Step 3: Calculation of main reinforcement (longitudinal or steel) 1. Design and detail a singly reinforced simply supported beam of Clear span 5.5 m to support a working live load of 60 KN/m. use Fe 460 steel and M30 grade concrete. Assume the support thickness as 220 mm. Over all depth of the beam is given as D=570 mm. 0.003+f/E,Y P = Pmax = Pb 0,008 d = D- 25 = A, = pbd AS fy Mu = 9 As fy {d - Step 1: Determination of bending moment and shear force 1.7f'c b As = Total ultimate load = (Factor of safety x Dead load) + (Factor of safety x Live load) Max value: Dead load = b XDX unit weight of concrete As No. of bars = O2) Simply supported beam bending moment = Mu = Wle?/8 = Ans: Number YDiameter. Step 2: Check for depth 600 600 +f, P = (0.85)B, 0.003 +f/E, Pmax = 0.008 Rumax = pPman, (1 Pmaxfy 1.7f - M. d = Rb

Structural Analysis
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Chapter2: Loads On Structures
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Name courses: (Design of structure -1)

EXPERIMENT :Designing and detailing of singly reinforced simply supported beam - ACI Code

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Step 1

Step 2

Step 3

Step 4

Step 3: Calculation of main reinforcement (longitudinal or steel):
1. Design and detail a singly reinforced simply supported beam of Clear span
5.5 m to support a working live load of 60 KN/m. use Fe 460 steel and M30
grade concrete. Assume the support thickness as 220 mm. Over all depth
of the beam is given as D=570 mm.
(0.003 + f,/E,
P = Pma =
Pb
0.008
d = D - 25 =
A = pbd
b =
AS fy
Mu =
Ф As fy {d -
Step 1: Determination of bending moment and shear force:
1.7f'c b
As =
2
Total ultimate load = (Factor of safety x Dead load) + (Factor of safety x Live
load)
Max value:
Dead load = b XDX unit weight of concrete
As
No. of bars =
Simply supported beam bending moment = Mu = Wle?/8 =
Ans: NumberYDiameter
Step 2: Check for depth:
600
(0.85)A, E(a
Ph =
600 +f,
( 0.003 +fy/E,
Pmax =
0.008
= P maxfy ( 1
Pmaxfy
1.7f.
Rumax
d =
V R„b
Transcribed Image Text:Step 3: Calculation of main reinforcement (longitudinal or steel): 1. Design and detail a singly reinforced simply supported beam of Clear span 5.5 m to support a working live load of 60 KN/m. use Fe 460 steel and M30 grade concrete. Assume the support thickness as 220 mm. Over all depth of the beam is given as D=570 mm. (0.003 + f,/E, P = Pma = Pb 0.008 d = D - 25 = A = pbd b = AS fy Mu = Ф As fy {d - Step 1: Determination of bending moment and shear force: 1.7f'c b As = 2 Total ultimate load = (Factor of safety x Dead load) + (Factor of safety x Live load) Max value: Dead load = b XDX unit weight of concrete As No. of bars = Simply supported beam bending moment = Mu = Wle?/8 = Ans: NumberYDiameter Step 2: Check for depth: 600 (0.85)A, E(a Ph = 600 +f, ( 0.003 +fy/E, Pmax = 0.008 = P maxfy ( 1 Pmaxfy 1.7f. Rumax d = V R„b
Step 4: Check for shear reinforcement:
Shear force of the simply supported beam = Vu = Wlet /2 =
Shear force Vu =
Maximum design shear at a distance d from the face of the support is
Vu =
The nominal shear strength provided by the concrete is
Vc = 0.17 F'c bd
Vu = + VG + + Vs
VG =
O VG = 0.75 x
O Vs = Vu - 4VG
Vs =
Choose 8 mm diameter 2 legged stirrups
Av = 2x 50 = 100 mm?.
Spacing S = Avfyd -
Vs
Check for maximum spacing:
for V, < (0.33V) bd
-{ for If V, > (0.33VĪAbod
Smax
Transcribed Image Text:Step 4: Check for shear reinforcement: Shear force of the simply supported beam = Vu = Wlet /2 = Shear force Vu = Maximum design shear at a distance d from the face of the support is Vu = The nominal shear strength provided by the concrete is Vc = 0.17 F'c bd Vu = + VG + + Vs VG = O VG = 0.75 x O Vs = Vu - 4VG Vs = Choose 8 mm diameter 2 legged stirrups Av = 2x 50 = 100 mm?. Spacing S = Avfyd - Vs Check for maximum spacing: for V, < (0.33V) bd -{ for If V, > (0.33VĪAbod Smax
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