PLEASE USE THE EXAMPLE AS GUIDE AND SHOW ALL THE PROCEDURES, OTHERWISE DON'T ANSWER THIS. Design a rectangular simply supported flexural member to be used as internal simply supported long beam. Slab has longer span of 8m and shorter span of 5m. Factored slab load is 12KN/m2. Use C-18 concrete and Grade 300 steel. Select SI reinforcement bars and keep maximum width to 300mm. The structure has to be used as commercial building. Architecturally maximum permitted depth should be 550mm.DataL: 8mW: 5mFactored Slab Load: 12 kN/m2Fc’: 18 MpaFy: 300 Mpab: 300 mmHa: 375 mmDensity of concrete: 2400 kg/m2Calculations need:Service dead loadFactored dead load:Equivalent width of slab supportedFactored slab load acting on beamTotal factored loadTotal factored bending momentEffective DepthDepth for deflectionMost general depthDead loadDepth of the beamMinimum steel ratioMaximum steel ratioReinforcement by hit and tryplease note that l= 8m as is the long beam that is need to be calculated additionally the Equivalent width of slab supported for a long beam needs to be calculated = (1- R²/3)lx the width needs to be increased by 10% thank you Example 3.6: Design a rectangular singly reinforced flexural member to be used as aninterior simply supported short beam. The slab system has a longer span of 5 m and ashorter span of 3.5 m. Factored stab load is 10.00 kN/m, 17.25 MPa and = 300MPa. Select SI reinforcing bars and keep the width "of member equal to 228 mm.Architecturally, maximum permitted depth is 375 mm.Solutton:Data(- 3,5 mfactored slab load 10.00 kN/m"S'- 15 MPaI,- 300 MPab- 228 mmhu= 375 mmApproximate Self Weight- 2400 x 2283.5 9.811000" 12 "1000Service dead load= 1.57 kN/mFactored dead load- 1.2 x 1.57 - 1.88 kN/mSelected Depth- 300 mm= h-60- 240 mmThe figure of 60 mm is corresponding to expected one layer of steel.EFactored Slab Load Acting On BeamMinimum Steel RatioFactored slab load on beam - width of slab x slab load per unit arca- 234 x 10.00 = 23.4 kN/mAnie - 141, - 14/300 - 0.00467Total Factored LoadMaximum Steel Ratio- 23.4 + 1.88- 25.3 kN/mAn - 0375 x 0.85 A17.25Total Factored Bending Moment- 0.375 x 0.85 x 0.85x300- 0.0156-M.Calculation Of Steel25.3x3.538.74 AN-mTrial 1: Assume a d/3- 240/3 - 80 mm8- 717 mm?M38.74 x10Minimum Effective Depth For Singly Reinforced Section0.9x 300 240-hat - dain + 60 (assuming one layer of steel)4,1,0.85 CaTrial 2:39.28.74 x10Vo.205 x17.25x 228M.+ 60-+ 60- 279 mmV0.205,77x 30064 mmDepth For Deflection Control0.85 x17.25x 22838.74 x 10"- 690 mmMinimum depth of beam for deflection control (A)e/ 20- 3500/ 200.9× 300 240-- 175 mmMaximum Architectural Depth690x 3000.85x17.25 x 228Trial 3:- 62 mmhamar - 375 mm38.74 x 10**- 687 mmMost General Depth0.9x300 240--// 12 - 3500/12 - 292 mmThe depth may be selected in multiples of the brick height, if possible.(sufficiently close to the previous answer)A reinforcement of 2 20 +1 IS provides the required arca of steel (Fig. 3.27).

Given : factored slab load = 10 kN/m2 fc = 18 MPa fy = 300 MPa b = 300 mm Ha = 550 mm

Answered: PLEASE USE THE EXAMPLE AS GUIDE AND…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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PLEASE USE THE EXAMPLE AS GUIDE AND SHOW ALL THE PROCEDURES, OTHERWISE DON'T ANSWER THIS.

Design a rectangular simply supported flexural member to be used as internal simply supported long beam. Slab has longer span of 8m and shorter span of 5m. Factored slab load is 12KN/m2. Use C-18 concrete and Grade 300 steel. Select SI reinforcement bars and keep maximum width to 300mm. The structure has to be used as commercial building. Architecturally maximum permitted depth should be 550mm.

Data

L: 8m

W: 5m

Factored Slab Load: 12 kN/m2

Fc’: 18 Mpa

Fy: 300 Mpa

b: 300 mm

Ha: 375 mm

Density of concrete: 2400 kg/m2

Calculations need:

Service dead load
Factored dead load:
Equivalent width of slab supported
Factored slab load acting on beam
Total factored load
Total factored bending moment
Effective Depth
Depth for deflection
Most general depth
Dead load
Depth of the beam
Minimum steel ratio
Maximum steel ratio
Reinforcement by hit and try

please note that l= 8m as is the long beam that is need to be calculated additionally the Equivalent width of slab supported for a long beam needs to be calculated = (1- R²/3)lx the width needs to be increased by 10% thank you

Example 3.6: Design a rectangular singly reinforced flexural member to be used as an
interior simply supported short beam. The slab system has a longer span of 5 m and a
shorter span of 3.5 m. Factored stab load is 10.00 kN/m, 17.25 MPa and = 300
MPa. Select SI reinforcing bars and keep the width "of member equal to 228 mm.
Architecturally, maximum permitted depth is 375 mm.
Solutton:
Data
(- 3,5 m
factored slab load 10.00 kN/m"
S'- 15 MPa
I,- 300 MPa
b- 228 mm
hu= 375 mm
Approximate Self Weight
- 2400 x 2283.5 9.81
1000" 12 "1000
Service dead load
= 1.57 kN/m
Factored dead load- 1.2 x 1.57 - 1.88 kN/m
Selected Depth
- 300 mm
= h-60
- 240 mm
The figure of 60 mm is corresponding to expected one layer of steel.
E
Factored Slab Load Acting On Beam
Minimum Steel Ratio
Factored slab load on beam - width of slab x slab load per unit arca
- 234 x 10.00 = 23.4 kN/m
Anie - 141, - 14/300 - 0.00467
Total Factored Load
Maximum Steel Ratio
- 23.4 + 1.88
- 25.3 kN/m
An - 0375 x 0.85 A
17.25
Total Factored Bending Moment
- 0.375 x 0.85 x 0.85
x
300
- 0.0156
-
M.
Calculation Of Steel
25.3x3.5
38.74 AN-m
Trial 1: Assume a d/3- 240/3 - 80 mm
8
- 717 mm?
M
38.74 x10
Minimum Effective Depth For Singly Reinforced Section
0.9x 300 240-
hat - dain + 60 (assuming one layer of steel)
4,1,
0.85 Ca
Trial 2:
39.28.74 x10
Vo.205 x17.25x 228
M.
+ 60-
+ 60- 279 mm
V0.205,
77x 300
64 mm
Depth For Deflection Control
0.85 x17.25x 228
38.74 x 10"
- 690 mm
Minimum depth of beam for deflection control (A)
e/ 20
- 3500/ 20
0.9× 300 240-
- 175 mm
Maximum Architectural Depth
690x 300
0.85x17.25 x 228
Trial 3:
- 62 mm
hamar - 375 mm
38.74 x 10**
- 687 mm
Most General Depth
0.9x300 240-
-// 12 - 3500/12 - 292 mm
The depth may be selected in multiples of the brick height, if possible.
(sufficiently close to the previous answer)
A reinforcement of 2 20 +1 IS provides the required arca of steel (Fig. 3.27).

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