Principles of Structural Design: A Comprehensive Report

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Principles of
Structural Design
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Contents
LO1: Bending Moments And Shear Forces For Simply Supported And Concrete Beams.......6
Numerical: Beam Arrangements And The Reactions At Support -1a...................................6
Diagram (Shear Force & Bending Moment)...................................................................................7
Numerical: Beam Arrangement And The Reactions At Both Supports-1b...........................8
Diagram (Shear Force & Bending Moment).................................................................................10
Numerical: Shear Force And Bending Moment Diagrams-1c.............................................11
Diagram (Shear Force & Bending Moment).................................................................................13
Numerical: Shear Force And Bending Moment Diagrams When The Load Is Additionally
Supported By Beam-1d........................................................................................................14
Diagram (Shear Force & Bending Moment).................................................................................16
Numerical: Shear Force And Bending Moment Diagrams When The Beam Is Carrying An
Additional Point Load-1e.....................................................................................................17
Diagram (Shear Force & Bending Moment) ................................................................................19
Statuary Requirements That Will Ensure The Safety In Structural Designs-1B.................20
 Fire Safety............................................................................................................................20
Other Legal Requirements Includes.............................................................................................20
The Important And Valid Factors For Live, Imposed And Dead Loads-1c........................21
Dead Loads..................................................................................................................................21
Live Loads OR Imposed Loads......................................................................................................21
Example-1 to illustrate dead load:...............................................................................................23
Example-2 illustration using Example-1.......................................................................................23
Steel Beam Selection By Using Maximum Bending Moments And Current Codes Of
Practice-1D...........................................................................................................................25
Example-3 for selecting an Economical A36 steel........................................................................25
LO2: Simply Supported Steel Beams Analyzation Using Deflection.....................................28
Numerical: Point Loads And A Uniformly Distributed Load Provided In Task 1 (A) To 1
(E)-2a....................................................................................................................................28
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Numerical: Structural Stability Affected By Deflection-2B................................................29
Numerical: Application Of Methods Of Supports For Structures To Steel And Concrete
Structures -2C.......................................................................................................................32
Lo2/Lo3: Assessment of Support Methods In Construction....................................................35
Numerical: Assessment Of The Most Effective Support Method For A Given Scenario-D
..............................................................................................................................................35
LO3: Capacity Of Steel And Concrete Columns Carrying Axial Loads.................................38
Numerical: Axial Load Capacity When The Column Is Short-3a.......................................38
Designing Of Steel And Concrete Columns Using Slenderness Ratio And Effective
Length-3A............................................................................................................................38
Axial Load Carrying Capacity Of Above Load-3B.............................................................39
Properties Of Different Materials Used For Beams And Structures In Fixed Structures-3C
..............................................................................................................................................39
Strength to weight ratio..............................................................................................................39
Ductility.......................................................................................................................................39
Corrosion Resistance...................................................................................................................39
Load Carrying Capacity................................................................................................................39
Size Comparing............................................................................................................................39
LO4: Analyzing The Design Methods For Steel, Concrete Beams And Columns..................40
Numerical: Designing And Comparing Solutions For Steel And Concrete Beams Also For
Column And Concrete Beams -4a........................................................................................40
Sketches and Stipulations in Support of Above Calculations..............................................42
Achievement Of Design Solution Using An Alternative Method-4c..................................42
Timber Structure..........................................................................................................................42
Thermal Property.........................................................................................................................42
Variation......................................................................................................................................42
Limitations Associated.................................................................................................................42
Using BIM For Production Of Accurate Structural Designs-4D.........................................43
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References................................................................................................................................44
List of Figures
Figure 1 ASCI manual.............................................................................................................21
Figure 2 Example 2 Readings..................................................................................................22
Figure 3 A36 steel beam..........................................................................................................24
Figure 4 internal share fore and bending moment diagram.....................................................25
Figure 5 A36 steel details from ASCI codes............................................................................25
Figure 6: Concrete and Steel size comparing...........................................................................38
Figure 7: Geometric and analytical designs BIM....................................................................42
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LO1: Bending Moments And Shear Forces For Simply Supported
And Concrete Beams.
Numerical: Beam Arrangements And The Reactions At Support -1a
Solution: W =150 KN
Given Data
A C B
7 m 7 m
l=14 m
length of span=l=14 m
Point load=150 KN
Required
Reaction Forces=R A ∧RB =?
Shear Force=S . F=?
Bending Moment=B . M =?
Reaction Forces
For the beam in equilibrium
a) ∑ of all theupwad forces=∑ of all thedownward forces
RA + RB =150 KN (1)
b) ∑ of all the moments=0
So, taking the moment about point A
RB × 14−150× 7=0
14 RB−1050=0
14 RB=1050
RB=1050 ÷ 14
RB=75 KN
Now, by putting the value of RB in EQ. 1 (Liang, 2005)
RA +75 KN =150 KN
RA =150 KN −70 KN
RA =75 KN
Shear Forces
At point A
( S . F ) A =W
2
( S . F ) A =150
2
( S . F ) A =75 KN
At point C
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( S . F ) c= W
2 −W
( S . F )c=−W
2
( S . F ) c=−150
2
( S . F )c=−75 KN
At Point B
( S . F )B = W
2
( S . F )B =−W
2 + W
2
( S . F )B =0
Bending Moment
At Point A
( B . M )A=0
At Point C
( B . M )C=R A × L
2
( B . M )C= W
2 × L
2
( B . M )C= WL
4
( B . M )C=(150 ×14)/ 4
(B . M )C=525 KN . m
At point B
(B . M )B =0
Diagram (Shear Force & Bending Moment)
W =150 KN
A C B
RA =75 KN 7 m 7 m RB=75 KN
l=14 m
Shear Force +VE
A C B
-VE
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Bending Moment
A C B
Numerical: Beam Arrangement And The Reactions At Both Supports-1b
Solution
Given Data
W =65 KN
A C B
a=2 m b=6 m
l=8 m
length of span=l=8 m
Point load=65 KN
Required
Reaction Forces=R A ∧RB =?
Shear Force=S . F=?
Bending Moment=B . M =?
Reaction Forces
At Point A
RA = W × b
l
RA = 65× 6
8
RA =48.75 KN
At Point B
RB= W ×a
l
RB= 65 ×2
8
RB=16.25 KN
Shear Forces
At Point A
( S . F . ) A =W × b
l
Now Putting the values of W, b and l in the above equation
( S . F . ) A =65 ×6
8
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( S . F . ) A =48.75 KN
At Point C
( S . F . )C= W ×b
l −W
( S . F . )C=W [ ( b
l )−1 ]
( S . F . ) C=W [ b−l
l ]
( S . F . ) C=−W [ l−b
l ]
Since we know that
l=a+b
So,
a=l−b
So, by putting the value of l−b in the corresponding equ. We get
( S . F . )C=−W ×a
l
Now Putting the values of W, a and l in the above equation
( S . F . )C=−65 ×2
8
( S . F . )C=−16.25
At Point B
( S . F . )B =−W × a
l +W × a
l
( S . F . )B =0
Bending Moment
At Point A
( B . M . ) A =0
At Point C
( B . M . )C=RA × a
Since RA = W × b
l So,
( B . M . ) C= W ×b
l ×a
Now Putting the values of W, a, b and l in the above equation, we get
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( B . M . )C= 65 ×6
8 ×2
( B . M . ) C=97.5 KN .m
At Point B
( B . M . ) B =0
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Diagram (Shear Force & Bending Moment)
W =65 KN
A C B
RA =48.75 a=2 m b=6 m RB=16.25 KN
l=8 m
Shear Force
A C B
Bending Moment
A C B
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Numerical: Shear Force And Bending Moment Diagrams-1c
Solution:
Given Data
W (UDL)=25 KN /m
A C B
12.5 m 12.5
l=25 m
length of span=l=25 m
Uniformly Distributed load =UDL=25 KN
Required
Reaction Forces=R A ∧RB =?
Shear Force=S . F=?
Bending Moment=B . M =?
Reaction Forces
For the beam in equilibrium
i. ∑ of all theupwad forces=∑ of all thedownward forces
RA + RB =25× 25 (1)
RA + RB =625 KN
ii. ∑ of all the moments=0
At Point B
So, taking a moment about point A
RB × 25=(25 ×25) 25
2
RB=312.5 KN
At Point A
By putting the value of RB in equ. (1). We get
RA + 312.5=625
RA =625−312.5
RA =312.5 KN
Shear Forces
At Point A
( S . F .)A =−W ×l
2
(S . F .)A =−25 ×25
2
(S . F .)A =−312.5 KN
At Point C
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(S . F .)C =−W × l
2 + W ×l
2
(S . F .)C =0
At Point B
(S . F .)B= W ×l
2
( S . F .)B= 25 ×25
2
(S . F .)B=312.5 KN
Bending Moment
Since we know that the maximum value of bending moment lies where shear force is zero.
So,
At Point A
(B . M )A=0
At Point B
(B . M )B =0
At Point C
Here, in this case, the bending moment will be equal to the area covered by sheer force
between point B and C.
So,
(B . M )C= 1
2 ×12.5 ×312.5
( B . M ) C=1953.12 KN
20 Principles of Structural Design