This document provides the details of an assignment to analyze a 48m concrete beam bridge with 5 spans using structural analysis software. It includes: 1) A description of the bridge geometry and assumptions made in the model. 2) 5 questions to calculate shear forces, reactions, and bending moments at various points on the bridge as a 30-tonne lorry moves across it. 3) The results of the calculations from the software in table and graph form, identifying the maximum values requested in each question. 4) Appendices with sample output from the software when the lorry is at a specific position.

1. 1
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CONTENT
ASSIGNMENT BREAF 02
QUESTION 03
CALCULATION 04
QUESTION 01 04
QUESTION 02 08
QUESTION 03 12
QUESTION 04 16
QUESTION 05 17
RESULTS 19
APPENDICES 20

2. 2
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ASSIGNMENT BREAF
The bridge shown above is a 48m long concrete beam bridge across the southern expressway, Sri
Lanka. The first interior support (B) is at 8m from the left embankment (A) while 2nd
and 3rd
interior
supports are at 24m and 40m from A respectively. It may be modelled as a continuous beam.
Assumptions
1- The bridge has one lane of width.
2- Support “A” on the left hand side is a pin support, while support “E” on the right hand side is a
roller.
3- All interior supports are pinned supports, but allows for the continuous beam action between the
two spans.
4- The geometrical cross-section of the bridge is the same from A to E.
5- The materials’ properties of the bridge is the same from A to E.
6- A lorry of 30 tonnes drives from A to E.
7- Assume that the lorry moves as a point load on the bridge.
8- Ignore the effect of self-weight of the bridge for the purpose of establishing the following

3. 3
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QUESTIONS:
Use the ISSD software of structural analysis to establish the following:
1- Consider point “P” which is at 32m from A. Create the following table:
Position of lorry from A (m) Shear force at P (kN)
0
4
8
12
16
20
24
28
32
36
40
44
48
Plot a graph showing the position of the lorry on the x-axis and the shear force at P on the y-axis.
From the graph find the position of the lorry from point A which creates the maximum shearing force
at P.
Comment on the difference between this graph and the shear force diagram.
Follow the same approach as point 1 above to establish the following:
2- The position of the lorry measured from support “A” which will create the maximum uplift reaction
at E. Find the value of this reaction.
3- The position of the lorry measured from support “A” which will create the maximum hogging
bending moment at C. Find the value of this moment.
4- The position of the lorry measured from support “A” which will create the maximum sagging
bending moment at point “Q” which is at 12m from A.
Find the value of this moment.
5- Find the maximum value of sagging bending moment which may develop on the bridge as the
lorry moves from A to E. Find the position of the lorry from support A which will generate this
maximum value.

4. 4
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CALCULATIONS:
Question 01: ISSD RESULTS SHEETS
Shear Force Diagrams obtained from ISSD software.
Fig. 1.3
Fig. 1.4
Fig. 1.2
Fig. 1.4
Fig. 1.1
Fig. 1.4

5. 5
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Fig. 1.6
Fig. 1.5
Fig. 1.4

6. 6
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Note:-
We cannot be obtained a result sheets, when applying the point loads at supports.
Because the point load at the support is act as an axial load for the support & Reaction
also in the same line of axial load. Therefore there is no Shear forces at the other supports
& spans, when load applying on a support.
Fig. 1.8
Fig. 1.7

7. 7
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Results table 01:
Position of lorry from "A" (m)
Shear Force at "P"
(kN)
0 0
4 3.75
8 0
12 15
16 30
20 29.99
24 0
28 68.437
32 157.5
36 57.187
40 0
44 13.125
48 0
Graph of table 01:
Graph 01
Comment: According to the results of ISSD in 4m intervals we obtain the Table 01.As shown in
the graph 01. There should be a maximum Shear force value in magnitude. From ISSD we can
obtain the results in positive & negative values. But when we design the element, we consider the
maximum value in magnitude. So the maximum Shear force at 32m is 157.5 kN in magnitude.
(According to the ISSD)
0 3.75 0
15
30 29.99
0
68.437
157.5
57.187
0
13.125
0
-20
0
20
40
60
80
100
120
140
160
180
0 10 20 30 40 50 60
ShearForceat"32m"(kN)
Possition of the Lorry (m)
Shear Force at 32m away from "A"

8. 8
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Question 02: ISSD RESULTS SHEETS
Reactions obtained from ISSD software.
Fig. 2.1
Fig. 2.2
Fig. 1.1
Fig. 2.3
Fig. 1.1

9. 9
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Fig. 2.4
Fig. 1.3
Fig. 1.1
Fig. 1.1
Fig. 2.5
Fig. 1.3
Fig. 1.1
Fig. 1.1
Fig. 2.6
Fig. 1.3
Fig. 1.1
Fig. 1.1

10. 10
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Note:-
We cannot be obtained a result sheets, when applying the point loads at supports.
Because the point load at the support is act as an axial load for the support & Reaction
also in the same line of axial load. Therefore there is no Reactions at the other supports,
when load applying on a support.
Fig. 2.7
Fig. 1.3
Fig. 1.1
Fig. 1.1
Fig. 2.8
Fig. 1.3
Fig. 1.1
Fig. 1.1

11. 11
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Results table 02:
Position of lorry from "A" (m)
Maximum Uplift
Reaction at "E" (kN)
0 0
4 -1.875
8 0
12 7.5
16 15
20 15
24 0
28 -31.875
32 -60
36 -58.125
40 0
44 129.375
48 0
Graph of table 02:
Graph 02 (Reaction Positive & Uplift reaction Negative)
Comment: According to the results of ISSD in 4m intervals we obtain the Table 02.As shown in
the graph 02 there should be a maximum uplift reaction. From ISSD we can obtain the results in
1m intervals. So the maximum uplift reaction will be occurred at 34m & its magnitude is
64.453kN (according to the ISSD).
0 -1.875 0
7.5
15 15
0
-31.875
-60 -58.125
0
129.375
0
-100
-50
0
50
100
150
0 10 20 30 40 50 60
Reactionsatsupports"E"(kN)
Possition of the Lorry (m)
Maximum Uplift Reaction at "E" (kN)

12. 12
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Question 03: ISSD RESULTS SHEETS
Bending Moments obtained from ISSD software.
Fig. 3.1
Fig. 1.3
Fig. 1.1
Fig. 1.1
Fig. 3.2
Fig. 1.3
Fig. 1.1
Fig. 1.1
Fig. 3.3
Fig. 1.3
Fig. 1.1
Fig. 1.1

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Fig. 3.4
Fig. 1.3
Fig. 1.1
Fig. 1.1
Fig. 3.5
Fig. 1.3
Fig. 1.1
Fig. 1.1
Fig. 3.6
Fig. 1.3
Fig. 1.1
Fig. 1.1

14. 14
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Note:-
We cannot be obtained a result sheets, when applying the point loads at supports.
Because the point load at the support is act as an axial load for the support & Reaction
also in the same line of axial load. Therefore there is no Bending moments at the other
supports & mid spans, when load applying on a support.
Fig. 3.7
Fig. 1.3
Fig. 1.1
Fig. 1.1
Fig. 3.8
Fig. 1.3
Fig. 1.1
Fig. 1.1

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Results table 03:
Position of lorry from "A" (m)
Maximum Hogging
Bending moment at
"C" (kNm)
0 0
4 44.99
8 0
12 -179.99
16 -359.99
20 -360
24 0
28 -359.99
32 -359.99
36 -179.99
40 0
44 44.99
48 0
Graph of table 03:
Graph 03 (Sagging Positive & Hogging Negative)
Comment: According to the results of ISSD in 4m intervals we obtain the Table 03. As shown in
the graph 03 there should be a maximum Hogging Bending moment. From ISSD we can obtain
the results in 1m intervals. So the maximum Hogging Bending moment will be occurred at
18m & 30m. Its magnitude is 393.750kNm (according to the ISSD).
0
44.99
0
-179.99
-359.99
-360
0
-359.99
-359.99
-179.99
0
44.99
0
-450
-400
-350
-300
-250
-200
-150
-100
-50
0
50
100
0 10 20 30 40 50 60
BendingMomentat"C"(kNm)
Possition of the Lorry (m)
Maximum Hogging Bending moment at "C" (kNm)

16. 16
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0
-112.499
0
506.249
149.999
18.749
0 0 0 0 0 0 0
-200
-100
0
100
200
300
400
500
600
0 10 20 30 40 50 60
BendingMomentat"Q"
position of the lorry (m)
Maximum Sagging Bending moment at "Q" (kNm)
Question 04:
Bending Moments obtained from ISSD software are as same as the question 03 in fig. 3.1, fig.
3.2, fig. 3.3, fig. 3.4, fig. 3.5, fig. 3.6, fig. 3.7, fig. 3.8.
According to the results;
Results table 04:
Position of lorry from “A” (m)
Maximum Sagging
Bending moment at
“Q” (kNm)
0 0
4 -112.499
8 0
12 506.249
16 149.999
20 18.749
24 0
28 0
32 0
36 0
40 0
44 0
48 0
Graph of table 04:
Graph 04 (Sagging Positive & Hogging Negative)

17. 17
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Question 05:
Bending Moments obtained from ISSD software are as same as the question 03 in fig. 3.1, fig.
3.2, fig. 3.3, fig. 3.4, fig. 3.5, fig. 3.6, fig. 3.7, fig. 3.8. According to the results the maximum
bending moment will occurred when the lorry is at 16m away from “A” & 32m away from “A”.
The Bending Moment Value is 779.99 kNm.
Fig 5.1
Fig 5.2

18. 18
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Results table 05:
Position of lorry from "A" (m)
Maximum Sagging
Bending moment
(kNm)
0 0
4 517.499
8 0
12 506.249
16 779.999
20 566.249
24 0
28 566.25
32 779.999
36 506.249
40 0
44 517.499
48 0
Graph of table 05:
Graph 05 (Sagging Positive)
0
517.499
0
506.249
779.999
566.249
0
566.25
779.999
506.249
0
517.499
0
0
100
200
300
400
500
600
700
800
900
0 10 20 30 40 50 60
MaximumBendingMoment(kNm)
position of the lorry (m)
Maximum Sagging Bending moment (kNm)

19. 19
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Results: Obtained from ISSD software in 4m intervals.
Question 01:
The position of the lorry from point “A” Which creates the maximum Shear force at “P” (“P” is 32m
away from point “A”)
Answer: When the lorry is at 32m from “A” creates the maximum Shear force & value is 157.5 kN.
Question 02:
The position of the lorry from point “A” Which creates the maximum Uplift Reaction at “E” (“E” is
48m away from point “A”)
Answer: When the lorry is at 32m from “A” creates the maximum Uplift Reaction & value is 60 kN.
Question 03:
The position of the lorry from point “A” Which creates the maximum Hogging Bending moment at
“C” (“C” is 24m away from point “A”)
Answer: When the lorry is at 16m, 20m, 28m & 32m from “A” creates the maximum Hogging
Bending moment & value is 360 kNm.
Question 04:
The position of the lorry from point “A” Which creates the maximum Sagging Bending moment at
“Q” (“Q” is 12m away from point “A”)
Answer: When the lorry is at 12m from “A” creates the maximum Sagging Bending moment &
value is 506.249 kNm.
Question 05:
The position of the lorry from point “A” Which creates the maximum Sagging Bending moment of
the bridge
Answer: When the lorry is at 16m, 32m from “A” creates the maximum Sagging Bending moment
& value is 779.999 kNm.

20. 20
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Appendices: 1) Lorry is at 4m from “A”
*************** NODE DISPLACEMENTS : ***************'
X [mm] Y [mm] Ang. [rad]
Node 1 : 0 0 62.72
Node 2 : 0 0 -48.64
Node 3 : 0 0 12.8
Node 4 : 0 0 -2.56
Node 5 : 0 0 1.28
Node 6 : 0 162559.975 -3.52
*************** MEMBER FORCES : ***************'
Mi [kNm] Mj [KNm] Ti [kN] Tj [kN] N [kN]
Member 1 (i= 2 , j= 3) : -164.999984 -44.999992 13.125 -44.999992 0
Member 2 (i= 3 , j= 4) : 44.999996 14.999998 -3.75 14.999998 0
Member 3 (i= 4 , j= 5) : -14.999999 0 1.875 0 0
Member 4 (i= 1 , j= 6) : 0 -517.499904 129.375 -517.499904 0
Member 5 (i= 2 , j= 6) : 165 517.499904 -170.625 517.499904 0
*************** VALUES OF THE MOMENTS AT MIDDLE DISTANCE FOR MEMBERS WITH UNIFORM LOADS : ***************'
Mi(x=0) [kNm] M(L/5) [KNm] M(2L/5) [KNm] M(3L/5) [KNm] M(4L/5) [KNm] Mj(x=L) [KNm] Mmin [KNm] M=0 for x=
[mm]
*************** REACTIONS : ***************'
Rhor [kN] Rvert [kN] RAng. [kNm]
Node 1 : 0 129.375 0
Node 2 : 0 183.75 0
Node 3 : 0 -16.875 0
Node 4 : 0 5.625 0

21. 21
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Node 5 : 0 -1.875 0
2) Lorry is at 12m from “A”
*************** NODE DISPLACEMENTS : ***************'
X [mm] Y [mm] Ang. [rad]
Node 1 : 0 0 -39.68
Node 2 : 0 0 79.36
Node 3 : 0 0 -51.2
Node 4 : 0 0 10.24
Node 5 : 0 0 -5.12
Node 6 : 0 389759.941 74.08
*************** MEMBER FORCES : ***************'
Mi [kNm] Mj [KNm] Ti [kN] Tj [kN] N [kN]
Member 1 (i= 1 , j= 2) : 0 464.999904 -58.125 464.999904 0
Member 2 (i= 3 , j= 4) : -179.999968 -59.999992 15 -59.999992 0
Member 3 (i= 4 , j= 5) : 59.999984 0 -7.5 0 0
Member 4 (i= 2 , j= 6) : -464.999808 -506.249728 242.812 -506.249728 0
Member 5 (i= 3 , j= 6) : 179.999952 506.249856 -57.187 506.249856 0
*************** VALUES OF THE MOMENTS AT MIDDLE DISTANCE FOR MEMBERS WITH UNIFORM LOADS : ***************'
Mi(x=0) [kNm] M(L/5) [KNm] M(2L/5) [KNm] M(3L/5) [KNm] M(4L/5) [KNm] Mj(x=L) [KNm] Mmin [KNm] M=0 for x=
[mm]
*************** REACTIONS : ***************'
Rhor [kN] Rvert [kN] RAng. [kNm]
Node 1 : 0 -58.125 0
Node 2 : 0 300.937 0
Node 3 : 0 72.187 0

22. 22
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Node 4 : 0 -22.5 0
Node 5 : 0 7.5 0
3) Lorry is at 16m from “A”
*************** NODE DISPLACEMENTS : ***************'
X [mm] Y [mm] Ang. [rad]
Node 1 : 0 0 -40.96
Node 2 : 0 0 81.92
Node 3 : 0 0 -102.4
Node 4 : 0 0 20.48
Node 5 : 0 0 -10.24
Node 6 : 0 778239.93 5.12
*************** MEMBER FORCES : ***************'
Mi [kNm] Mj [KNm] Ti [kN] Tj [kN] N [kN]
Member 1 (i= 1 , j= 2) : 0 479.999936 -60 479.999936 0
Member 2 (i= 3 , j= 4) : -359.999968 -119.999992 30 -119.999992 0
Member 3 (i= 4 , j= 5) : 119.999984 0 -15 0 0
Member 4 (i= 2 , j= 6) : -480 -779.999936 157.5 -779.999936 0
Member 5 (i= 3 , j= 6) : 359.999936 779.999872 -142.5 779.999872 0
*************** VALUES OF THE MOMENTS AT MIDDLE DISTANCE FOR MEMBERS WITH UNIFORM LOADS : ***************'
Mi(x=0) [kNm] M(L/5) [KNm] M(2L/5) [KNm] M(3L/5) [KNm] M(4L/5) [KNm] Mj(x=L) [KNm] Mmin [KNm] M=0 for x=
[mm]
*************** REACTIONS : ***************'
Rhor [kN] Rvert [kN] RAng. [kNm]
Node 1 : 0 -60 0
Node 2 : 0 217.5 0
Node 3 : 0 172.5 0

23. 23
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Node 4 : 0 -45 0
Node 5 : 0 15 0
4) Lorry is at 20m from “A”
*************** NODE DISPLACEMENTS : ***************'
X [mm] Y [mm] Ang. [rad]
Node 1 : 0 0 -21.76
Node 2 : 0 0 43.52
Node 3 : 0 0 -102.4
Node 4 : 0 0 20.48
Node 5 : 0 0 -10.24
Node 6 : 0 435839.925 -76
*************** MEMBER FORCES : ***************'
Mi [kNm] Mj [KNm] Ti [kN] Tj [kN] N [kN]
Member 1 (i= 1 , j= 2) : 0 254.999952 -31.875 254.999952 0
Member 2 (i= 3 , j= 4) : -359.999904 -119.999968 30 -119.999968 0
Member 3 (i= 4 , j= 5) : 119.999968 0 -15 0 0
Member 4 (i= 2 , j= 6) : -254.999968 -566.24992 68.437 -566.24992 0
Member 5 (i= 3 , j= 6) : 360 566.249856 -231.562 566.249856 0
*************** VALUES OF THE MOMENTS AT MIDDLE DISTANCE FOR MEMBERS WITH UNIFORM LOADS : ***************'
Mi(x=0) [kNm] M(L/5) [KNm] M(2L/5) [KNm] M(3L/5) [KNm] M(4L/5) [KNm] Mj(x=L) [KNm] Mmin [KNm] M=0 for x=
[mm]
*************** REACTIONS : ***************'
Rhor [kN] Rvert [kN] RAng. [kNm]
Node 1 : 0 -31.875 0
Node 2 : 0 100.313 0

24. 24
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Node 3 : 0 261.562 0
Node 4 : 0 -45 0
Node 5 : 0 15 0
5) Lorry is at 28m from “A”
*************** NODE DISPLACEMENTS : ***************'
X [mm] Y [mm] Ang. [rad]
Node 1 : 0 0 10.24
Node 2 : 0 0 -20.48
Node 3 : 0 0 102.4
Node 4 : 0 0 -43.52
Node 5 : 0 0 21.76
Node 6 : 0 435839.925 76
*************** MEMBER FORCES : ***************'
Mi [kNm] Mj [KNm] Ti [kN] Tj [kN] N [kN]
Member 1 (i= 1 , j= 2) : 0 -119.999968 15 -119.999968 0
Member 2 (i= 2 , j= 3) : 119.999968 359.999904 -30 359.999904 0
Member 3 (i= 4 , j= 5) : -254.999952 0 31.875 0 0
Member 4 (i= 3 , j= 6) : -360 -566.249856 231.562 -566.249856 0
Member 5 (i= 4 , j= 6) : 254.999968 566.24992 -68.437 566.24992 0
*************** VALUES OF THE MOMENTS AT MIDDLE DISTANCE FOR MEMBERS WITH UNIFORM LOADS : ***************'
Mi(x=0) [kNm] M(L/5) [KNm] M(2L/5) [KNm] M(3L/5) [KNm] M(4L/5) [KNm] Mj(x=L) [KNm] Mmin [KNm] M=0 for x=
[mm]
*************** REACTIONS : ***************'
Rhor [kN] Rvert [kN] RAng. [kNm]
Node 1 : 0 15 0

25. 25
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Node 2 : 0 -45 0
Node 3 : 0 261.562 0
Node 4 : 0 100.313 0
Node 5 : 0 -31.875 0
6) Lorry is at 32m from “A”
*************** NODE DISPLACEMENTS : ***************'
X [mm] Y [mm] Ang. [rad]
Node 1 : 0 0 10.24
Node 2 : 0 0 -20.48
Node 3 : 0 0 102.4
Node 4 : 0 0 -81.92
Node 5 : 0 0 40.96
Node 6 : 0 778239.93 -5.12
*************** MEMBER FORCES : ***************'
Mi [kNm] Mj [KNm] Ti [kN] Tj [kN] N [kN]
Member 1 (i= 1 , j= 2) : 0 -119.999984 15 -119.999984 0
Member 2 (i= 2 , j= 3) : 119.999992 359.999968 -30 359.999968 0
Member 3 (i= 4 , j= 5) : -479.999936 0 60 0 0
Member 4 (i= 3 , j= 6) : -359.999936 -779.999872 142.5 -779.999872 0
Member 5 (i= 4 , j= 6) : 480 779.999936 -157.5 779.999936 0
*************** VALUES OF THE MOMENTS AT MIDDLE DISTANCE FOR MEMBERS WITH UNIFORM LOADS : ***************'
Mi(x=0) [kNm] M(L/5) [KNm] M(2L/5) [KNm] M(3L/5) [KNm] M(4L/5) [KNm] Mj(x=L) [KNm] Mmin [KNm] M=0 for x=
[mm]
*************** REACTIONS : ***************'
Rhor [kN] Rvert [kN] RAng. [kNm]

26. 26
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Node 1 : 0 15 0
Node 2 : 0 -45 0
Node 3 : 0 172.5 0
Node 4 : 0 217.5 0
Node 5 : 0 -60 0
7) Lorry is at 36m from “A”
*************** NODE DISPLACEMENTS : ***************'
X [mm] Y [mm] Ang. [rad]
Node 1 : 0 0 5.12
Node 2 : 0 0 -10.24
Node 3 : 0 0 51.2
Node 4 : 0 0 -79.36
Node 5 : 0 0 39.68
Node 6 : 0 389759.941 -74.08
*************** MEMBER FORCES : ***************'
Mi [kNm] Mj [KNm] Ti [kN] Tj [kN] N [kN]
Member 1 (i= 1 , j= 2) : 0 -59.999984 7.5 -59.999984 0
Member 2 (i= 2 , j= 3) : 59.999992 179.999968 -15 179.999968 0
Member 3 (i= 4 , j= 5) : -464.999904 0 58.125 0 0
Member 4 (i= 3 , j= 6) : -179.999952 -506.249856 57.187 -506.249856 0
Member 5 (i= 4 , j= 6) : 464.999808 506.249728 -242.812 506.249728 0
*************** VALUES OF THE MOMENTS AT MIDDLE DISTANCE FOR MEMBERS WITH UNIFORM LOADS : ***************'
Mi(x=0) [kNm] M(L/5) [KNm] M(2L/5) [KNm] M(3L/5) [KNm] M(4L/5) [KNm] Mj(x=L) [KNm] Mmin [KNm] M=0 for x=
[mm]
*************** REACTIONS : ***************'

27. 27
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Rhor [kN] Rvert [kN] RAng. [kNm]
Node 1 : 0 7.5 0
Node 2 : 0 -22.5 0
Node 3 : 0 72.187 0
Node 4 : 0 300.937 0
Node 5 : 0 -58.125 0
8) Lorry is at 44m from “A”
*************** NODE DISPLACEMENTS : ***************'
X [mm] Y [mm] Ang. [rad]
Node 1 : 0 0 -1.28
Node 2 : 0 0 2.56
Node 3 : 0 0 -12.8
Node 4 : 0 0 48.64
Node 5 : 0 0 -62.72
Node 6 : 0 162559.975 3.52
*************** MEMBER FORCES : ***************'
Mi [kNm] Mj [KNm] Ti [kN] Tj [kN] N [kN]
Member 1 (i= 1 , j= 2) : 0 14.999998 -1.875 14.999998 0
Member 2 (i= 2 , j= 3) : -14.999997 -44.999992 3.75 -44.999992 0
Member 3 (i= 3 , j= 4) : 44.999988 164.999968 -13.125 164.999968 0
Member 4 (i= 4 , j= 6) : -164.999936 -517.49984 170.625 -517.49984 0
Member 5 (i= 5 , j= 6) : 0 517.499872 -129.375 517.499872 0
*************** VALUES OF THE MOMENTS AT MIDDLE DISTANCE FOR MEMBERS WITH UNIFORM LOADS : ***************'
Mi(x=0) [kNm] M(L/5) [KNm] M(2L/5) [KNm] M(3L/5) [KNm] M(4L/5) [KNm] Mj(x=L) [KNm] Mmin [KNm] M=0 for x=
[mm]
*************** REACTIONS : ***************'

28. 28
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Rhor [kN] Rvert [kN] RAng. [kNm]
Node 1 : 0 -1.875 0
Node 2 : 0 5.625 0
Node 3 : 0 -16.875 0
Node 4 : 0 183.75 0
Node 5 : 0 129.375 0