Table of Contents 1.0 Design 6

1.1 Introduction 6

2.0 Calculations 7 2.1 Arch Frame 7

2.1.0 Funicular Geometry of Long Span Arch Frame 8 2.1.1 Funicular Geometry of Short Span Arch Frame 17 2.1.2 Maximum Compression in the Short Span Arch 22 2.1.3 Buckling Check at Supports of Short Span Arch 23 2.1.4 Selecting a member size for the Short Span Bridge 24 2.1.5 Selecting a cable size for the Short Span Bridge 25

2.2 Beam Calculations 26 2.2.0 Normal, Shear, and Bending Moment Diagrams 27 2.2.1 Principle of Virtual Work: Determining MidSpan Deflection of Beam 28 2.2.2 Principle of Virtual Work: Determining Rotation at One End of a Beam 32 2.2.3 Stresses MidSpan of the Beam 35 2.2.4 Application of Mohr’s Circle 38 2.2.4 Failure of Materials Application 39

2.0 Pier Calculations 40 2.3.0 Determining the Pier Cross-Sectional Dimensions 41 2.3.1 Buckling Check on Established Pier Dimensions 45 2.3.2 Shear and Bending Stress on Pier 46

3.0 Conclusion 53

4.0 References 55

5.0 Appendix 56

2

List of Tables Table 1: Material properties used in calculations 7 Table 2: Properties of Selected Round HSS Member from Steel Institute of North America 24 Table 3: Properties of Selected Square HSS Member from Steel Institute of North America 32

3

List of Figures Figure 1: Elevation of car and pedestrian intended uses 7 Figure 2: Load path diagram of the overall bridge 8 Figure 3: Tributary area for cable loading on long span arch 9 Figure 4: FBD of long span arch 9 Figure 5: FBD of long span arch left side 10 Figure 6: FBD of long span arch right side 11 Figure 7: FBD of equivalent cable system for long span arch 13 Figure 8: FBD of analogous beam system for long span arch (General Cable Theorem) 13 Figure 9: Analogous beam system for long span arch shear diagram 14 Figure 10: Analogous beam system for long span arch moment diagram 15 Figure 11: Final geometric dimensions of long span arch 16 Figure 12: Tributary area for cable loading on short span arch 17 Figure 13: FBD of short span arch 18 Figure 14: FBD of equivalent cable system for short span arch 19 Figure 15: FBD of analogous beam system for short span arch (General Cable Theorem) 19 Figure 16: Analogous beam system for short span arch shear diagram 20 Figure 17: Analogous beam system for short span arch moment diagram 21 Figure 18: Final geometric dimensions of short span arch 22 Figure 19: Plan of the edge girder supporting the cable deck to be analyzed 26 Figure 20: P-system FBD elevation of the edge girder supporting the cable deck. 27 Figure 21: P-system of edge girder normal force diagram 27 Figure 22: P-system of edge girder shear force diagram 28 Figure 23: P-system of edge girder moment diagram 28 Figure 24: Q-system FBD of the edge girder supporting the cable deck 29 Figure 25: Q-system of edge girder shear force diagram 30 Figure 26: Q-system of edge girder moment diagram 30 Figure 27:Q-system FBD of the edge girder for solving rotation at right end 32 Figure 28: Q-system of edge girder shear force diagram (for rotation) 33 Figure 29: Q-system of edge girder moment diagram (for rotation) 33 Figure 30: Main deck support beam loading diagram 35 Figure 31: X-section of beam that will be analyzed for planar state of stress at midspan 36 Figure 32: Mohr’s circle for deck girder beam 39 Figure 33: Tresca’s criterion analysis of girder beam 40 Figure 34: Elevation of the column with loads from the arches transferring to the middle pier 41 Figure 35:Cross-section dimensions of the column 42 Figure 36: Section of the bridge deck with the pier supporting arches beyond 42

4

Figure 37: Final established cross-section dimensions of the column 45 Figure 38: Combined effect of axial and lateral load as a result of symmetry on pier 47 Figure 39: Center column FBD and cross section details 48 Figure 40: Center column normal force and shear force diagrams 49 Figure 41: Center column moment diagram 49 Figure 42: Center column Mohr’s criterion analysis for point ‘A’ 51 Figure 43: Center column shear analysis at point ‘B’ 51 Figure 44:Center column Mohr’s criterion analysis for point ‘B’ 53

5

1.0 Design

The following section contains the objective, scope, and background information about the

design intention and analysis of the bridge.

1.1 Introduction The objective of this project is to design and propose the reconstruction of the pedestrian bridge

located between Carl A. Pollock Hall and parking lot ‘A’ over University Avenue. The proposed

design is inspired by the beauty of funicular arches which can efficiently span large distances by

transferring loads using a system of single force members, all in compression. The bridge will

have two such arches on each side, one entirely spanning University Avenue and the other

entirely spanning Ring Road on the University campus.

One goal of the new design is to create a bridge which is seemingly gently placed above traffic.

Using the ability of the arches to span great distances, two of the existing three columns

supporting the bridge will be removed. This will clear the vision line of drivers on the roads and

create a safer area for pedestrians and cars to coexist since this is a busy location. Only one

vertical support at the center span of the bridge will remain, supporting the arches at their base

intersection point.

A cable system will suspend the deck of the bridge 0.8m above the arch base support to provide

an aesthetic that simulates a levitation. By using short members with connections, the members

used in the design can be small in cross-section and reduce the length of material required to

build the bridge. As shown in figure 1, this design not only creates a safer driving experience for

the heavy traffic that graces University Ave, but is also pleasing to the eye.

6

Figure 1: Elevation of car and pedestrian intended uses

2.0 Calculations For the following calculation package, the dead load of the structure will be neglected. Since the deck

must support people and snow, it will be assumed that it must support 10.2kPa. The cable is to be made of

high strength steel while all other steel members are to be made from regular strength steel. The common

material properties will be assumed as follows:

Table 1: Material properties used in calculations

Note that the tensile strength of concrete was retrieved from a table by the Engineering Toolbox

and the table can be found in the appendix [1].

2.1 Arch Frame

The center support of the existing bridge is located slightly off of the center of the span. In order

to keep the roads clear, the new support will be located at the same spot, resulting in a slight

difference in span of 0.1m for the two bridges. This will not be a difference identifiable visually,

but it will still require both arches to be calculated since the longer spanning arch will have

greater thrust for the same predefined height. Throughout this report the longer spanning arch

7

Material Yield Strength (MPa) Young’s Modulus (GPa)

Steel 350 200

High Strength Steel (Cable) 600

Concrete (Compression) 50 17

Concrete (Tension) 5

will be referred to as the ‘long span arch’ and the shorter spanning arch will be referred to as the

‘short span arch’.

The main load of the bridge is the deck which is designed to support 10.2kPa which includes the

weight of people and snow since this deck is not covered. The load of the deck is supported by

cables which are supported by the arches from above. The arches transfer the load to the supports

at the ends and center of the bridge which transfers the loads down to the ground. The structures

at the ends of the bridge act as buttresses, dispersing the lateral load the arches exert. The

existing stair tower and CPH attaching structure is to be designed by others and therefore, the

extreme ends of the bridge will be considered simply as pin supports (figure 2). Figure 2

illustrates the load path through the bridge:

Figure 2: Load path diagram of the overall bridge

2.1.0 Funicular Geometry of Long Span Arch Frame

The first arch geometry to be calculated is that of the longest spanning arch, which is located on

the West side of the bridge. It spans 32.6m with cable connections to the deck every 2.508m. The

desired maximum height for the arch is 5.8m and this parameter will be used to help calculate the

funicular shape.

8

9

Tension from Tributary Area

Arch overall FBD:

Figure 4: FBD of long span arch

The arch frame shown in figure 4 is supported by two pin connections on either end. The

overall structure is indeterminate, therefore in order to solve for the reactions, the overall FBD

will be sectioned into two pieces about the hinge.

10.2kPa (1.1m )(2.508m) = T max, long For human and snow loads, 10.2kPa

must be able to be supported. The

tributary area on deck is taken as

shown in the diagram.

Figure 3: Tributary area for cable loading

28.15 kN = T max, long

10

Arch FBD Left Side

Figure 5: FBD of long span arch left side

∑MI+↺= 0 = −28.15kN(2.508m) −28.15kN(2*2.508m) −28.15kN(3*2.508m)

−28.15kN(4*2.508m)−28.15kN(5*2.508m)−28.15kN(6*2.508m)+

Qy(6*2.508 + (½) 2.508m)−Qx(5.8m)

5.8Qx+1482.60 = 16.302 Qy (1)

11

Arch FBD Right Side

Set equation (1) = equation (2)

Substitute Qx into equation (1):

Figure 6: FBD of long span arch right side ∑MJ+↺= 0 = 28.15kN(2.508m) +28.15kN(2*2.508m)

+28.15kN(3*2.508m)

+28.15kN(4*2.508m)+28.15kN(5*2.508m)+28.15kN(6*2.5

08m)+ Qy(6*2.508 + (½) 2.508m)+Qx(5.8m)

−5.8Qx−1482.60 = 16.302 Qy (2)

5.8Qx+1482.60 = −5.8Qx−1482.60

Qx= H = −255.62kN

5.8(−255.62kN)+1482.60 = 16.302Qy Qy = 0

12

Arch FBD Left Side - Reactions

Arch FBD Right Side - Reactions

↑∑FY+= 0 = −28.15kN(7) + IY + QY

197.05kN − 0= IY

197.05kN = IY

→FX+= 0 = IX + QX

− IX = QX

−(−255.62kN) = IX

255.62kN = IX

↑∑FY+= 0 = −28.15kN(7) + JY − QY

197.05kN − 0= JY

197.05kN = JY

→FX+= 0 = −JX − QX

−(−255.62kN) = JX

255.62kN = JX

13

Apply General Cable Theorem

The arch frame is intended to be funicular. Therefore, the geometry based on the equivalent

cable structure needs to be determined (figure 7):

Figure 7: FBD of equivalent cable system for long span arch Analogous Beam

Figure 8: FBD of analogous beam system for long span arch (General Cable Theorem)

The reactions for the analogous beam are the following:

∑MI+↺= 0 = −28.15kN(2.508m) −28.15kN(2*2.508m) −28.15kN(3*2.508m) −28.15kN(4*2.508m)

−28.15kN(5*2.508m) −28.15kN(6*2.508m) −28.15kN(7*2.508m)

−28.15kN(8*2.508m)−28.15kN(9*2.508m) −28.15kN(10*2.508m) −28.15kN(11*2.508m)

−28.15kN(12*2.508m)+ Jy, beam(13*2.508m)

5506.811 = Jy, beam (32.604)

168.90kN = Jy, beam

14

Likewise, due to symmetry:

Check:

Shear and Moment Diagram of Beam

The shear and moment diagrams are essential to calculating the geometric height of the frame

at various points. Based on the point loads on the beam, the resulting shear diagram was

derived:

Figure 9: Analogous beam system for long span arch shear diagram

The areas, A1-A6, were then calculated under the shear diagram to obtain the moment:

168.90kN = IY, beam

↑FY+= 0 = −28.15kN(12) + 168.90 + IY, beam

168.90 = IY, beam Therefore, the reaction matches the

expectation.

A1 = 168.90kN(2.508m) = 423.6 kNm

A2 = 140.75kN(2.508m) = 353 kNm

A3 = 112.6kN(2.508m) = 282.4 kNm

A4 = 84.45kN(2.508m) = 211.80 kNm

A5 = 56.3kN(2.508m) = 141.20 kNm

A6 = 28.15kN(2.508m) = 70.60 kNm

15

Figure 10: Analogous beam system for long span arch moment diagram MQ = ∑ A1-A6

Check: The desired height of 5.8m at hQ is

used to evaluate the horizontal thrust H and

compare if it is the same as the result

derived earlier in the arch frame and

indeed the value of H (or Qx) is the same.

MQ = M max = 1482.60 kNm

Funicular Geometry

H = hQM Q = 5.8 m

1482.60 kNm = 255.62 kN

hK = HM K = 255.62 kN

423.6 kNm = 1.657m Check: At this step it is imperative to

check that the relative height of the node

hK exceeds 0.8m to ensure that the node is

not under or at the height of the

deck—otherwise a cable cannot be

attached to support the deck below.

Clearly, 1.657 m > 0.8m so the result is ok.

hL = HM L = 255.62 kN

776.6 kNm = 3.038m

It is important to note that the heights obtained at the nodes of the frame are referenced relative

to the support nodes I and J which are below the deck. Thus, to obtain the length of the cables,

0.8m must be deducted from each cable nodal height. Figure 11 shows the final updated

geometric dimensions of the arch relative to the supports:

Figure 11: Final geometric dimensions of long span arch

2.1.1 Funicular Geometry of Short Span Arch Frame

The second arch geometry to be calculated is that of the shortest spanning arch, which is located

on the East side of the bridge. It spans 32.5m with cable connections to the deck every 2.5m. The

16

hM = HM M = 1059 kNm

255.62 kN = 4.140m

hN = HM N = 255.62 kN

1270.8 kNm = 4.970m

hO = HM O = 1412 kNm

255.62 kN = 5.520m

hP = HM P = 255.62 kN

1482.6 kNm = 5.800m

desired maximum height for the arch is 5.8m and this parameter will be used to help calculate the

funicular shape.

17

Tension from Tributary Area

Arch overall FBD:

10.2kPa (1.1m )(2.50m) = T max, short For human and snow loads, 10.2kPa

must be able to be supported. The

tributary area on the deck is taken as

shown in the diagram.

Figure 12: Tributary area for cable loading on short span arch

28.05 kN = T max, short

18

Figure 13: FBD of short span arch

The arch frame shown in figure 13 is supported by two pins on either end. The overall

structure is indeterminate, therefore in order to solve for the reactions, the overall FBD will be

sectioned into two pieces about the hinge.

Due to symmetry, AY = BY

Arch FBD Left Side

↑∑FY+= 0 = 28.05kN(7) = AY = BY

196.35kN = AY

196.35kN = BY

∑MA+↺= 0

= −28.05kN(2.50m) −28.05kN(2*2.50m) −28.05kN(3*2.50m)

19

−28.05kN(4*2.50m) −28.05kN(5*2.50m)−28.05kN(6*2.50m) +H(5.8m)

H = 253.9 kN

Apply General Cable Theorem

The arch frame is intended to be funicular. Therefore, the geometry based on the equivalent

cable structure needs to be determined:

Figure 14: FBD of equivalent cable system for short span arch Analogous Beam

Figure 15: FBD of analogous beam system for short span arch (General Cable Theorem)

The reactions for the analogous beam are the following:

Due to symmetry, AY = BY

↑∑FY+= 0

= −28.05kN(12)+ AY, beam(2)

336.6 = AY, beam(2)

20

Shear and Moment Diagram of Beam

The shear and moment diagrams are essential to calculating the geometric height of the frame

at various points. Based on the point loads on the beam, the resulting shear diagram was

derived:

Figure 16: Analogous beam system for short span arch shear diagram

The areas, A1-A6, were then calculated under the shear diagram to obtain the moment:

168.30kN = Ay, beam

A1 = 168.30kN(2.50m) = 420.75 kNm

A2 = 140.25kN(2.50m) = 350.625 kNm

A3 = 112.2kN(2.50m) = 280.500 kNm

A4 = 84.15kN(2.50m) = 210.375 kNm

A5 = 56.1kN(2.50m) = 140.250 kNm

A6 = 28.05kN(2.50m) = 70.125 kNm

Figure 17: Analogous beam system for short span arch moment diagram

21

MQ = ∑ A1-A6

Check: The desired height of 5.8m at hH is

checked to ensure the desired requirement

was indeed met.

MQ = M max = 1472.625 kNm Funicular Geometry

hH = HM H = 253.9 kN

1472.625 kNm = 5.8m

hC = HM C = 253.9 kN

420.75 kNm = 1.657m Check: At this step it is imperative to

check that the relative height of the node

hK exceeds 0.8m to ensure that the node is

not under or at the height of the

deck—otherwise a cable cannot be

attached to support the deck below.

Clearly, 1.657 m > 0.8m so the result is ok.

hD = HM D = 253.9 kN

771.375 kNm = 3.038m

hE = HM E = 253.9 kN

1051.875 kNm = 4.143m

hF = HM F = 253.9 kN

1262.25 kNm = 4.971m

hO = HM O = 253.9 kN

1412 kNm = 5.524m

hP = HM P = 253.9 kN

1482.6 kNm = 5.800m

Once again, it is important to note that the heights obtained at the nodes of the frame are

referenced relative to the supports A and B which are 0.8m below the deck. Thus, to obtain the

length of the cables, 0.8m must be deducted from each arch nodal height. Figure 18 shows the

final updated geometric dimensions of the arch relative to the supports:

Figure 18: Final geometric dimensions of short span arch

2.1.2 Maximum Compression in the Short Span Arch

Since the thrust and funicular geometry results between the 32.5m short span arch and 32.6m

long span arch were very similar with only a 0.677% difference in H (equation 3), the remaining

calculations will be resumed for only the short spanned arch for conciseness.

22

Percent Difference = *100%Hshort span

H − Hlong span short span

Percent Difference = *100%253.9 kN255.62 kN − 253.9 kN

Percent Difference = 0.677% (3)

The location of maximum thrust occurs in member AC, the steepest member of the arch and closest member to the supports:

2.1.3 Buckling Check at Supports of Short Span Arch

Likewise, buckling will be checked at the supports of the short span arch by determining the

minimum value of moment of inertia required for the member.

23

Minimum Cross-Sectional Area:

max T = √A 2y + H

2

max T = √(168.3) 253.9) 2 + (

2

max 304.61kN T =

σy = AF

350 MPa = A mm2304.61kN (10 )* 3 N

kN

A = 870.31 mm2

Joint AC is a pin to pin connection, therefore an effective length factor of K=1 will be used.

Substituting these values into the concentric buckling equation:

Pcr = 304.61kN

Lac = = 2.999m√1.662 + 2.52

Esteel = 200*109 N/m2

Pcr = π E I2* *(K L ) *

2

I = π E2*(K L ) P cr* 2* Rearranging and solving for the

required moment of inertia.

2.1.4 Selecting a member size for the Short Span Bridge

Using the minimum required area obtained from the axial stress and the minimum required

moment of inertia derived from the critical buckling load, a member size for round HSS can be

selected for the short span bridge. The referenced sizing chart for dimensions and sections

properties can be found in figure A in the appendix.

24

I = π (200 10 )2* *

9(1 2.999m ) (304610N )* 2*

I = 1.3879* m4 = 1.3879*106 mm410−6

Given:

The moment of inertia and cross-sectional area are scanned on the sizing chart to find the

smallest member that meets or exceeds the minimum required values of both. In figure A of

the appendix, the moment of inertia is the governing value, as expected with such members.

Thus, the sectional dimensions and properties of the selected round HSS steel member are

shown in table 2.

Table 2: Properties of Selected Round HSS Member from Steel Institute of North America (figure A - Appendix 1)

Imin = 1.3879*106 mm4 = 3.33 in4 .

Amin = 870.31 mm2 = 1.349 in2

Property Imperial Size Metric Size

Outside Diameter 3.5 in 88.9 mm

2.1.5 Selecting a cable size for the Short Span Bridge

A high strength steel cable size is selected for the short span bridge based on the maximum

expected tension, Tmax and yield stress.

25

Wall thickness 0.28 in 7.11 mm

Cross-sectional Area 2.83 in2 1825.8 mm2

Moment of Inertia 3.70 in4 1.54*106 mm4

Weight per Unit Length 10.26 lb/ft 1.42 kg/m

Given:

Tmax = 28.05 kN Derived in section 2.1.1 based on the tributary area.

steel = 600 MPa σy

Cable Size from Tensile Stress

= σy AF

600 MPa = A mm228050 N

Amin = 46.75 mm2

A = r2π

r = √ π46.75 mm2

r = 3.86 mm

d = 8 mm Therefore, the minimum diameter of the steel cables

rounding up to the nearest millimeter is 8 mm for

the short span bridge.

2.2 Beam Calculations The deck of the bridge is supported by beams running the width of the deck with welded cable

connections that suspend it from the arches. The clear width of the deck is 2.2m with the beams

being slightly longer (2.4m) to allow for the connection of the cables at the ends. These main

beams are supplemented by small stringers which connect to the sides of the beams, creating a

grid. These smaller members are crucial for rigidity in lateral movement of the deck, but we will

assume they offer no support to the main support beams in the analysis.

Figure 19: Plan of the edge girder supporting the cable deck to be analyzed

2.2.0 Normal, Shear, and Bending Moment Diagrams

A free body diagram of the edge girder to be analyzed is shown in figure 20. The beam was

taken slightly longer than the deck width of 2.2m to 2.4m to allow for cable connections.

Therefore, the 25.5 kN/m distributed load ends 0.1m before either end of the beam. The

26

distributed load was derived from the worst case scenario loading of 10.2 kPa being multiplied

by the 2.5m total unit span of the deck the beam is supporting shown earlier in figure 19.

Figure 20: P-system FBD elevation of the edge girder supporting the cable deck.

27

P-system FBD Cable Tension

P-system Diagrams

Figure 21: P-system of edge girder normal force diagram

↑∑FY+= 0 = 2Tcable −2.2m *25.5 kN/m

Tcable = 28.05 kN

2.2.1 Principle of Virtual Work: Determining MidSpan Deflection of Beam

The shear diagram generated in 2.2.0 can be used to determine moment equations for Mp along

the length of the beam. Subsequently, the same procedure can be adopted for the Q-system.

28

Figure 22: P-system of edge girder shear force diagram

To obtain the moment, the area under the shear was calculated:

Figure 23: P-system of edge girder moment diagram

A1 = 0.1(28.05 kN) = 2.805 kNm

A2 = (28.05 kN) (1.1) = 15.4275 kNm21

P-system Moment Equations Beam Span (m) Moment Equation (kNm )

0 .1≤ x1 ≤ 0 Mp( = 28.05x)x1

0.1 < .2x2 ≤ 1 Mp( = 2.805 + )x2 (28.05)(x .1)21 − 0

To determine the deflection of the beam at mid-span, a1 kN load is correspondingly placed

mid-span in the Q-system (figure 24).

Figure 24: Q-system FBD of the edge girder supporting the cable deck

29

Mp( = 14.025x + 1.4025)x2

1.2 < .3x3 ≤ 2 Mp( = 18.2335 - )x3 (28.05)(x .2)21 − 1

Mp( = 35.0625 - 14.025x)x3

2.3 < 2.4x3 ≤ Mp( = 2.805 − )x4 8.05(x .3)2 − 2

Mp( = 67.32 − )x4 8.05x2

Q-system FBD Cable Tension ↑∑FY

+= 0 = 2Tcable − 1 kN

Tcable = 0.5 kN

30

Q-system Diagrams

Figure 25: Q-system of edge girder shear force diagram

Figure 26: Q-system of edge girder moment diagram

Q-system Moment Equations

Deflection Mid-Span of P-System Beam Using the equation,

a square cross section size can be calculated for the steel beam using the moment of inertia. The general, the acceptable deflection for a floor is . Thus, the max deflection at δ = L

250 midspan is:

Beam Span (m) Moment Equation (kNm )

0 .2≤ x1 ≤ 1 Mp( = .05x)x1

1.2 < .4x2 ≤ 2 Mp( = 0.6 − )x2 0.5)(x .2)( − 1

Mp( = 1.2 − 0.5x)x2

δ M dx∑

Q P = 1

EI ∫L

0M Q P

(4)

31

Substituting these values and the moment equations into equation (4),

= = 9.6 mmδmax 2502400 mm

= 9.6*10-3mδmax

E = 2*108 m2

kN

Q = 1kN

δ Q P = [ (28.05x)(0.5x) dx (14.025x .4025)(0.5x) dx (35.0625 4.025x1

EI ∫0.1

0+ ∫

1.2

0.1+ 1 + ∫

2.3

1.2− 1

(1.2-0.5x) + xd (67.32 8.05x)(1.2 .5x) dx]∫2.4

2.3− 2 − 0

I = [ (14.025x ) dx (7.0125x .70125x) dx (7.0125x 4.36125x 1

EQ δP∫0.1

0

2 + ∫1.2

0.1

2 + 0 + ∫2.3

1.2

2 − 3

+ 2.075)4 xd (14.025x 67.32x 80.784) dx ]∫2.4

2.3

2 − +

I = + { [4.675x ] (2.3375x .350625x ] 2.3375x 7.180625x1EQ δP

300.1 + [ 3 + 0 2 1.2

0.1 + [ 3 − 1 2

+ 42.075x]2.31.2 4.675x 33.66x 0.784x][ 3 − 2 + 8 2.4

2.3

I = { [4.675 ] 4.5441 5.84375 )] 34.3274 29.7891] 1EQ δP

* 10−3 + [ − ( * 10−3 + [ − +

64.6274 4.6225] } [ − 6

I = [9.08593125 kN2m3]1( 2 10 kN /m )(1 kN )(9.6 10 m)*

8 2 *−3

I = 4.7323*10-6 m4

I = 4.7323*106 mm4

or

I = 11.37 in4

2.2.2 Principle of Virtual Work: Determining Rotation at One End of a Beam

To calculate the rotation of the beam at its end, a new Q-system FBD must be generated to be

able to calculate new MQ moment equations (figure 27). The P-system and moment equations

calculated in section 2.2.0 and 2.2.1, however, can be reused.

Figure 27: Q-system FBD of the edge girder supporting the cable deck to calculate rotation at right end

32

Based on this minimum moment of inertia, we can size the beam appropriately using the table

of square HSS members and properties.

Table 3: Properties of Selected Square HSS Member from Steel Institute of North America (figure B - Appendix 1)

Property Imperial Size Metric Size

Moment of Inertia 11.9 in4 4.953*106 mm4

Nominal Size 4” x 4” x ½” 100 mm x 100 mm x 12.7 mm

Wall Thickness 0.465 in 11.811 mm

Cross-Sectional Area 6.02 in2 3883.863 mm2

Q-system FBD Cable Tension ∑MA

+↺ = 0 = By (2.4) + 1 kN

By = -0.4167 kN

33

By = Ay = -0.4167 kN

Q-system Diagrams

Figure 28: Q-system of edge girder shear force diagram (for rotation)

Figure 29: Q-system of edge girder moment diagram (for rotation)

Q-system Moment Equations

Recall P-system Moment Equations The moment equations for the P-system were derived in section 2.2.1:

Beam Span (m) Moment Equation (kNm)

0 .4≤ x1 ≤ 2 MQ( = 0.4167x)x1

Beam Span (m) Moment Equation (kNm )

34

Rotation at the End of P-System Beam Using the equation,

0 .1≤ x1 ≤ 0 Mp( = 28.05x)x1

0.1 < .2x2 ≤ 1 Mp( = 14.025x + 1.4025)x2

1.2 < .3x3 ≤ 2 Mp( = 35.0625 - 14.025x)x3

2.3 < 2.4x3 ≤ Mp( = 67.32 − )x4 8.05x2

M dx∑

Q θP = 1

EI ∫L

0M Q P

(5)

= Q θP [ (28.05x)(0.4167x) dx (14.025x .4025)(0.4167x) dx (35.06251

EI ∫0.1

0+ ∫

1.2

0.1+ 1 + ∫

2.3

1.2

x)(0.4167x) + 4.025− 1 xd (67.32 8.05x)(0.4167x) dx]∫2.4

2.3− 2

= θP + [ (11.688x ) dx (5.844x .584x) dx (14.611x .844x )1

EIQ ∫0.1

0

2 + ∫1.2

0.1

2 + 0 + ∫2.3

1.2− 5 2

(28.052x 1.688x ) dx]∫2.4

2.3− 1 2

= θP

+ { [3.896x ] 1.948x .292x ] 7.3055x .948x1EIQ

300.1 + [ 3 + 0 2 1.2

0.1 + [ 2 − 1 3 ]2.31.2

14.026x 3.896x ][ 2 − 3 2.42.3

= θP

+ { [3.896 0 ] 3.787 4.868 14.945 .1541EIQ * 1 −3

+ [ − ( * 10 )]−3 + [ − 7 ]

}26.926 26.790][ −

= θP [11.713028 kN2m3]1( 2 10 kN /m )(1 kN )(4.953 10 m)*

8 2 *−6

2.2.3 Stresses MidSpan of the Beam

The planar state of stress will be analyzed for the following deck support beam 44 mm below the

neutral axis at midspan:

Figure 30: Main deck support beam loading diagram

Figure 31: X-section of beam that will be analyzed for planar state of stress at midspan

35

= θP

0.012 rad

Therefore, the rotation at the end of the beam is 0.012 rad.

Details about the properties of this square HSS beam were discussed earlier in section 2.2.1 and

further properties of the selected beam can be found in figure B of appendix.

36

Given

At the midspan of the beam, in section 2.2.0 the internal moment and shear were found to be

18.2325 kNm and 0 kN respectively.

The shear modulus can be derived from the modulus of elasticity and poisson’s ratio:

Bending Stress at Midspan

Shear Stress

v = 0.3

E = 200 GPa

I = 4.953*106mm4

G = E2(1+v)

G = 2(1+0.3)200 GP a

G = 76.923 GPa

= σx I

Mc

= σx 4.953 10 mm*

6 4

(1.82325 10 Nmm) (44 mm)* 7

= σx 161.97 MPa Tensile stress

= σy 0

= σz 0

37

Generalized Hooke’s Law: Determining Normal and Shear Strain Assumptions:

1) A linear elastic response 2) An isotropic response

3) Assume plane stress for calculations since t = 11.811 mm (thin)

=τ It

V Q

=τ xy 0 MPa

=τ xz 0 MPa

=τ yz 0 MPa

=εx =(σ (σ ))1E x − v y + σz (161.97 =1

200 10 MP a*3 .3(0))− 0 8.0985* 10−4

=εy

=(σ (σ ))1E y − v x + σz =1

200 10 MP a*3 0 .3(161.97))( − 0 -2.4296* 10−4

=εz =(σ (σ ))1E z − v x + σz

=1

200 10 MP a*3 0 .3(161.97))( − 0 -2.4296* 10−4

=γxy = = 0γyz γzx

=γyx = = 0γzy γxz

=τ max ± √( ) )2σ +σx y 2 + (τ xy

2

=τ max 80.985 MPa± Max shear

=, σ σ1 2 2

σ +σx y ± √( ) )2σ +σx y 2 + (τ xy

2

= 2

161.97 ± √( )2161.97 2

= σ1 161.97 MPa Principle stress

= σ2 0 MPa Principle stress

2.2.4 Application of Mohr’s Circle

Using the results obtained in the previous section, 2.2.3, a graph of Mohr’s circle can be established to

conveniently show the relationship between principal stresses and maximum and minimum shear stress.

38

The particle is already in the rotation of the principal normal stresses since = 0.τ xy

τ xy = 0 MPa

Figure 32: Mohr’s circle for deck girder beam

σavg = 2σ +σx y

σavg = = 80.985 MPa2161.97 MP a

C = (80.985, 0)

R = = 80.985 MPa √( ) )2σ +σx y 2 + (τ xy

2

A = (161.97, 0)

G = (0,0)

2.2.4 Failure of Materials Application

To evaluate the failure of the beam, Tresca’s criterion will be used since steel is a ductile

material and Von Mises surface is a more liberal method.

39

Given

Using the principle stresses Trsescon’s criteria can be applied to evaluate if the material will

fail under the subjected conditions.

Tresca’s Criterion

Figure 33: Tresca’s criterion analysis of girder beam

According to Tresca’s hexagon, the material is within the safe zone and will not yield.

350 MP a σy =

161.87 MP a σ1 =

0 MP a σ2 =

2.0 Pier Calculations In this section the pier of the bridge will be analyzed. The pier rests at the middle of the span of

the bridge and carries part of the forces from the two resting arches (figure 34). The pier will be

made out of concrete and for the simplicity of this analysis, the effect of any rebar or reinforcing

will not be taken into consideration.

Figure 34: Elevation of the column with loads from the arches transferring to the middle pier

2.3.0 Determining the Pier Cross-Sectional Dimensions

One of the aesthetic goals of this project was to have the bridge appear as close to free spanning as

possible. To help achieve this appearance, the length of the cross-section was designated as 2200 mm to

allow the width in elevation to appear slimmer and less stocky (figure 35). However, as the width of the

cross-section becomes slimmer in relation to the length of the cross-section, the potential for buckling to

occur prior to yielding in the event of failure increases. It is desired for the column to fail due to yielding

prior to buckling since buckling can occur very rapidly with little evidence and time for mitigation.

40

Therefore, the first objective with the pier calculations is to determine the smallest cross-section

dimension ‘w’ that allows the failure mode to transition from buckling to yielding.

Figure 35: Cross-section dimensions of the column

Figure 36: Section of the bridge deck with the pier supporting arches beyond

41

42

Given

Calculate Required Properties for Buckling

Check for eccentricity,

Therefore, as a result of symmetry in the loading distance, magnitude, and direction, the

combination of the vertical loads can be considered concentric at the centroid.

Solve for Width ‘w’

Let Pyield = Pcr to calculate the transition width for buckling failure to yielding failure.

K = 0.7 The column is assumed fixed at its base and

pinned at the top.

E = 17 GPa The material selected is concrete.

f’c = 50 MPa

Le = K*L

Le = 0.7*(5000 mm)

Le = 3500 m

A = 2200(w) mm2

Iy = w(22003) = 8.87*108w mm4112 Referring to figure 36 from above.

Ix = 2200(w3) = 183.33w3112 Ix governs because w < 2200, therefore a

lower critical buckling load is allowed.

e = load 1 +load 2load 1(distance to centroid) − load 2 (distance to centroid)

e = = 02(393.4 kN )393.4kN (1m) − 393.4kN (1m)

43

Check:

Substitute the calculated minimum width into the buckling equation to ensure the critical

buckling load is greater than the axial loads the column is currently subjected to

[i.e Pcr > 2(196.3)+2(197.05)].

Therefore the minimum width required for yielding to govern before buckling in this column is

w = 209.30 mm.

Pyield = Pcr

A σy = (L )e

2π EI2

50 MPa (2200w mm2) = (3500 mm)2

π (17 10 MP a)(183.33w )2 * 3 3

w (1.1* )105 = 2.511 w3

(1.1* )105 = 2.511 w2

w = 209.30 mm

Pcr = (3500 mm)2

π (17 10 MP a)(183.33(209.30) )2 * 3 3

Pcr = 2.302577 * N or 23,022.57 kN107

Likewise Pyield= Pcr , therefore, Pcr > 786.8 kN so the column is safe.

2.3.1 Buckling Check on Established Pier Dimensions

In the previous section 2.3.0 it was determined that a minimum width of 209.3 mm is required

for the cross-section width to prevent buckling from governing first. In this section the column

width is rounded to nearest 50th mm and the final Pcr is calculated for the established dimensions.

Figure 37, shows the new established dimensions of the column rounding the width.

Figure 37: Final established cross-section dimensions of the column

44

Given The same values derived in the previous section, 2.3.0 , will be used in the following

calculations:

K = 0.7

Le = 3500 m

E = 17 GPa

f’c = 50 MPa

Iy = 8.87*108w mm4 = 8.87*108(250) = 2.2175*1011mm4

Ix = 183.33w3 = 183.33(250)3 = 2.864*109mm4 Check:

Ix governs as expected since

w<2200 mm.

2.3.2 Shear and Bending Stress on Pier

Earlier in the introduction of section 2.0 it was discussed and shown in figure 36, that the middle

pier supports part of the long and short span arch loads. Due to the difference in span between

the arches, the net horizontal reactions do not exactly cancel each other out at the pier. As a

45

Final Critical Load

Final Yielding Load

Pcr = (L )e2

π EI2

Pcr = (3500 mm)2π (17 10 MP a)(2.864 10 )2 * 3 * 9

Pcr = 3.923432 * N or 39,234.32 kN107 Check:

As expected Pcr > 786.8 kN, safe for buckling.

Pyield = A σy

Pyield = 50 MP a) (2200 mm 50 mm) ( * 2

Pyield = 2.75 * N or 27,500 kN107 Check:

Pcr > Pyield which is good because it

reinforces that yielding will occur before

buckling. Additionally, Pyield> 786.8 kN, so

the column is also safe for not yielding under

the current subjected axial loads.

result, the lateral force introduces shear and bending stress into the column in addition to the

axial stress (figure 38).

Figure 38: Combined effect of axial and lateral load as a result of symmetry on pier

For the pier, the principal stresses and in plane shear stresses will be evaluated at points A and B.

The points to be analyzed were selected at the bottom of the beam since the bottom is fixed and

bending will be at a maximum there. The top of the column is being analyzed here under the

assumption it is a free end, but with point loads of the forces exerted by the arches. The required

material fracture stress derived using failure criterions will be compared to the associated failure

stresses of concrete.

46

47

FBD of the Column

Figure 39: Center column FBD and cross section details

↑FΣ y+ = 0

= 786.8 kNP y

→FΣ x+ = 0

= 3.44 kNP x

M ↺Σ +P = 0

= −3.44 kN(5m)P M

= −17.2 kNmP M

48

Figure 40: Center column normal force and shear force diagrams

Figure 41: Center column moment diagram

49

As stated earlier, the moment diagram reinforces that the maximum moment occurs at the base

of the column because the lever arm to the lateral force is greatest.

Analyzing Point A At point ‘A’ because since the A = 0 when calculating Q . In contrast, bending stress0 τ xy =

is at a maximum here because the point is farthest from the neutral axis.

Since it’s already in its principle orientation .0 τ xy = Θp = 0°

Principal Stresses at ‘A’

The principal stresses and can be obtained as follows: σ1 σ2

Using Mohr’s criterion to evaluate failure:

= − σz AP + I

Mc

= − σz7.868 10 N*

5

(250 2200 mm )* 2 +(2200 (250 ) mm )*

3 4

(−1.72 10 Nmm)(125 mm)* 7

Both cause compression at

point ‘A.’

= − 1.430545 MPa - 0.06269 MPa σz

= − 1.493235 MPa σz

= 0 MPa σx

=, σ σ1 2 2σ +σx z ± √( ) )2

σ +σx y 2 + (τ xy2

=, σ σ1 2 20−1.493235 ± √( ) 0)2

0−1.493235 2 + ( 2

= − 0.7466175 MPa, σ σ1 2 .7466175 ± 0

= 0 MPa σ1

= -1.493235 MPa σ2

50

Figure 42: Center column Mohr’s criterion analysis for point ‘A’

Analyzing Point B

Figure 43: Center column shear analysis at point ‘B’

At point ‘B’ the bending stress is 0 MPa because the distance to the neutral axis is y = 0. In

contrast, shear stress is greater here because the area of cross-section considered is greater.

The tensile strength of portland

cement concrete is taken from

engineering toolbox as = 5 σtensile

MPa (figure C in the appendix).

The ultimate compressive strength

being considered is 50 MPa.

Therefore, the principal stresses

fall within the safe zone.

51

τ B yz = ItV Q

τ B yz = ( 2200 250 mm )(2200 mm)1

12 * *3 4

(3440 N )(2200 125 mm 62.5 mm)* 2*

0.009382 MPa τ B yz =

= − 1.430545 MPa σz No bending stress, only axial

stress at the neutral axis.

Principal Stresses at ‘B’

=, σ σ1 2 2σ +σx z ± √( ) )2

σ +σx y 2 + (τ xy2

=, σ σ1 2 20−1.430545 ± √( ) 0.009382)2

0−1.430545 2 + ( 2

= − 0.71525 MPa, σ σ1 2 .71533 ± 0

= MPa σ1 .48 * 10−5

= MPa σ2 .43058− 1

= τ max ± √( ) )2σ +σx y 2 + (τ xy

2

= 0.71533 MPaτ max

3.0 Conclusion Overall, the double arched, funicular design allows a long spanning bridge with few vertical supports. The

leftmost and rightmost columns were chosen to be eliminated, clearing the vision path of drivers on

University Ave and Ring road, making a safer space for pedestrians and traffic. The overall thrust and

nodal heights were very similar between the small and long spanning arches. The small differences in

magnitude were logical because the span between the two bridges deferred only by 10 cm. Consequently,

the net lateral force on the pier created only minor bending and shear stresses that did not exceed the

failure criterion.

To prevent buckling or yielding, the arches were sized to be made out of 90 mm round HSS. Selecting

steel provided the benefit of high strength to weight ratio; thus, allowing the cross-section of the arch

members to be smaller. This helped enhance the lightweight, dainty aesthetic of the design. Similarly, the

52

Using Mohr’s criterion to evaluate failure:

Figure 44: Center column Mohr’s criterion analysis for point ‘B’

The tensile strength of

portland cement concrete is

taken from engineering

toolbox as = 5 MPa. σtensile

The ultimate compressive

strength being considered is

50 MPa. Therefore, the

principal stresses fall within

the safe zone.

8 mm cables enhanced the weightless aesthetic of the bridge by allowing the deck to be suspended 0.8m

above the arch base supports.

Overall this simple, yet effective bridge design allowed long spans to be achieved with small members.

The design requires less manpower during construction as the longest member in the bridge is only 3m.

This makes transportation of materials to site significantly easier with the potential to even prefabricate

part of the arches at a plant and transport them to site; therefore, facilitating quick construction on this

busy University road.

53

4.0 References “Product Overview & Benefits - Hollow Structural Sections,” Steel Tube Institute, 27-Jul-2020. [Online]. Available: https://steeltubeinstitute.org/hollow-structural-sections/product-overview-benefits/. [Accessed: 18-Dec-2020].

“Concrete - Properties,” Engineering ToolBox. [Online]. Available: https://www.engineeringtoolbox.com/concrete-properties-d_1223.html. [Accessed: 18-Dec-2020].

54

5.0 Appendix

Figure A: Dimensions and section properties of round HSS

Figure B: Dimensions and section properties of square HSS

55

Figure C: Engineering toolbox regular strength concrete material properties

56