assignment course work bridge design

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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

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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

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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.

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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

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Fig. 1.6

Fig. 1.5

Fig. 1.4

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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

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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

Shea

r Fo

rce

at "

32

m"

(kN

)

Possition of the Lorry (m)

Shear Force at 32m away from "A"

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Reactions obtained from ISSD software.

Fig. 2.1

Fig. 2.2

Fig. 1.1

Fig. 2.3

Fig. 1.1

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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

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Note:-



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

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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 07.5

15 150

-31.875

-60 -58.125

0

129.375

0

-100

-50

0

50

100

150

0 10 20 30 40 50 60

Rea

ctio

ns

at s

up

po

rts

"E"

(kN

)


Maximum Uplift Reaction at "E" (kN)

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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

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Note:-



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

Ben

din

g M

om

ent

at "

C"

(kN

m)


Maximum Hogging Bending moment at "C" (kNm)

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0

-112.499

0

506.249

149.999

18.7490 0 0 0 0 0 0

-200

-100

0

100

200

300

400

500

600

0 10 20 30 40 50 60

Ben

din

g M

om

ent

at "

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)

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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

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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

Max

imu

m B

end

ing

Mo

men

t (k

Nm

)

position of the lorry (m)

Maximum Sagging Bending moment (kNm)

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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.

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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

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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

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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

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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

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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

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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]

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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 : ***************'

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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 : ***************'

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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

assignment course work bridge design

Engineering