Capital University of Science and
Technology, Islamabad
Department of Civil Engineering
Plain and Reinforced Concrete-I
Lab Manual
Design Aid
Structural
Member
ACI Code 318-14 Guideline
Thumb Rule
Suggestions
Estimate
the
minimum
thickness of
slab for all
support
conditions.
Then Select
Max value
One
Way
Slab
h=
perimeter/180
Use Thumb
Rule
h= L/12
h:b = 2-2.5
Use thumb
rule if spans
are
symmetrical
Two
Way
Slab
h should
preferably be
multiple of 3”
Beams
Columns
Footing
A=P/q where P is total load on footing and q is bearing
capacity of soil.
122
W=h=L/(2x12)
Q=1tsf
=2.204ksf
Capital University of Science and
Technology, Islamabad
Department of Civil Engineering
Plain and Reinforced Concrete-I
Lab Manual
Step 1: Decide grids layout.
Column were placed in such a way that they make 25’ x 20’ panel starting from corner.
Hence max of 6 columns are provided in the Hall.
Step 2: Get rough estimate of slab thickness, beam and column size using Design Aid
(Annex N1- B)
i) Slab Thickness
Since it is a two-way slab (as 25/20<2), thumb rule provided in Annex N1-b
was used. h=
perimeter/180
= (25+25+20+20)/180
=0.5’ = 6”
ii) Beam Size
a) For L=25 ft.
h = l/12 = (25*12)/12
= 25” b = h/2.5 =
10”
Beam 1 size is 9” x 24”
b) For L=20 ft.
h = l/12 = (20*12)/12
= 20” b = h/2.5 = 8”
Beam 1 size is 9” x 21”
iii) Column Size
w=h=L/(2x12)
L = length of longest beam transferring load on column
w = h = (25 x 12) / (2 x 12) = 12.5”
Column size selected is 12” x 12”
Step 3: Estimate foundation sizes considering load and bearing capacity of soil.
For columns at corner without stairs.
Description
Ground Floor
Slab
Wall 1
Wall 2
12 x 12.5 x .75
x .12
=
13.5k
12 x 10 x .75 x
.12
=
11k
First Floor
10 x 12.5 x 0.5) x
0.15 =
9.4 k
Roof
** 9.4 k
Total
18.4k
*13.5k
27k
*11k
22k
123
Remarks
Capital University of Science and
Technology, Islamabad
Department of Civil Engineering
Beam 1
supporting
wall
Beam 2
supporting
wall
Beam 1
supporting
truss
0.75 x 2 x 12.5
x 0.15
=
2.81 k
(0.75 x
1.67x10) x
0.15 = 1.88k
Plain and Reinforced Concrete-I
Lab Manual
*2.81 k
5.62k
*1.88k
3.76k
0.75 x 2 x
12.5 x 0.15
=
2.81 k
(0.75 x
1.67x10) x
0.15 =
1.88k
Beam 2
supporting
truss
Supper
imposed
dead load
Column
0.05 x 10 x 12.5=
6.25k
(1 x 1x12) x
0.15
=
1.8 k
Live load
2.81k
1.88k
6.25k
*1.8 k
=1.8/2 =
0.9k
4.5k
0.1 x 10 x 12.5 =
12.5 k
0.015 x 10
x 12.5 =
1.9 k
13.4k
TOTAL LOAD P
* Assuming symmetry i.e both floors will have same size and loads.
** Assuming truss weight equals to slab weight
A = (P x Factor for moment) /q
A = 117.22 x 1.3 / 2.204 =
69.14 C1 = C2 = sqrt (69.14) =
8.31’
≈ 8.-‘-6”
Footing thickness = Total thickness of slabs = 12”
Footing size is 8.5’ x 8.5’ x 1’
B) Similarly Calculate Footing size for column on boundary& centre.
Repeat Step 2 for first floor framing plan.
124
117.22k