 =218.236<2 h=320 Ok
593.809

Use 10 mm ∅ bars @200 mm O .C .

b) Compute Rebars for Middle Strip.

b=3750 mm
d=165−25=140 mm
M u=91.075 kN . m

1.7 ( 91.075 x 10 6 )
ω=0.85− 0.85 −
ω=0.068306
√ 2
(0.9)(21)(3750)¿ ¿
¿

f c'
ρ=ω
fy
21
ρ=0.068306 x
275
ρ=0.0052161

A s=ρ bd=0.0052161 ( 3750 ) (140 )

A s=2738.453 mm2

A smin=0.002 bd=0.002 ( 3750 ) ( 140 )

A smin=1050 m m2(NSCP 2015, Section 408.6.1.1
 A s ≥ A smin Ok

Ab
S=3750
As
π
( x 1 02)
4
S=3750 =107.551<2 h=320 Ok
2738.453

Use 10 mm ∅ bars @100 mm O .C .

B. Consider positive moment at midspan

M u= 193.835 kN.m, at continuous edge
l 2 7.35
= =0.919
l1 8

α 2=5.757

l2
α 2 = (5,757 ) 0.919=5.291>1.0
l1
l βt
α 2
l 1 Portion of positive moment at midspan Mu in Column Strip
Table 408.10.5.2
l2
l1
0.5 1 2
0 1
0 ≥ 2.5
0.75
0 1
≥ 1.0
≥ 2.5
0.9 0.75 0.45

By Interpolation:

X 1−0.75 0.45−0.75
=
1.01−1 2−1

X 1 =0.747

Description Percent Moment Factored Moment Unit

Column Strip Moment 74.70% 193.835 144.794745 kN.m
Beam Moment 85% 144.7947 123.0755333 kN.m
Column Strip Slab Moment 15% 144.7947 21.71921175 kN.m
Middle Strip Slab Moment - 193.835 49.040255 kN.m

c) Compute Rebars for Column Strip.

b=1650 mm
d=165−25=140 mm
 21.719
M u= =10.860 kN . m
2

1.7 ( 10.860 x 1 06 )
ω=0.85− 0.85 −
ω=0.017957
√ 2
(0.9)(21)(1650)¿ ¿
¿

f c'
ρ=ω
fy
21
ρ=0.017957
275
ρ=0.0013713

A s=ρ bd=0.0013713 ( 1650 ) ( 140 )

A s=316.770 mm 2

A smin=0.002 bd=0.002 ( 1650 ) ( 140 )

A smin=462m m 2(NSCP 2015, Section 408.6.1.1

A s ≤ A smin N . G. use A s=462m m2

Ab
S=1650
As
π
( x 1 02 )
4
S=1650 =280.499<2 h=320 Ok
462

Use 10 mm ∅ bars @250 mm O .C .

d) Compute Rebars for Middle Strip.

b=3750 mm
d=165−25=140 mm
M u=49.040 kN . m

1.7 ( 49.040 x 10 6 )
ω=0.85− 0.85 −
ω=0.036067
√ 2
(0.9)(21)(3750)¿ ¿
¿

f c'
ρ=ω
fy
21
ρ=0.036067 x
275
ρ=0.0027542

A s=ρ bd=0.0027542 ( 3750 ) (140 )

A s=1445.955 mm 2
A smin=0.002 bd=0.002 ( 3750 ) ( 140 )
A smin=1050 m m2(NSCP 2015, Section 408.6.1.1

A s ≥ A smin Ok

Ab
S=3750
As
π
( x 1 02)
4
S=3750 =203.688<2 h=320 Ok
1445.955

Use 10 mm ∅ bars @200 mm O .C .