Steel and Concrete Beam Analysis: Civil Engineering

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Principles of Structural Design
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Contents
LO1 Calculate bending moment and shear forces for simply supported steel and concrete beams.
.........................................................................................................................................................4
Task1............................................................................................................................................4
(i) Calculation of the Reaction Force in the diagram..................................................................7
(ii) Calculation of the Shear Force..........................................................................................7
(c).................................................................................................................................................9
1(B)............................................................................................................................................16
1(C)............................................................................................................................................18
1(D)............................................................................................................................................19
LO2 Determine deflection for simply supported steel beams.......................................................21
Task 2.............................................................................................................................................21
2(A)............................................................................................................................................21
2(B)............................................................................................................................................22
2(C)............................................................................................................................................24
LO2/LO3: To Assess Support Methods in Construction...............................................................26
2(D)............................................................................................................................................26
LO3 Calculate the axial load carrying capacity of steel and reinforced concrete columns...........28
Task 3.............................................................................................................................................28
3(A)............................................................................................................................................28
3(B)............................................................................................................................................29
3(C)............................................................................................................................................29
LO4 Explore design methods for steel, reinforced concrete beams and columns.........................31
Task 4.............................................................................................................................................31
4(A)............................................................................................................................................31
4(B)............................................................................................................................................32
4(C)............................................................................................................................................32
4(D)............................................................................................................................................33
Referrences....................................................................................................................................34
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List of Figures
Figure 1: Point Load acting at the centre.........................................................................................4
Figure 2: Bending Moment Diagram...............................................................................................6
Figure 3: Point load at 2 meters from left........................................................................................7
Figure 4: Combined Diagram..........................................................................................................8
Figure 5: Uniform Load acting on the beam...................................................................................9
Figure 6: Bending Moment Diagram.............................................................................................11
Figure 7: Uniform Load combined with a point load at the centre...............................................11
Figure 8: Bending Moment Diagram.............................................................................................13
Figure 9: Uniform load & point load at 10 m from right...............................................................14
Figure 10: Bending Moment Diagram...........................................................................................16
Figure 11: Reference Diagram.......................................................................................................19
Figure 12: Uniform Load with point Load at the center................................................................21
Figure 13: Point Load at 2 meters from left..................................................................................22
Figure 14: Diagram of a beam.......................................................................................................24
Figure 15: Uniform Load on the beam..........................................................................................24
Figure 16: Uniform Load with a Point load at the centre..............................................................26
Figure 17: Structural Design Solution...........................................................................................32
Figure 18: Work on BIM...............................................................................................................33
List of Tables
Table 1: Properties of Steel............................................................................................................28
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LO1 Calculate bending moment and shear forces for simply supported steel
and concrete beams.
Task1
1(A)
(a)
As given in the question
Length of Span (LOS) = 14 m
Point load = 150 KN
As per the given data, the following interpretations are made:
Figure 1: Point Load acting at the centre
Source: (Author, 2019)
(i) The calculation for the Reaction Force acting on the span.
For maintaining the vertical equilibrium, the sum of all forces acting upward should be equal
to the sum of all forces acting downwards. As the span is in equilibrium the reaction forces
can be identified.
i.e. RA + RC = RB -----------------------------------(i)
Also, as the span is stable the sum of all the moment acting on an object is equal to 0 (zero).
At point A,
RC x 14 = Load at point B x perpendicular distance
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= 150 x 7
RC = 150x7/14 = 75 KN
Now, taking point C
RA + RC = 150
RA = 150-75 = 75 KN
(ii) Calculation for Shear Forces:
At point A;
Shear Force acting at A = Point Load (150KN)/2
= 150 KN/2 = 75 KN
Now, at point C:
Shear Force at C + Point Load at B(150KN)/2 = Point Load at B (150 KN)/2
Shear Force at C = 0 KN
Now, at point B;
Shear Force acting at B = Point load(150KN)/2 – Point Load(150KN)
= Point Load(150KN)-2Point Load(150KN)/2
= -75 KN
(iii) Calculation of Bending Moment from the above diagram:
Take point A;
Bending Moment on point A = 0 (Zero)
Take point C;
Bending Moment on C = 0 (Zero)
Now at point B;
Bending Moment= RA x (Half-length of span)
= (75KN) x 7mtr.
= 525KNmtr.
By combining all the results from the above equations, the shear force and bending moments are
known at point A, B & C. These results can be used for the shear force diagram & bending
moment diagram.
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Figure 2: Bending Moment Diagram
Source: (Author, 2019)
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(b)
The given details are:
Length of Span (LOS) = 8 meters
Point Load = 65 KN
Load’s Position = 2 meters from the left support
With the data the following diagram can be drawn;
Figure 3: Point load at 2 meters from left
Source: (Author, 2019)
(i) Calculation of the Reaction Force in the diagram:
At point C;
RA x 8 = 65 x 6
RA = (65x6)/8 = 48.75 KN
At point A;
RC x 8 = 65 x 2
RC = (65 x 2)/8 = 16.25 KN
(ii) Calculation of the Shear Force;
At point A;
Shear Force at A = Point Load (65 KN) x 6/8
=48.75 KN
At point C;
Shear Force at C + (Point Load(65KN) x 2)/8 = (65 KN x 2)/8
Shear Force at C = 0 KN
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Now, at point B
Shear Force at B = Point load (65 KN)/8 – Point Load(65KN)
= (Point Load (65 KN) – 2Point load (65 KN))/8
= -16.25KN
(iii) Calculation of the Bending Moment
At point A;
Bending Moment at A = 0 (zero)
At point C;
Bending Moment at R = 0 (zero)
Now, At point B;
Bending Moment at Q = (65 x 6/8) x2 = 97.50KN
From the above solutions the shear force diagram and the bending moment diagram can be a plot
on the graph as follows:
Figure 4: Combined Diagram
Source: (Author, 2019)
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(c):
The given details are:
Length of Span (LOS) = 25 meters
Point Load = 25 KN
With the data the following diagram can be drawn;
Figure 5: Uniform Load acting on the beam
Source: (Author, 2019)
(i) The calculation for the Reaction Force acting on the span.
For maintaining the vertical equilibrium, the sum of all forces acting upward should be equal
to the sum of all forces acting downwards. As the span is in equilibrium the reaction forces
can be identified.
i.e. RA + RC = 25*25 = 625-----------------------------------(i)
Also, as the span is stable the sum of all the moment acting on an object is equal to 0 (zero).
At point A,
RC x 25 = Load at point B x perpendicular distance
= 625 x 12.5
RC = 625x12.5/25 = 312.50 KN
Now, taking point C
RA + RC = 625
RA = 625- RC = 625-312.50 = 312.50 KN
(ii) Calculation for Shear Forces:
At point A;
Shear Force acting at A = - [Point Load (25KN) *25/2]
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= - 312.50 KN
Now, at point B:
Shear Force at B + Load (25 KN) *25/2 = Load (25 KN) *25/2
Shear Force at B = 0 KN
Now, at point C;
Shear Force acting at C = Point load(25KN) *25/2
= 312.50 KN
(iii) Calculation of Bending Moment from the above diagram:
Take point A;
Bending Moment on point A = 0 (Zero)
Take point C;
Bending Moment on C = 0 (Zero)
Now at point B;
Bending Moment= RC x (Half-length of span)
= (312.50 KN) x (12.5/2)
= 1953.125 KNM
By combining all the results from the above equations, the shear force and bending moments are
known at point A, B & C. These results can be used for the shear force diagram & bending
moment diagram.
10

Figure 6: Bending Moment Diagram
Source: (Author, 2019)
(d):
The given details are:
Length of Span (LOS) = 14 meters
Uniform Load = 20 KN per meter
With the data the following diagram can be drawn;
Figure 7: Uniform Load combined with a point load at the centre
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(i) The calculation for the Reaction Force acting on the span.
For maintaining the vertical equilibrium, the sum of all forces acting upward should be equal
to the sum of all forces acting downwards. As the span is in equilibrium the reaction forces
can be identified.
i.e. RA + RC = 150 + (20*14) = 430KN-----------------------(i)
Also, as the span is stable the sum of all the moment acting on an object is equal to 0 (zero).
At point A,
RC x 14 = Load at point B x perpendicular distance
= 150 x 7+ (20*14*7) = 1050+1960
RC = (1050+1960)/14 = 215 KN
Now, taking point C
RA + RC = 430
RA = 430- RC = 215 KN
(ii) Calculation for Shear Forces:
At point A;
Shear Force acting at A = - ((1050+1960)/14)
= - 215 KN
Now, at point B:
Shear Force at B = 215- Acting Force (150) – (20*7)
Shear Force at B = 215-150-140 = -75KN
Now, at point C;
Shear Force acting at C = ((1050+1960)/14)
= 215 KN
(iii) Calculation of Bending Moment from the above diagram:
Take point A;
Bending Moment on point A = 0 (Zero)
Take point C;
Bending Moment on C = 0 (Zero)
Now at point B;
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