By: Numan Shaikh SAHANI ASSOCIATES Client: Antony lara

Consulting Structural Engineers

Date: 26-06-2019 Renewable Energy private
Checked by NPS Job No.: 1122 limited
Date: 26-06-2019 Subject: Typical Design
Calculations
Sh. No. 1 Rev : P0

TYPICAL STRUCTURAL DESIGN CALCULATION

PROJECT: Processing Treatment Public Disposal Municipal Soiled Waste with State Of Art Waste
Energy Plant.

LOCATION: Moshi, Chinchwad.

Brief Description:

Construction of RCC weigh bridge with ramp and also a gate cabin as a ground structure.

GENERAL LOAD DATA:

➢ Unit load of concrete =25 KN/m3

➢ Unit load of floor finish/Bricks =21 KN/m3

➢ Live load assumed for design of typical floor =2 KN/m2

➢ Floor finish considered for typical floor = 50 mm

➢ Waterproofing considered for Terrace floor = 50 mm

➢ Live load assumed for Weigh bridge = 600 KN

MATERIAL USED

Concrete grade used – M25

Reinforcement steel grade used – Fe 500

SAHANI ASSOCIATES
By: Numan Shaikh SAHANI ASSOCIATES Client: Antony lara
Consulting Structural Engineers
Date: 26-06-2019 Renewable Energy private
Checked by NPS Job No.: 1122 limited
Date: 26-06-2019 Subject: Typical Design
Calculations
Sh. No. 1 Rev : P0

DESIGN OF GATE HOUSE CABIN

DESIGN OF TYPICAL SLAB RS1

(Reference Drawing No.:1122-S13A)

Dimensions :

Short span = 2.04 m

Long span = 4.54 m

Load Data :

Live load = 2 Kn/m2

Water proofing = 150 mm
Slab thickness = 120 mm

Load Calculation :

Self weight = 0.12X25 = 3. Kn/m2

Floor finish load = 0.15X21 = 3.15Kn/m2
Live load = 2 Kn/m2

Total =8.15 Kn/m2

Factored Load = 1.5 X 8.15 =12.225 Kn/m2

SAHANI ASSOCIATES
By: Numan Shaikh SAHANI ASSOCIATES Client: Antony lara
Consulting Structural Engineers
Date: 26-06-2019 Renewable Energy private
Checked by NPS Job No.: 1122 limited
Date: 26-06-2019 Subject: Typical Design
Calculations
Sh. No. 1 Rev : P0

Bending Moment in main direction

M= 12.225∗2.270²
8
= 7.87 Kn.m

With 120 mm Total Thickness of slab

Provide 8Ø @ 120mm C/C In shorter Direction and Provide 8Ø @ 200mm C/C in Longer
Direction
Design of Beam RB1 :

Span (L): 4.77 m

Section and materials:

a) Depth (D) = L = 4770

10 10
= 477mm
Provide overall depth of beam as 450mm
b) Effective cover = 40mm
c) effective depth (d) = D-d= 410mm

Loads:
Uniformly distributed load = (w) kn/m
i. Slab load: =8.313 Kn/m
ii. Self weight = 0.23X0.45X25= 2.58 Kn/m
iii. Total factored load= 1.5X 10.893 = 16.34 Kn/m

Calculation of shear force and bending moment

Moment= W*L²/8 =16.34 *4.77²/8= 46.47 Kn.m

Shear =Wu*L/2 = 16.34 *4.77/2 = 38.97 Kn

SAHANI ASSOCIATES
By: Numan Shaikh SAHANI ASSOCIATES Client: Antony lara
Consulting Structural Engineers
Date: 26-06-2019 Renewable Energy private
Checked by NPS Job No.: 1122 limited
Date: 26-06-2019 Subject: Typical Design
Calculations
Sh. No. 1 Rev : P0

Calculation of reinforcement,
Mu= 69.7 kn.m, b=230mm, d=420mm

Ast_required = (0.5.fck.b.d)/Fy*(1-(√1-(4.6*Mu/Fck.b.d²))= 300 mm²

Ast(min)= 0.85.b.d/Fy
= 160.31 mm²

Provide (2-16Ø + 2-12Ø ) at bottom and(2-12Ø) as anchor bar

Design of shear reinforcement,

Max. Shear force = Vu = 38.97 KN

Vu
Nominal shear stress =  V = =0.42 MPa
bd

100  Ast
Percentage of steel reinforcement, Pt = = 0.42 %
bd

From IS 456-2000 Table No – 19

Shear stress in concrete c = 0.440 N/mm2

  v <  c <  c max

It means shear reinforcement is required. Let us use 8 mm-2 legged bars

Asv = 2 x 50.2 = 100.53 mm2

Vus=Vu

Spacing = (0.87*Fy*Asv*d)/Vus = 325.45 mm

Sv = (Asv*0.87*Fy)/0.4*b = 394.52mm

Provided 8 mm  bars at 150 mm c/c 2 legged Stirrups

SAHANI ASSOCIATES
 By: Numan Shaikh SAHANI ASSOCIATES Client: Antony lara
Consulting Structural Engineers
Date: 26-06-2019 Renewable Energy private
Checked by NPS Job No.: 1122 limited
Date: 26-06-2019 Subject: Typical Design
Calculations
Sh. No. 1 Rev : P0

Design of Beam PB3 :

Span (L): 4.77 m

Section and materials:

d) Breadth ( b > breadth of wall ) = 230mm

e) Depth (D) = L = 4770
10 10
= 477mm
Provide overall depth of beam as 450mm
f) Effective cover = 40mm
g) effective depth ( d) = D-d= 410mm

Loads:
Uniformly distributed load = (w) kn/m

i. Wall load = 0.23X3X21=14.5 Kn/m

ii. Self-weight = 0.23X0.45X25=2.58 Kn/m
iii. Total factored load= 1.5 X 17.0875=25.63 Kn/m

Calculation of shear force and bending moment

Moment= W*L²/8 =25.63*4.77²/8= 72.9 Kn.m

Shear =Wu*L/2 = 25.63*4.77/2=61.12 Kn

Calculation of reinforcement,
Mu= 72.9 KNm, b=230mm, d=410mm

Ast required=(0.5.fck.b.d)/Fy*(1-(√1-(4.6*Mu/Fck.b.d²))= 452.34mm²

Ast(min)= 0.85.b.d/Fy = 160.31 mm²
Provide (2-16Ø + 2-12Ø ) at bottom and (2-12Ø) as anchor bar

SAHANI ASSOCIATES
 By: Numan Shaikh SAHANI ASSOCIATES Client: Antony lara
Consulting Structural Engineers
Date: 26-06-2019 Renewable Energy private
Checked by NPS Job No.: 1122 limited
Date: 26-06-2019 Subject: Typical Design
Calculations
Sh. No. 1 Rev : P0

Design of shear reinforcement,

Max. Shear force = Vu = 61.12 KN

Vu
Nominal shear stress =  V = = 0.648 MPa
bd

100  Ast
Percentage of steel reinforcement, Pt = = 0.47 %
bd

From IS 456-2000 Table No – 19

Shear stress in concrete  c = 0.474 N/mm2

  v >  c <  c max

It means shear reinforcement is required.

Vus =Vu-  c x b x d = 16.42 Kn

Let us use 8 mm-2 legged bars

Asv = 2 x 50.2 = 100.53 mm2

Vus =Vu

Spacing = (0.87*Fy*Asv*d) / Vus = 1091.93 mm

Sv = (Asv*0.87*Fy) / 0.4*b = 394.52mm

Provided 8 mm  bars at 150 mm c/c 2 legged Stirrups

SAHANI ASSOCIATES
 By: Numan Shaikh SAHANI ASSOCIATES Client: Antony lara
Consulting Structural Engineers
Date: 26-06-2019 Renewable Energy private
Checked by NPS Job No.: 1122 limited
Date: 26-06-2019 Subject: Typical Design
Calculations
Sh. No. 1 Rev : P0

DESIGN OF COLUMN C7 (TYPICAL):

LOAD ON COLUMN C7 DUE TO BEAMS = 200.18 Kn

SELF WEIGHT OF COLUMN = 0.23 X 0.3 X 4.8 X 25 X 1.5

= 12.42 KN

HENCE TOTAL LOAD = 212.6 KN

LOAD ON COLUMN, Pu = 0.4 x Fck x Ac + 0.67 x Fy x Asc

Pu = 200.18 Kn
Asc = 1 % of Ag
Asc = 690 mm2
No. of bars = Asc/ area of steel per bar

Provided 230 X 300 Column with 4-16 Ø main bar and with 8mm column links at 150 c/c.

DESIGN OF FOOTING F1 (TYPICAL):

CUMULATIVE AXIAL LOAD FROM C7 + SELF WEIGH OF FOOTING = 214.53 KN

SBC ASSUMED = 250 Kn/m2

Size of footing required = 250 / 214.53 = 1.16 m2

Provided 1200 X 1200 Footing with 10@120mm c/c main bar and with 10@120mm c/c as
distribution steel WITH 400mm overall depth.

Due to space restriction provide combined footing for columns C9 & C1, C10 & C2, C8 & C6

With 450mm overall depth and reinforcement as follows

Bottom reinforcement= (Long bar) Y-10mm bottom bar at 120mmc/c and (Short bar) as Y-10mm
at 120mmc/c.
Top reinforcement = (Long bar) Y-10mm bottom bar at 120mmc/c and as Y-10mm bottom bar at
120mmc/c.

SAHANI ASSOCIATES
By: Numan Shaikh SAHANI ASSOCIATES Client: Antony lara
Consulting Structural Engineers
Date: 26-06-2019 Renewable Energy private
Checked by NPS Job No.: 1122 limited
Date: 26-06-2019 Subject: Typical Design
Calculations
Sh. No. 1 Rev : P0

DESIGN OF WEIGH BRIDGE

Design of Beam PB1 :

Span (L): 4.83 m

Section and materials:

h) Breadth ( b > breadth of wall ) = 230mm

i) Depth (D) = L = 4830
10 10
= 483mm
Provide overall depth of beam as 450mm
j) Effective cover = 40mm
k) effective depth ( d) = D-d= 410mm

Loads:
Uniformly distributed load = (w) kn/m

iv. slab load = 24 Kn/m

v. Self-weight = 0.23X0.45X25=2.58 Kn/m
vi. Total factored load= 1.5 X 26.587=40 Kn/m

Calculation of shear force and bending moment

Moment= W*L²/8 =40*4.83²/8= 116.64 Kn.m

Shear =Wu*L/2 = 40*4.83/2=97 Kn

Calculation of reinforcement,
Mu= 116.64 KNm, b=230mm, d=410mm

Ast required=(0.5.fck.b.d)/Fy*(1-(√1-(4.6*Mu/Fck.b.d²))= 796.835mm²

Ast(min)= 0.85.b.d/Fy = 160.31 mm²
SAHANI ASSOCIATES
 By: Numan Shaikh SAHANI ASSOCIATES Client: Antony lara
Consulting Structural Engineers
Date: 26-06-2019 Renewable Energy private
Checked by NPS Job No.: 1122 limited
Date: 26-06-2019 Subject: Typical Design
Calculations
Sh. No. 1 Rev : P0

Provide (4-16Ø) at bottom and (2-12Ø) as anchor bar

Design of shear reinforcement,

Max. Shear force = Vu = 97 KN

Vu
Nominal shear stress =  V = = 1.023 MPa
bd

100  Ast
Percentage of steel reinforcement, Pt = = 0.845 %
bd

From IS 456-2000 Table No – 19

Shear stress in concrete  c = 0.6 N/mm2

  v >  c <  c max

It means shear reinforcement is required.

Vus =Vu-  c x b x d = 40.42 Kn

Let us use 8 mm-2 legged bars

Asv = 2 x 50.2 = 100.53 mm2

Vus =Vu

Spacing = (0.87*Fy*Asv*d) / Vus = 368.17 mm

Sv = (Asv*0.87*Fy) / 0.4*b = 394.52mm

Provided 8 mm  bars at 150 mm c/c 2 legged Stirrups

SAHANI ASSOCIATES
 By: Numan Shaikh SAHANI ASSOCIATES Client: Antony lara
Consulting Structural Engineers
Date: 26-06-2019 Renewable Energy private
Checked by NPS Job No.: 1122 limited
Date: 26-06-2019 Subject: Typical Design
Calculations
Sh. No. 1 Rev : P0

DESIGN OF COLUMN C1 (TYPICAL):

LOAD ON COLUMN C1 DUE TO BEAMS = 135 Kn

SELF WEIGH OF COLUMN = 0.6 X 0.6 X 1 X 25 X 1.5

= 13.5 KN

HENCE TOTAL LOAD = 150 KN

LOAD ON COLUMN, Pu = 0.4 x Fck x Ac + 0.67 x Fy x Asc

Pu = 150 Kn
Asc = 0.8 % of Ag
Asc = 2880 mm2
No. of bars = Asc/ area of steel per bar

Provided 600 X 600 Column with 8-16 Ø main bar and with 8mm column links at 150 c/c.

DESIGN OF FOOTING F1 (TYPICAL):

CUMULATIVE AXIAL LOAD FROM C1 + SELF WEIGHT OF FOOTING = 171 KN

SBC ASSUMED = 250 Kn/m2

Size of footing required = 250 / 171 = 1.46 m2

Provide 1200 X 1200 Footing with 10@120mm c/c main bar and with 10@120mm c/c as
distribution steel top and bottom with 400mm overall depth.

SAHANI ASSOCIATES