Structural Analysis: Simply Supported Beams and Column Design

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Contents
LO1 Calculate bending moments and shear forces for simply supported steel and concrete
beams..........................................................................................................................................2
Sol.1 (A) (a)...........................................................................................................................2
Sol.1 (A) (b)...........................................................................................................................4
Sol.1 (A) (c)...........................................................................................................................7
Sol.1 (A) (d).........................................................................................................................10
Sol.1 (A) (e).........................................................................................................................13
Sol.1 (B)...............................................................................................................................16
Sol.1 (C)...............................................................................................................................17
Sol.1 (D)...............................................................................................................................19
LO2 Determine deflection for simply supported steel beams..................................................21
Sol. 2 (A)..............................................................................................................................21
2 (B).....................................................................................................................................23
2 (C).....................................................................................................................................24
2 (C).....................................................................................................................................25
LO2/LO3: To Assess Support Methods in Construction.........................................................28
2 (D).....................................................................................................................................28
LO3 Calculate the axial load carrying capacity for steel & reinforced concrete columns.......30
Point (A)...............................................................................................................................30
Point (B) (a):........................................................................................................................30
Point (B) (b).........................................................................................................................31
Point (C)...............................................................................................................................32
LO4 Explore design methods for steel, reinforced concrete beams and columns...................33
Task no. 4 (A)......................................................................................................................33
Point 4(B).............................................................................................................................34
4(C)......................................................................................................................................35

4(D)......................................................................................................................................35
References................................................................................................................................37
List of Figures
Figure 1: (a) FBD Arrangement.................................................................................................3
Figure 2: Shear Force Graph and Bending Moment Graph.......................................................5
Figure 3: FBD Arrangement......................................................................................................6
Figure 4: Shear Force Graph and Bending Moment Graph of 1(A)(b)......................................8
Figure 5: (c) FBD Arrangement.................................................................................................9
Figure 6: Shear Force Graph and Bending Moment Graph of 1(A)(c)....................................11
Figure 7: (d) FBD Arrangement...............................................................................................12
Figure 8: Shear Force Graph and Bending Moment Graph of 1(A)(d)....................................14
Figure 9: (e) FBD Arrangement...............................................................................................15
Figure 10: Shear Force Graph and Bending Moment Graph of 1(A)(e)..................................17
Figure 11: Load calculation structure arrangement.................................................................20
Figure 12: (a) FBD Arrangement.............................................................................................22
Figure 13: (b) FBD Arrangement.............................................................................................24
Figure 14: Deflection effect on beams.....................................................................................26
Figure 15: (c) FBD Arrangement.............................................................................................26
Figure 16: (d) FBD Arrangement.............................................................................................29
Figure 17: Developed Design...................................................................................................35
Figure 18: BIM Design............................................................................................................37

LO1 Calculate bending moments and shear forces for simply supported
steel and concrete beams.
Sol.1 (A) (a)
Given is,
Length or L is = 14m
Point Load or W is = 150KN
FBD (Free Body Diagram) of given situation is
Figure 1: (a) FBD Arrangement
According to std. rule
Upward reacting forces sum = downward reacting forces sum
Or,
RY = RX + RZ … … eq. (i)
Total Momentum = 0
Rf or Reaction force on X
Rz= 150 ×7
14
And
Rz=75 KN
Reaction force on Y

RX =150−75
And
RX =75 KN
Shear Force on X
SX = 150
2
Or
SX =75 KN
Shear Force on Z
SZ =150
2
Or
SZ =75 KN
Shear Force on Y
SY =−150
2
Or
SY =−75 KN
Bending Moment on X
BX =0
Bending Moment on Z
BZ=0
Bending Moment on Y
BY =RX + ( 1
2 + Span length )

BY =(75 KN )+ ( 7 m )
BY =525 KNm
Figure 2: Shear Force Graph and Bending Moment Graph
Sol.1 (A) (b)
Given is,
Length or L is = 8m
Point Load or W is = 65KN
FBD (Free Body Diagram) of given situation is

Figure 3: FBD Arrangement
Rf or Reaction force on X
Rz= 65+6
8
And
Rz=48.750 KN
Reaction force on Z
RX = 65+2
8
And
RX =16.250 KN
Shear Force on X
SX = 65+6
8
Or
SX =48.750 KN
Shear Force on Z
SZ =150
8 − 150
8
Or
SZ =0 KN

Shear Force on Y
SY = ( 65 KN ) −2+ ( 65 KN )
8
Or
SY =−16.250 KN
Bending Moment on X
BX =0
Bending Moment on Z
BZ=0
Bending Moment on Y
BY =( 65 ×6
8 )×2
Hence,
BY =97.50 KN

Figure 4: Shear Force Graph and Bending Moment Graph of 1(A)(b)
Sol.1 (A) (c)
Given is,
Length or L is = 25m
Uniform load is = 25KN/m
FBD (Free Body Diagram) of given situation is

Figure 5: (c) FBD Arrangement
According to std. rule
Upward reacting forces sum = downward reacting forces sum
Or,
RY = RX + RZ = 252
Or
RY = RX + RZ = 625
Total Momentum = 0
Rf or Reaction force on X
Rz=(625 × 12.5)
25
And
Rz=312.50 KN
Reaction force on Z
RX =625−312.50
And
RX =312.50 KN
Shear Force on X
SX =− ( 25 KN ×25
2 )

Or
SX =−312.50 KN
Shear Force on Z
SZ =( 25 KN × 25
2 )
Or
SZ =312.50 KN
Shear Force on Y
SY = ( ( 25 KN ×25 )
2 ) −( ( 25 KN × 25 )
2 )
Or
SY =0 KN
Bending Moment on X
BX =0
Bending Moment on Z
BZ=0
Bending Moment on Y
BY =312.50 × 12.5
2
Hence,
BY =1953.1250 KN

Figure 6: Shear Force Graph and Bending Moment Graph of 1(A)(c)
Sol.1 (A) (d)
Given is,
Length or L is = 14m
Uniform Load is = 125KN
FBD (Free Body Diagram) of given situation is

Figure 7: (d) FBD Arrangement
According to std. rule
Upward reacting forces sum = downward reacting forces sum
Or,
RY = RX + RZ = 150+280
Or
RY = RX + RZ = 430KN
Total Momentum = 0
Rf or Reaction force on X
Rz=(1050+1960)
14
And
Rz=215 KN
Reaction force on Z
RX =430−215
And
RX =215 KN
Shear Force on X