Analysis and Design of Slabs “One Way”

Analysis and Design of Slabs “One Way”

Slabs

• In reinforced concrete construction, slabs are used to flat, useful surfaces..
• A reinforced slab is a broad, flat plate, usually horizontal, with top and bottom surfaces parallel or nearly so.
• It may be supported by reinforced concrete beams (and is usually cast monolithically with such beams), by masonry or by reinforced concrete walls, by steel structural members, directly by columns, or continuously by ground.

One-Way Slab

“The slab which resists the entire/major part of applied load by bending only in one direction”

• If slab is supported on all four sides and R = (Shorter side / Longer side) < 0.5 it behaves as one-way slab.
• Slabs having supports on less than four sides can be designed as one-way.
• Two edge supported slab is always one-way.
• Cantilever slab is always one-way.
• Main steel is only provided parallel to span
• One-way slab is designed as singly reinforced rectangula

L = Effective Span
Lesser of the following:
L= Ln + h/2 + h/2
= Ln + h
h = depth of slab and
L = c/c distance between supports.

Examples of One-Way Slab

• Shades in the roofing system (cantilever)
• Slab of stairs
• Cantilever retaining walls
• Footings

Bar Spacing Cover For Slabs

1. 3 x h               (local practice is 2 x h)
2. 450 mm         (local practice is 300 mm)
3. (158300/fy) -2.5Cc
4. 12600/fy

Cc = Clear Cover

Distribution, Temperature & Shrinkage Steel for Slabs (ACI-318-7.12)

• Shrinkage and temperature reinforcement is required at right angle to main reinforcement to minimize cracking and to tie the structure together to ensure its acting as assumed in design
•  Top and bottom reinforcements are both effective in controlling the cracks

s(max) shall be lesser of following

1.    5 x h             (field practice is 2 x h)
2.   450 mm       (field practice is 2 x h)

Minimum Steel For Slabs

Same as the distribution steel

Design Procedure for One-Way Slab

1. Check whether the slab is one-way or two-way.
2. Calculate hmin and round it to higher 10mm multiple.
   i.Not less than 110 mm for rooms
   ii.Not less than 75 mm for sunshades.
3. Calculate dead load acting on the slab.
   Dead Load = Load per unit area x 1m width.
4. Calculate live load acting on the slab.
   Live load = Load per unit area x 1m width.
5. Calculate total factored load per unit strip (kN/m)
6. Calculate the moments either directly (simply supported) or by using coefficient for continuous slabs
7. Calculate effective depth.
   d  = h – (20 + (½)db)
   db = 10, 13, 15 generally used
8. Check that   d ≥ dmin
9. Calculate As required for 1m width
10. Calculate minimum/distribution/temperature & shrinkage steel
11. Select diameter and spacing for main steel
12. Check the spacing for max. and min. spacing      smin  ≈ 90mm
    if spacing is less than minimum increase the diameter of bar
13. For continuous slabs, curtail or bent up the +ve steel. For -ve steel see how much steel is already available. Provide remaining amount of steel.
14. Calculate the amount of distribution steel. Decide its dia. & spacing like main steel.
15. Check the slab for shear.
    ΦvVc ≥ Vu
16. Carry out detailing and show results on the drawings
17. Prepare bar bending schedule, if required

Approximate Steel for Estimate

Approximate amount of steel in slab
= 0.07 kg/mm/m2
If slab thickness = 100 mm
steel = 0.07 x 100 = 7kg /m2

Example

Design a cantilever projecting out from a room slab extending 1.0m and to be used as balcony (LL = 300 kg/m2). A brick wall of 250 mm thickness including plaster of 1.0m height is provided at the end of cantilever  fc’ = 17.25 MPa,  fy = 300 MPa, Slab thickness of room = 125 mm. Slab bottom steel in the direction of cantilever is # 13 @ 190 mm c/c

Solution

Slab Load

Self Weight of Slab = (125/1000)*2400 = 300kg/m2
75 mm brick ballast/ screed  = (75/1000)*1800 = 135kg/m2
60 mm floor finishes  = (60/1000)*2300 = 138kg/m2
Total dead load = 300+135+138 = 573kg/m2

Live Load = 300kg/m2

Factored Load = (1.2×573+1.6×300)x9.81/1000
= 11.46 KN/m2 = 11.46KN/m for a unit strip
P = 1.2x(0.25x1x1)x1930x9.81/1000
P = 5.65 KN

Remaining steel required at the top = 510-342 168mm2    (#10@400 c/c)
use #10 @ 380mm c/c

Distribution steel

=0.002x1000x125=250mm2  use # 10 @ 280mm 

Analysis and Design of Slabs “One Way”

Analysis and Design of Slabs “One Way”

Slabs

• In reinforced concrete construction, slabs are used to flat, useful surfaces..
• A reinforced slab is a broad, flat plate, usually horizontal, with top and bottom surfaces parallel or nearly so.
• It may be supported by reinforced concrete beams (and is usually cast monolithically with such beams), by masonry or by reinforced concrete walls, by steel structural members, directly by columns, or continuously by ground.

One-Way Slab

“The slab which resists the entire/major part of applied load by bending only in one direction”

• If slab is supported on all four sides and R = (Shorter side / Longer side) < 0.5 it behaves as one-way slab.
• Slabs having supports on less than four sides can be designed as one-way.
• Two edge supported slab is always one-way.
• Cantilever slab is always one-way.
• Main steel is only provided parallel to span
• One-way slab is designed as singly reinforced rectangula

L = Effective Span
Lesser of the following:
L= Ln + h/2 + h/2
= Ln + h
h = depth of slab and
L = c/c distance between supports.

Examples of One-Way Slab

• Shades in the roofing system (cantilever)
• Slab of stairs
• Cantilever retaining walls
• Footings

Bar Spacing Cover For Slabs

1. 3 x h               (local practice is 2 x h)
2. 450 mm         (local practice is 300 mm)
3. (158300/fy) -2.5Cc
4. 12600/fy

Cc = Clear Cover

Distribution, Temperature & Shrinkage Steel for Slabs (ACI-318-7.12)

• Shrinkage and temperature reinforcement is required at right angle to main reinforcement to minimize cracking and to tie the structure together to ensure its acting as assumed in design
•  Top and bottom reinforcements are both effective in controlling the cracks

s(max) shall be lesser of following

1.    5 x h             (field practice is 2 x h)
2.   450 mm       (field practice is 2 x h)

Minimum Steel For Slabs

Same as the distribution steel

Design Procedure for One-Way Slab

1. Check whether the slab is one-way or two-way.
2. Calculate hmin and round it to higher 10mm multiple.
   i.Not less than 110 mm for rooms
   ii.Not less than 75 mm for sunshades.
3. Calculate dead load acting on the slab.
   Dead Load = Load per unit area x 1m width.
4. Calculate live load acting on the slab.
   Live load = Load per unit area x 1m width.
5. Calculate total factored load per unit strip (kN/m)
6. Calculate the moments either directly (simply supported) or by using coefficient for continuous slabs
7. Calculate effective depth.
   d  = h – (20 + (½)db)
   db = 10, 13, 15 generally used
8. Check that   d ≥ dmin
9. Calculate As required for 1m width
10. Calculate minimum/distribution/temperature & shrinkage steel
11. Select diameter and spacing for main steel
12. Check the spacing for max. and min. spacing      smin  ≈ 90mm
    if spacing is less than minimum increase the diameter of bar
13. For continuous slabs, curtail or bent up the +ve steel. For -ve steel see how much steel is already available. Provide remaining amount of steel.
14. Calculate the amount of distribution steel. Decide its dia. & spacing like main steel.
15. Check the slab for shear.
    ΦvVc ≥ Vu
16. Carry out detailing and show results on the drawings
17. Prepare bar bending schedule, if required

Approximate Steel for Estimate

Approximate amount of steel in slab
= 0.07 kg/mm/m2
If slab thickness = 100 mm
steel = 0.07 x 100 = 7kg /m2

Example

Design a cantilever projecting out from a room slab extending 1.0m and to be used as balcony (LL = 300 kg/m2). A brick wall of 250 mm thickness including plaster of 1.0m height is provided at the end of cantilever  fc’ = 17.25 MPa,  fy = 300 MPa, Slab thickness of room = 125 mm. Slab bottom steel in the direction of cantilever is # 13 @ 190 mm c/c

Solution

Slab Load

Self Weight of Slab = (125/1000)*2400 = 300kg/m2
75 mm brick ballast/ screed  = (75/1000)*1800 = 135kg/m2
60 mm floor finishes  = (60/1000)*2300 = 138kg/m2
Total dead load = 300+135+138 = 573kg/m2

Live Load = 300kg/m2

Factored Load = (1.2×573+1.6×300)x9.81/1000
= 11.46 KN/m2 = 11.46KN/m for a unit strip
P = 1.2x(0.25x1x1)x1930x9.81/1000
P = 5.65 KN

Remaining steel required at the top = 510-342 168mm2    (#10@400 c/c)
use #10 @ 380mm c/c

Distribution steel

=0.002x1000x125=250mm2  use # 10 @ 280mm 

DRIVEN CAST IN-SITU CONCRETE PILES CONSTRUCTION PROCESS

Driven cast in-situ concrete piles are constructed by driving a closed-ended hollow steel or concrete casing into the ground and then filling it with concrete. The casing may be left in position to form part of the pile, or withdrawn for reuse as the concrete is placed. The details of driven cast in-situ piles are shown in figure below.

Add caption

Fig.: Driven f In-Situ Concrete Piles Details

The concrete is then rammed into position by a hammer as the casing is withdrawn ensuring firm contact with the soil and the compaction of concrete. Care must be taken to see that the concrete is not over-rammed or the casing withdrawn too quickly.

There is a danger that as the liner tube is withdrawn it will lift the upper portion of the in-situ concrete, thus leaving a void or necking in the upper portion of the pile. This can be avoided by good quality control of the concrete and slow withdrawal of the casing.

Driven cast in-situ concrete piles can prove to be economic for sand, loose gravels, soft silts and clays, particularly when large numbers of piles are required. For small numbers of piles the on-site costs can prove expensive.

Driven Cast In-Situ Concrete Piles Construction Process

The process of construction of driven cast in-situ concrete piles are shown in figures below:

1. Permanent pile casing first made ready for driving

2. Pile casing is then driven into the soil

Fig.: Driving of Pile Casing

3. Reinforcement cage is then lowered into the pile core

Fig.: Reinforcement Cage placement in pile core

4. Then the concrete is poured in the pile core to complete the construction of driven cast in-situ piles.

Fig.: Concreting of Driven Cast In-Situ Pile

Unit Weight of Materials Used at Construction Site

S.No	Material	Theoretical Weight in(KG/M³)	Approx Weight at Site in		Remarks
			Kg	Per
1	Cement	1440	50	Bag
2	Steel	7850	d²/162		d -dia in mm
3	Sand-
	Dry	1600	50 to 55	farma	1 farma=1.25cft
	River	1840	57 to 63	farma	1 farma=1.25cft
4	Stone(basalt)	2850 to 2960	48 to 52	farma	metal 12mm to 20mm
5	Water	1000	1	liter
6	PCC	2240	8.24 to 8.5	Cube mould	cube mould size=15x15x15cm
7	RCC 2% Steel	2420
8	Bricks	1600 to 1920	1.9 to 2	no	9x4x2 3/4″
			4.8 to 4.9	no	9x6x3 3/4”
9	Brick Masonry	1920
10	Soil(damp)	1760	50 to 55	cft	Black cotton
11	Cement concrete block(solid)	1800	18 to 20	cft	30x15x20 cm
			10 to 11	no	30x10x20 cm
12	Cement Mortar	2080	57 to 62	cft
13	Lime Mortar	1760	48 to 52	cft
14	Lime	640	30	bag
15	Glass	2530	0.9 to 0.95	sft	4mm tk plain
16	Teak Wood	670 to 830	18 to 20	cft
17	Sal Wood	990	22 to 24	cft
18	Marble mosaic tile		2.8 to 3.2	no	25x25x22mm
			4.8 to 5.2	no	30x30x25mm
19	Chequered tile		2.5 to 2.8	no	25x25x22mm
20	Glazed tile15x15cm		0.20 to 0.25	no	5mm tk
21	Marble Stone	2620	5.1	sft	3/4″tk
22	Granite Stone	2460-2800	5.35	sft	3/4″tk
23	Coddappa	2720	6.4	sft	1 1/4″tk
24	A.C.sheet corrugated	16	1.2	sft
25	Bitumen	1040	220	Drum	200liter drum
26	Window frame (simple design)		1.9 to2.1	sft
27	Door Frame
	a)3’00×7’0		25 to 27	no	section 4″x2 1/2″
	b)2’6″x7’0		24 to 26	no	section 4″x2 1/2″