Civil Engineering Department

​

TOPIC-BAR BENDING FABRICATION OF REINFORCEMENT FOR A SLAB

SEMESTER-6^TH

​

BY

Mrs. MANDAKINI MOHANTA / MR BIKASH DAS

(LAB ASST. Civil Engineering Department)

AY:2021-2022

AIM OF THE EXPERIMENT :-

Bar bending and fabrication of reinforcement for a slab.

INSTRUMENTS REQUIRED:-

​

BAR BENDER

TAPE

STEEL BSRS

BINDING WIRES

BAR CUTTING MACHINE

Slabs

Slab is an important structural element which is constructed to create flat and useful surfaces such as floors, roofs, and ceilings. It is a horizontal structural component, with top and bottom surfaces parallel or near so. Commonly, slabs are supported by beams, columns (concrete or steel), walls, or the ground. The depth of a concrete slab floor is very small compared to its span.

What Is a One Way Slab?

According to IS 456:2000, The ratio of longer span(l) to shorter span(b) which is (L/B) greater than 2 is known as One way slab. In practical, One way slab is supported by only two parallel beams or walls. Normally we don’t use one way slabs often.

What Is a Two Way Slab?

A two way street is a street that allows vehicles to travel in both directions. On most two way streets, especially main streets, a line is painted down the middle of the road to remind drivers to stay on their side of the road. Sometimes one portion of a street is two way and the other portion is one way.

One Way Slab Reinforcement Details

​

To support the bending moment, the main bars are provided in a shorter span.

• Slab Size = 6000 x 2500 x 150
• 10mm Main Bars @ 150mm c/c spacing
• 8mm Distribution Bars @ 150mm c/c Spacing
• Slab thickness is 150 mm
• Top Extra Bars – 8mm @ 150 c/c
• Beam Cover – 25 mm

Bar Bending Schedule for One Way Slab

Step 1 – Find out the total Number of Main Bars & Distribution Bars

• Number of Main Bars = (Length of Longer Side / Spacing) + 1 = (6000/150)+1 = 41 Numbers
• Number of Distribution Bars = (Length of Shorter Side / Spacing) + 1 = (2500/150)+1 = 18 Numbers

Step 2 – Find Out the Cutting Length of Bars (Main bar & Distribution bar)

Therefore, Inclined Length D = 0.42 x d

Where, D = Slab thickness – 2 x Clear Cover – Bar Dia = 150-(2*25)-10 = 90 mm

Inclined Length = 0.42 X 90 = 37.8 mm

• Cutting Length of Main Bar =  Clear Span (Lclear) + (2 x Beam Width) – (2 x Con. Cover) + (1 x Inclined Length) – (45° bend x 2)

= 2500+(2*240)-(2*25)+37.8-20 = 2947.8 mm = 2.95 m

• Cutting Length of Distribution Bar =  Clear Span (Lclear) – (2 x Con. Cover)= 6000 – 50 = 5950 mm or 5.95 m

​

• Number of Top Bars  = Crank Length (L/4) / Spacing + 1 = ((2500/4)/150)+1 x 2 side = 10 Numbers
• Cutting Length of Top Bars = Same cutting length as distribution bars (5.95 m)

Step 3 – Find out the number of Top Extra Rods (for L/4 length)

Two Way Slab Reinforcement Details

From Drawing

• Slab size – 6000 x 5400 x 150
• 10mm Main Bars @ 150mm C/C spacing
• 10mm Distribution Bars @ 200mm C/C Spacing
• Development Length L_d = 40d
• Top Extra Bar 8mm @ 150 mm c/c (both direction)

Bar Bending Schedule for Two Way Slab

Step 1 – Find out the Total Number of Main Bars & Distribution Bars

• Number of Main Bars = (Length of Longer Side / Spacing) + 1 = (6000/150)+1 = 41 Numbers
• Number of Distribution Bars = (Length of Shorter Side / Spacing) + 1 = (5400/150)+1 = 28 Numbers

Step 2 – Find Out the Cutting Length of Bars

• Cutting Length of Bar = Clear Span (Lclear) + (2 x Beam Width) – (2 x Con. Cover) + (1 x Inclined Length) – (45° bend x 2)

Therefore, Inclined Length D = 0.42 x d

Where, D = Slab thickness – 2 Side Clear Cover – Bar Dia = 150 – (2*25) – 10 = 90 mm

Inclined Length = 0.42 X 90 = 37.8 mm

• Cutting Length of Main Bar = Clear Span (Lclear) + (2 x Beam Width) – (2 x Con. Cover) + (1 x Inclined Length) – (45° bend x 2)

where 45° bend = 1 d

= 5400 + (2*240) -(2*25) + 37.8 – (10*2) = 5847.8 mm = 5.85 m

• Cutting Length of Distribution Bar = Clear Span (Lclear) + (2 x Beam Width) – (2 x Con. Cover) + (1 x Inclined Length) – (45° bend x 2)

= 6000 + (2*240) -(2*25) + 37.8 – (10*2) = 6447.8 mm or 6.45 m

Step 3 – Top Extra Rods

a) Number of Top Bars (Longer & Shorter Side)

• Number of Top Bars (Longer Side) = Crank Length (L/5)/ Spacing = (5400/5)/150 = 7.2 = 7 No.s x 2 side = 14 No.s
• Number of Top Bars (Shorter Side) = Crank Length (L/5)/ Spacing = (6000/5)/150 = 8 = 8 No.s x 2 side = 16 No.s

b) Cutting length of Top Bars (Longer & Shorter Side)

• Cutting Length of Top Bars (Longer Side) = 6000 + (2 x L_d) = 6000 + (2 x 40 x 10) = 6800 mm or 6.8 m
• Cutting Length of Top Bars (shorter Side) = 5400 + (2 x L_d) = 5400 + (2 x 40 x 10) = 6200 mm or 6.2 m