Slab & Stair Reinforcement, Lecture notes of Engineering Mathematics

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IN SITU CONCRETE WORKS:

REINFORCEMENT

IN SITU CONCRETE WORKS:

REINFORCEMENT

• Measurement of slab reinforcement

• There are two general types of slabs with regards to its load carrying

capacity:

• One-way slab^ – slab supported by beam / wall support (parallel to

the long span) and whose reinforcement runs in one

direction only, i.e. from support to support.

S/L < 0.50 : S = SHORT SPAN, L = LONG SPAN

IN SITU CONCRETE WORKS:

REINFORCEMENT

• ONE WAY SLAB

IN SITU CONCRETE WORKS:

REINFORCEMENT

• Measurement of slab reinforcement

• There are two general types of slabs with regards to its load carrying

capacity:

• two-way slab^ – slabs which are supported on four sides and are

reinforced in two directions. Reinforcement is

placed in such a way to transmit loads to the four

supporting beams.

S/L > 0.50 : S = SHORT SPAN, L = LONG SPAN

IN SITU CONCRETE WORKS:

REINFORCEMENT

• TWO WAY SLAB

IN SITU CONCRETE WORKS:

REINFORCEMENT

IN SITU CONCRETE WORKS:

REINFORCEMENT

• Measurement of slab reinforcement

• Important notes for measuring slab reinforcement:

• Always check if distribution bars have been provided, especially for top

bars. Unless top bars are continuous from support to support, there will

always be distribution bars to main top bars.

• Pay careful attention that spacing for bars are not interchanged between

the two directions.

• Pay attention for slabs that have drops in them. Extra details may be

provided or adjustments are to be made.

Sample problem no. 3

• COMPUTE FOR

THE CONCRETE,

FORMWORK AND

REINFORCEMENT

OF SLABS S-

AND S-

• Given:

C1A = 200X400MM

CG1A = 200X400MM

G1 = 200X400MM

G1A = 200X400MM

G2 = 200X450MM

G6 = 250X600MM

G7 = 250X600MM

B2 = 200X450MM

CONCRETE COVER = 20MM

• Sample problem no.
• Sample problem no.
• Sample problem no.

Sample problem no. 3

SOLUTION : GIVEN: C1A: 200X400MM G6: 250X600MM CG1A: 200X400MM G7: 250X600MM G1: 200X400MM CONCRETE COVER = 20MM CLEAR LENGTH (CL): CL@LONG SPAN = 4.2 + 1.75 – CG1A WIDTH = 4.2 + 1.75 – 0.2 = 5.75M CL@SHORT SPAN = 1.675 – C1A WIDTH = 1.675 – 0.2 = 1.475M = = 0.26 < 0.5 : ONE-WAY SLAB LONG SPAN: NO. OF BARS = ROUND[ ] + 1 = ROUND [ ] + 1 = ROUND(12.68) + 1 = 13 +1 = 14 BARS TOP BARS (Φ10mm) @LEFT END = NO. OF BARS X TOTAL LENGTH X WEIGHT FACTOR = 14 X (+ ANCHORAGE LENGTH) X 0. = 14 X ( + 0.50) X 0.617 = 8.566 KGS TOP BARS (Φ10mm) @RIGHT END = NO. OF BARS X TOTAL LENGTH X WEIGHT FACTOR = 14 X (+ ) X 0. = 14 X ( + ) X 0.617 = 5.327 KGS

Sample problem no. 3

SOLUTION:

BOTTOMS BARS (Φ10mm, CONTINUOUS FROM LEFT TO RIGHT END) = NO. OF BARS X TOTAL LENGTH X WEIGHT FACTOR = 14 X ( CL@SHORT SPAN + ANCHORAGE LENGTH + ) X 0. = 14 X (1.475 + 0.50 + ) X 0.617 = 18.14 KG TOTAL BARS @LONG SPAN (Φ10mm) = 8.566 + 5.327 + 18.14 = 32.03 KGS SHORT SPAN: FOR TOP BARS (LEFT & RIGHT END): NO. OF BARS = ROUND [ ] + 1 = ROUND [ ] + 1 = ROUND (1.57) + 1 = 2 + 1 = 3 BARS TOP BARS, Φ10mm @LEFT END & RIGHT END = NO. OF BARS X TOTAL LENGTH X WEIGHT FACTOR = 3 X ( CLEAR LENGTH@LONG SPAN +

• ) X 0.617 X 2 = 3 X (5.75 + + ) X 0.617 X 2 = 22.12 KGS

Sample problem no. 3

SOLUTION:

FOR BOTTOM BARS@MIDDLE:

NO. OF BARS = ROUND [ ] - 1

= ROUND [ ] - 1

= ROUND(3.23) - 1 = 3 – 1 = 2 BARS

BOTTOM BARS, Φ10mm @ MIDDLE = NO. OF BARS X TOTAL LENGTH X WEIGHT FACTOR = 2 X (CLEAR LENGTH@LONG SPAN +

• ) X 0. = 2 X (5.75 + + ) X 0.617 = 7.38 KGS TOTAL BARS @SHORT SPAN (Φ10mm) = 22.12 + 3.69 + 7.38 + 7.38 = 40.57 KGS

TOTAL REINFORCEMENT @SLAB S-1 (Φ10mm) = 32.03 KGS + 40.57 KGS = 72.6 KGS

Sample problem no. 3

S-4(TWO-WAY):

SS = SHORT SPAN

SS^ Clear length @SS = 3.60m

LS

LS = LONG SPAN

Clear length @LS = 6.20m