M Modern University For Information and Technology

Civil Engineering Department

Lectures Notes of

Structure Steel

Design 3

CENG 312

Prepared By

Dr: Mohamed Osman Zakaria

(First Edition 2021)

Structure Steel Design 3

Civil Engineering Department

By

Prof. Ass. Dr. Mohamed Osman Zakaria

Faculty of Engineering

MTI University

Contents

Introduction 3

Chapter one ; Layout 4

Chapter two ; Loading and maximum straining actions 14

Chapter three; Design of beams 45

Introduction

Bridges

Types of Bridges

1-With Respect to Purpose of Bridge;

a Rail way Bridges for trains

b Road Way Bridges for cars

c Foot Bridges (for Pedestrians) .

d Combined Bridges for supporting

trains & cars OR cars & pedestrians

2- With Respect to Type of the Main system

- Frames (plate girder) less than 30 m span

- Trusses greater than 30 m span

3- With respect to the statical system of main beam

- Simple beam

- Continuous beam

- Cantilever beam

4- With Respect to the Position of Bridge Floor

- Deck bridge where floor is in the upper most position and structure

elements are not seen

- Pony bridge where road way floor is in lower position and traffic is through

structure main element

- Semi Deck bridge which is in between the upper 2 cases.

Chapter One

Steel Bridges

Layout of Plate Girder Road Way Bridge

for cars '& trucks

1-Beams

'

....._

....._

2-Bracing

......_

Main Cross Stringer girders girders

Hz. Bracing Vl. Bracing

3-Stiff'eners

Hz. Stiff.

......_

Vl. Stiff.

Steps of constructing bridge

Bridges are constructed to cross a river or canal as shown in figure below.

+ Direction of Car Motion

Side1 L Bridge length _I Side2

(L)

Elevation of bridge

Step 1

Choose the type of bridge deck or pony or semi pony, let us assume the bridge is deck (same

procedure for other types)

Step 2; Beams

Step 2-a Main girder (M.G.)

In the figure below shown the 2 MG of the bridge

Step 2-b: Cross girder (X.G.)

In colour red the XG are shown in next figure. Spacing between XG is noted S

Plan of the bridge showing MG and XG

(L)

Step 2- c: Stringers

Stringer beams are shown in green, spacing between them (a);

Thus all beams are now completed.

9

· er

length J

(L)

Step 3 : Bracing;

Step 3-a; Horizontal Bracing

10

Upper Wind Bracing plan

Step 3-b; Vertical Bracing

L Bridge length J (L)

Step 4 :Stiffeners:

Step 4- a; vertical stiffeners

Step 4- b; horizontal stiffeners

11

When to use stiffeners??

if h< 1 rn . . . . No stiff. needed

if h= 1-2rn .... use Vl. stiff. Only vl. stiff.

if h=2-2.8 m use Vl. stiff. & Hz.@h/5

if h more then 2.8m .... use Vl. stiff., and Hz.@h/5 & Hz.@h/2

Stiffeners do not appears in plan

Lower plan of the bridge after the completion of all elements

NOTES

It is important to note that:

- Spacing between stringers (a) is taken between 1.75 – 2.25 m

- Spacing between XG (S) is taken between 4 – 6 m

- Total depth of MG may be assumed L/(8 – 12) m

- Breadth of bridge is B for deck bridge but for pony and semi deck it is

B’, where B’ = B + 2 Lc , (Lc are the length of side walk)

12

How to determine the type of Bridge

In this section we present the method of determining the type of Bridge (deck or pony or semi deck).

To this end we define the height of construction (H) as;

H (deck) = Span L/(8-12) + 2x0.03(as upper and lower flange thickness) + 0.05(asphalt thickness) +0.25 (R.C. slab) + L/600 (deflection) + 2cm(safety)

But for pony bridge height of construction is;

H (pony) = breadth B’/(7-9) + 2x0.03(as upper and lower flange thickness) + 0.05(asphalt thickness) +0.25 (R.C. slab) + L/600 (deflection) + 2cm(safety)

We define also the available height of construction Havailable denotedHa;

Ha = required road level – sail level

The following example illustrates the calculation procedure.

Example 1:

Road Way Plate Girder bridge having a Span of 21.0m and available height of Construction is 1.55m and Width of Road is 5.0m and two side walk of 1.50m each.

Solution

Height Of construction

Web of M.G. = L/10 (210 cm) + two flanges of M.G. 2*3(6cm) + R.C. slab (25 cm) + asphalt layer (5cm) + deflection of M.G.=L/600 (3.5cm) + safety (2cm) = 251.5 cm

251.5 > 1.55 therefore deck bridge is not possible

Try semi deck bridge

X = 1.55 – 0.3 – 1.0 – 0.055 = 0.19 m< 0.5 m

Also semi deck is not possible.

Therefore use pony bridge

13

Example 2:

Road Way Plate Girder bridge having a Span of 30.0m and available height of Construction is 3.45m and Width of Road is 12.0m and two side walk of 2.00m each.

Solution

Height Of construction

Web of M.G. = L/10 (300 cm) + two flanges of M.G. 2*3(6cm) + R.C. slab (25 cm) + asphalt layer (5cm) + deflection of M.G.=L/600 (5cm) + safety (2cm) = 343 cm

343 < 345 Use deck bridge

X- bracing and V- bracing are not possible because of the small difference (only 2 cm). Therefore use inverted U- frame as bracing.

14

Chapter Two

Loading and maximum straining action

In this chapter we introduce the position and values of loading to obtain the maximum straining actions.

In this chapter we use the concept of Influence Line (IL), the reader should be familiar with this concept.

This concept is introduced briefly;

For any simple beam of span L loaded by a unit moving load as shown in figure below ;

1 KN

α

A B

L

The influence line of both reaction at support A and B is as shown in the following figure;

1 KN

IL. RA

1 KN

IL. RB

Maximum shear is also at sections at support a; thus Q maximum is shown below

15

1 KN

Maximum shear IL. Qmax.

The maximum moment is for simple beam at the middle of the span and equal to (1KN)L/4;

L/2

L/4

IL. Mmax

Above influence line are crucial in the calculation of maximum straining actions.

To demonstrate the use of above influence lines let us solve the following numerical example.

Calculate the BM and SF for the shown beam:

3 KN

3m 2KN/m

A C B

4 m

6 m

Let us first draw the influence line of max. BM as;

2m

A2 A1

L/4 = 1.5 influence line for Mc

16

First we should calculate the area A1 (between right arrow and right support) as

A1 = (½) 1.5 x 3 = 2.25 and A2 (area between the 2 arrows)

The ordinate at 2m is equal to one

A2 = 1x (1 + 1.5)/2 = 1.25

Mc = 3 KN ( L/4 =1.5) + (A1 + A2) 2 KN/m = 4.5 + 7 = 11.5 KN.m

Now let us draw the influence line for shear at support A;

2m

1 KN

Maximum shear IL. Qmax.

Now it is required to calculate the 2 ordinates indicated by the arrows as;

2/3 and ½ from left to right

Now the area between left arrow and right support (A) = ½ (4) 2/3 = 4/3

Qmax at left support = 3KN (1/2) + (A) 2KN/m = 1.5 + 8/3 = 4.167 KN

The reader may calculate the maximum shear at right support to obtain the value 6.83 KN.

Obviously loads for bridges are moving not static as the previous example, the following example show how to use IL concept with moving loads.

Example:

For the shown system of moving loads obtain the maximum BM & SF when applied to the simple beam below.

A B

6 m

17

15 KN 15KN 15 KN

1.5m 1m

For Moment

L/4 = 1.5

Above the influence line of BM. It is clear that we should put loads to be as near as possible from the maximum ordinate, that is the middle of the beam.

Trail 1: 15 KN 15KN 15 KN

L/4 = 1.5

Trail 2: 15 KN 15KN 15 KN

0.75 1.5 1

Trail 3: 15 KN 15KN 15 KN

L/4 = 1.5

After calculating, trail 2 will gave the maximum value of BM;

18

Mmax = 15 (0.75 + 1.5 + 1) = 48.75 KN.m

For Shear :

1 KN

IL. QA

1 KN

IL. QB

Trail 1:

15 KN 15KN 15 KN

1 KN

IL for QA

Trail 2:

15KN 15KN 15

IL. For QB

19

Qmax is obtained for trail 2;

Qmax = 15 (1 + 5/6 + 3.5/6) = 36.25 KN

Loads on Stringer

1- Dead load

2- Live load

Calculation is always done for a strip of 1meter width perpendicular to the

direction of the stringer. See the plan in next figure.

1 – Dead Load:

Wdead = o.w. + (F.C.(asphalt) + tav γc ) a

Where: own weight of stringer may be assumed from 0.1 – 0.15 t/m

and FC (asphalt) is assumed 0.15 – 0.2 t/m2 , a is as defined before the distance

between two stringers, thus;

Wdead = 0.15 + ( 0.2 + 0.21x2.5) a

The moment of D.L. is the well known (WS2/8) where S is the length of the

stringer equal to the spacing between XG. Also the shear is WS/2.

20

3- Live Load :

The LL is calculated from the pattern of the Egyptian code for roadway steel

bridges;

See this pattern in the next figure;

The loads in this pattern are:

1- Three heavy concentrated loads of 60t, 40t and 20 ton. One in each

lane of 3meters. Each of these loads are distributed in 4 wheels.

Distance between wheels is shown in next figure

2- Uniformly distributed load of 0.9 t/m2 with the same lane of the 60

ton concentrated load, while in the other 2 lanes the distributed load

is of 0.25 t/m2 .

3- Side walks are covered with distributed load of 0.25 t/m2 if they have

a width up to 1.5m. if more the intensity will be 0.5 t/m2.

21

NOTES :

- It is important to underline that above trucks of 60, 40, 20 t are placed in order to obtain maximum straining actions (60, 40,20 OR 40, 60, 20 or …… any combination to obtain maximum moment and shear)

- If the width of bridge is less than 9m the three concentrated loads are reduced to 2 (60 and 40). For less than 6m only 60t is applied.

- It is not allowed to put half truck

Steps to obtain maximum straining actions

1- Define the stringer to be studied (usually the one in the middle)

2- Put loads in the x-section to be the nearest possible to studied stringer

3- Take strips of 1 m width perpendicular to the studied stringer ( in this case strips are in the lateral direction of the bridge). From these strips we calculate the reactions that our studied stringer will support

4- Put those reactions in the position giving maximum moment and shear

In the case of stringer we have only 2 strips - For concentrated load from which we obtain the reaction P as; see next

figure.

P = Σ concentrated load x corresponding value from IL

15 ton 15 ton 10 ton 10 ton

A C B

a P a

Strip 1

22

1 y1

Influence line for RC

- For distributed load from which we obtain the reaction W as;

0.9t/m2 0.5m 0.25t/m2

A C B

a W a

Strip 2

A1 A2

Influence line for RC

W = Σ uniform load x corresponding area from IL

For the intensity 0.9 the area is A1 between left support and second arrow, while for intensity 0.25 it will be A2 from second arrow till right support.

Now we have the following structure system for the studied stringer;

W P P 1.2m

S

23

To obtain maximum moment we proceed for UDL and then concentrated load of above system to get the following;

W P P 1.2m

0.3 0.9m

S

Above system is equal to the following;

P P 1.2m

0.3 0.9m

+

W

S

MLL total = Mconcentrated from influence line + WS2/8

And for maximum shear;

P P 1.2m

24

1 KN

IL for Qmax

QLL total = Qconcentrated from influence line + WS/2

The following example illustrate the procedure to obtain maximum straining action for any stringer.

Example :

Calculate the maximum straining actions acting on an intermediate stringer of a deck roadway bridge having the following data; a =2m, S=5m, B = 12m and 2 side walk of 1m each.

Solution

1- Dead Load

Wdead = ow + (FC +tav x γc ) a = 0.1 + (0.175 + 0.2 x 2.5) 2 = 1.5 t/m

Mdead = WS2 /8 = 1.5 x 25 /8 = 4.7 t.m

Qdead = WS/2 = 1.5x5/2 = 3.75 ton

2- Live Load

The stringer to be design is the one in the middle; Taking a 2 strips (one for concentrated load and the second for UDL) of one meter width perpendicular to the direction of stringer and then get the required P and W reactions (see next figure). 15 ton 15 ton 10 ton 10 ton

2 m 1 m

A C B

a = 2m P a = 2m

Strip 1

25

1 y1 = 0.5

Influence line for RC

P = 15 x 1 + 10 x 0.5 = 20 ton

0.9t/m2 0.5m 0.25t/m2

A C B

a = 2m W a = 2m

Strip 2

A1 A2 A3

1 0.75

Influence line for RC

W = 0.9(0.5 x 1 x 2 + 0.5 (1 + 0.75) /2) + 0.25 (0.5 x 0.75 x 1.5) = 1.43 t/m

To obtain maximum BM, we should have the following system of loading;

W P P 1.2m

0.3 0.9m

S

Witch is split into 2 systems ( concentrated and UDL loading) see next figure;

P = 20 P = 20t 1.2m

0.3 0.9m

26

+

W = 1.43 t/m

S = 5m

And from the following influence line we calculate MLL;

0.3m 0.9m

1.1 0.8

S/ 4 = 1.25

MLL = Mconcentrated + Mdistributed = 20 (1.1 + 0.8) + 1.43 (0.5 x 5 x1.25)

= 38 + 4.47 = 42.47 t.m = 42.5 t.m

Mtotal = MDL + MLL = 4.7 + 42.5 = 47.2 t.m

To obtain maximum SF, we should have the following system of loading;

P = 20 P = 20 ton

1.2m

W = 1.43 t/m

1 KN 0.76

QLL = Qconcentrated + Qdistributed = 20 (1 + 0.76) + 1.43 (0.5 x 1 x 5)

= 35.2 + 3.6 = 38.8 ton

27

Loads on cross girder (XG)

The reader may note that direction of XG is in the lateral direction of bridge that is perpendicular to the direction of motion and its length is equal to the breadth of the bridge (B for deck bridge and B’ for pony and semi deck).

IMPORTANT NOTE:

The strips taken for calculation of loads to be carried for any element are always perpendicular to the studied element.

Thus for the case of XG they are in the direction of motion. We always calculate straining actions for the two cases of loading;

1- Dead load

2- Live load

As mentioned above the span of XG is B or B’ ( B’ = B + 2 sidewalks) and it is always a simple beam except for the case of deck bridge having 2 sidewalks we get the double over hanging beam.

A B

B for deck

OR B’ for pony and semi deck

A B

Lc B Lc

Case of deck bridge only

The concept of influence line is also used in the study of XG. We should add the influence line for the case of double over hanging beam;

28

-0.5 Lc - 0.5 Lc

+

B/ 4

Influence line for BM (double over hanging beam)

-Lc /B 1

-Lc / B

A B

Lc B Lc

Influence line for QA

1 + Lc / B

1

-Lc / B

A B

Influence line for RA

1- Dead Load

- O.W. of XG (assume 0.4 t/m)

- FC as before 0.15 – 0.2 t/m2

- Concrete tav x γc t/m2

- OW of stringer 0.15 t/m

29

Wdead = 0.4 t/m + (owstringer/a + FC + tav x γc ) S = …. t/m

For simple beam (as for stringer)

Mdead = Wdead (B or B’)2 /8

Qdead = Wdead (B or B’)/2

New here is the case of double over hanging beam

Mdead = Wdead (B or B’)2 /8 – Wdead Lc

2 /2

Qdead = Wdead (B or B’)/2 shear is as for the case of simple beam

2- Live Load :

The same pattern explained for stringer is used, for convenience it is shown

below;

30

However the reader must remark that XG is perpendicular to the direction of motion, therefore number of strips will be more than 2 which were done for stringer.

Steps to obtain maximum straining actions;

After the choice of the studied XG (usually intermediate one) the following steps are done;

- Put loads near studied XG

- Take strips perpendicular on XG ( 5 strips as explained hereafter); the first for concentrated load of 60 ton (15 t each wheel) , the second for 10 t each wheel and the third for 5 t. two strips for UDL of 0.9t/m2 & 0.25t/m2. Finally if we have sidewalks more than 1.5m a six strip is taken for an intensity of 0.5 t/m2. See below figure for details.

15 t 15 t

A C B

S P1 S

Strip 1

10 t 10 t

A C B

S P2 S

Strip 2

5 t 5 t

A C B

S P3 S

Strip 3

31

1 y1

Above the 3 Strips for concentrated load and influence line for reactions P1, P2 & P3

0.9 t/m2

A C B

S W1 S Strip 4

0.25 t/m2

A C B

S W2 S Strip 5

W1 = 0.9 x S = ….. t/m

W2 = 0.25 x S = ….. t/m

As mentioned above strip number 6 is taken if the sidewalks are greater than 1.5 m.

0.5 t/m2

A C B

S W3 S Strip 6

W3 = 0.5 x S = …. t/m

To obtain maximum BM, we should have the heavy loads near the middle of the span;

32

P1 P1 2m

W1

0.5 0.5 m

B or B’

After that we put the load of 40 ton associated with 0.25 t/m2 the nearest possible to the middle, that is to the left of above case.

P2 P2 P1 P1 2m

W2 W1

0.5 0.5 m

B or B’

Finally we put the third reaction P3 associated with W2 in the right side.

P2 P2 P1 P1 P3 P3 2m

W2 W1 W2

0.5 0.5 m

B or B’

MLL = Mdistributed + Mconcentrated

Calculated from the moment influence line as previously done for stringer.

If we have XG as an overhanging beam above loading is also applied (same loading as simple beam) because the two cantilevers will produce negative moment resulting a reduction of M maximum.

33

For maximum shear ;

To obtain maximum SF, we should have the heavy loads near the support as shown in figure below;

P3 P3 P2 P2 P1 P1 2m

W2 W1

0.5 0.5 m

B or B’

Loading for maximum shear

+ 1

A B

I.L. for QB

Now if we have cantilevers only the right one is loaded.

Example

Obtain the maximum straining actions of a deck road way bridge with 2 cantilever sidewalks having the following data;

S = 5m, B = 10m (5 x 2) , Lc = 2m, a = 2m

Solution

1- Dead Loads

Wdead = 0.4 t/m + ( owstringer/a + FC + tav x γc ) S = … t/m

34

Wdead = 0.4 t/m + (0.15/2 +0.175 + 0.21 x 2.5 ) 5 = 4.274 t/m

Mdead = Wdead (B or B’)2 /8 – Wdead Lc2 /2

= 53.43 – 8.55 = 44.88 t.m

Qdead = Wdead B / 2 = 21.375 ton

2- Live Loads

- It is obvious that we are going to study an intermediate XG.

- Put loads near studied XG

- Take the 6 strips explained above (sidewalks are 2 m which is greater than 1.5 m) see details in figure below;

15 t 15 t

A C B

S = 5 P1 S= 5m

Strip 1

10 t 10 t

A C B

5 P2 5 m

Strip 2

5 t 5 t

A C B

5m P3 5m

Strip 3

35

2 0.76

P1 = 15 (1 + 0.76) = 26.4 ton P2 = 10 (1 + 0.76) = 17.6 ton P1 = 5 (1 + 0.76) =26.4 ton

0.9 t/m2 x 5 = 4.5 t/m

A C B

5m W1 =4.5t/m 5m Strip 4

0.25 t/m2 x 5 = 1.25 t/m

A C B

5m W2 = 1.25 5m Strip 5

W1 = 0.9 x S = 4.5 t/m

W2 = 0.25 x S = 1.25 t/m

It is clear that strip6 will produce W3 = 2.5 t/m (twice strip 5)

In order to get Mmax we have B = 10m ,therefore the following system of loading will take place;

P2 P2 P1 P1 P3 P3 2m

W2 W1 W2

0.5 0.5 m

B = 10 m

The reader may note the shift of P1 from the center to the lift in order to fit

36

the breadth of 10 m.

-0.5 Lc = -1 - 1

1.25 0.75

2.25 1.75

MLL concentrated = 26.4 (1.75 + 2.25) + 2x17.6 (1.25) + 2x8.8 (0.75) = 162.8 t.m For convenience we draw once again the IL in order to use it for distributed loads -1 - 1

2 + 1.5

2.5

MLL distributed = 4.5 ((2.5 + 1.5)2/2 + (2.5 + 1)1/2) + 1.25 (0.5x3x1.5 + 0.5x4x2) = 35.95 t.m

Mmax LL= MLL concentrated + MLL distributed = 162.8 + 35.94 = 198.74 t.m

To get Qmax the following system of loading is adopted;

P3 P3 P2 P2 P1 P1 2m

W2 W1 w3

0.5 0.5 m

B = 10 m

37

0.85 1

0.25 0.55 0.7 + 0.2

A 0.2 B

I.L. for shear at b (QB)

Qmax LL = 2x26.4x0.85 + 2x17.6x0.55 + 2x8.8x0.25 + 4.5x0.85x3 + 2.5x0.5x0.2x2 + 1.25x0.5x7x0.7 = 83.68 ton

Loads on main girder (MG)

The reader may note that direction of MG is in the longitudinal direction of bridge that is n the direction of motion and its length is equal to the length of the bridge L. Therefore it will be similar to the stringer (as far as the number and direction of strips is concerned). In a bridge we have 2 MG each carry half of the bridge.

1- Dead Loads

The dead load is given by;

Wdead = (FC + tav x γc + Wss inside ) B/2 + (FC + tav x γc + Wss outside )Lc

From code Wss inside and Wss outside are defined by the following equations;

Wss inside = 150 + 4L + 0.03L2 = … kg/m2

Wss outside = 100 + 3L = … kg/m2

The term inside is used for loads between the 2 MG, while outside mean the sidewalks. Above equation of dead load is only applicable for deck bridges

38

with cantilever. In case we do not have cantilevers (Sidewalks are less than

1m) the following equation is applied;

Wdead = (FC + tav x γc + Wss inside ) B/2 + (FC + tav x γc )Lc

But for pony or semi deck bridges, we use the following relation;

Wdead = (FC + tav x γc + Wss inside ) B’/2

Mdead = Wdead L2 / 8

Qdead = Wdead L / 2

2- Live Loads

Again the same pattern is applied for the calculation of maximum straining

actions in MG.

39

This pattern has been explained previously and is typical for all structure elements of roadway bridges.

Steps of Calculation

- We have in this case only 2 MG and they are typical therefore we choose any of them say MG at B in the next figure

- Put the heavy load near the studied MG

- In this case we have only 2 strips ; one for concentrated load and the second for distributed (same as stringer)

- Below concentrated loads which gave maximum reaction at main girder B. it is important to underline that we must leave a distance of 0.5m from B.

5t 5t 10t 10t 15t 15t 2m

A Breadth of bridge B

Concentrated Loads to get maximum reaction at B

- To obtain maximum reaction at B from distributed load we should have;

0.9t/m2

0.25t/m2 0.25 or 0.5t/m2

A Breadth of the bridge B

40

Reaction at B from concentrated load (P) is given by;

P = Σ concentrated load x corresponding value from IL

1 1 + Lc/B

+

A B

I.L. for reaction at B (RB)

While the reaction at B from distributed load (W) is given by;

W = Σ uniform load x corresponding area from IL

Having the two reactions P and W , we should put them in the position giving the maximum straining actions.

To get maximum moment (Mmax)

For maximum moment from the two concentrated load (P) put them at a distance 1.2/4 = 0.3 m and 1.2x3/4 = 0.9 m from the middle of the span. While for distributed load (W) it will be all over the beam.

P P 1.2m

0.3 0.9m

+

W

L

41

And from the influence below we can obtain Mmax.

L/2 L/2

L/4

IL. Mmax

Mmax LL = Mconcentrated + Mdistributed

For = Mdistributed = wL2/8 directly (however same result is obtained if we apply the IL)

To get maximum shear (Qmax)

P P 1.2m

W

1 KN

IL for Q

QLL total = Qconcentrated from influence line + WL/2

The following example demonstrate the procedure in detail

Example

A roadway deck bridge of length 30m and breadth B = 10m (5x2) and 2 sidewalks; the left is 3m while the right is only 2m as shown in figure. Calculate the maximum straining actions required for design.

42

Wss outside FC + tc γc + Wss inside Wss outside +FC +tcγc

2 m B = 10m 2m

WDL

1- Dead Load

Wss inside = 150 +4L + 0.03L2 = 150 + 4x30 + 0.03(30)2 = 297 kg/m2 Wss outside = 100 + 3L = 100 + 3x30 = 190 kg/m2

WDL = (FC + tc γc + Wss inside ) B/2 + (FC + tc γc + Wss outside) Lc

WDL = (0.175 + 0.21x2.5 + 0.297) 10/2 + (0.175 + 0.21x2.5 + 0.19)2 = 6.765 t/m

MDL = Wdead L2/8 = 6.765 (30)2 / 8 = 761 t.m

QDL = Wdead L / 2 = 6.765 x 30 /2 = 101.5 ton

2- Live Load

Let us study the right MG (B) because it is more critical than (A)

We have now 2 strips to study for concentrated load and then distributed load.

5t 5t 10t 10t 15t 15t 2m

0.5m

2m breadth of bridge B

Strip 1 Concentrated Loads to get maximum reaction at B

0.9t/m2

0.5 0.25t/m2 0.5t/m2

3m breadth of the bridge B

Strip 2 Distributed Loads to get maximum reaction at B

43

We can now calculate the reaction from both concentrated (P) and distributed (W) loads using the IL at reaction B

0.85 1 1.2

0.25 0.55 +

0.1

A B

I.L. for reaction at B (RB)

P = 2x15x0.85 + 2x10x0.55 + 2x5x0.25 = 39 ton

W =0.5x2x1.1 + 0.9x3x0.85 + 0.25x6x0.4 +0.5x1x0.5 = 4.245 t/m

Now we can obtain maximum straining actions;

Starting with maximum moment Mmax and using the system of loads below;

39t 39t 1.2m

0.3 0.9m

+

4.245t/m

30m

The influence line for Mmax is shown in next figure

MLL = Mconcentrated + Mdistributed = Mconcentrated (from IL) + WL2/8

44

L/2 = 15 L/2 = 15m

7.35 7.05

IL. Mmax 7.5

MLL = 39 (7.35 + 7.05) + 4.245 (30)2/ 8 = 561.6 + 474.56 = 1039.2 t.m

And for Qmax we have the following system of loading;

39t 39t 1.2m 4.245t/m

1 0.98

QLL = Qconcentrated + Qdistributed = 2 x 39 x 0.98 + 4.245 (30) /2 =

76.44 + 63.68 = 140.12 ton

Finally ;

Mtotal = MDL + MLL = 761 + 1039.2 = 1800.2 t.m

Qtotal = QDL + QLL = 101.5 + 140.12 = 241.62 ton

45

Chapter Three

Design of Beams

Beams who carry the bridge are; stringers, cross girders (XG) and main girders (MG). Design of those beams means gave all dimensions and we should satisfy the code. A preliminary design was given in the first chapter which gave an estimation of dimension. In this chapter we determine exact and final dimensions taking into account the maximum straining actions calculated in chapter two. Obviously all code requirements will be satisfied.

Steps of designing beams:

1- Calculate design values for;

- Moment; Mmax = MDL + MLL

- Shear; Qmax = QDL + QLL

Mmax and Qmax are explained in detail in chapter two - Fatigue moment which is necessary in bridges (should be calculated for

moving loads)

Mfatigue = Mmax – Mmin this will be explained in this chapter

2- Choice of section

- Hot rolled sections ( given in steel tables) IPE, IPN,…

Here we must have Fact (actual stress) = Fall (allowable stress) Fact = M/S = 0.64 Fy From this equation we can obtain (Sx) and from table the section is obtained

- Built-up sections

Here we have four unknowns (hweb, tweb, bflange, tflange) see figure below bf (3)

tf (4)

tw

hw (1) (2)

46

from the preliminary estimation shown in chapter one we can calculate the four unknowns and then check our estimation. We start by

1- hw = span / (8 – 12) for MG

hw = span / (7 – 9) for XG

hw = span / (10) for stringer

2- Tw will be taken as the maximum of the following 4 cases

- Minimum thick for plates tw1 = 8mm

- Minimum thick for webs, (hw / tw2) = 830/fy get tw2

- Shear requirement 0.35 fy = Qmax / (hw tw3) get tw3

- Buckling requirement is obtained one from the 4 following cases;

(a) If hw < 1m (Qmax / hw tw4) = (119 x 0.35 fy) / ((dw/tw4) /√fy) get tw4

(b) If hw = 1 – 2m hw/tw4 = 190 / √fy get tw4

(c) If hw = 2 – 2.8m hw/tw4 = 320 / √fy get tw4

(d) If hw > 2.8m hw/tw4 = 365 / √fy get tw4

Take tw = max. of tw1, tw2, tw3, tw4 not more than 1.6 cm, if more take tw = 1.6cm

After calculating the web dimensions we can proceed for the flange;

From structure analysis it is known that moment is resisted by the 2 flanges, we add to them 1/6 of the web area and by converting moment to C & T we get;

C = T = M /(0.98 hw)

Allowable stresses = force / resisting area

0.58 Fy = C / (Af + Aw/6) from which we obtain Af

Af = bf x tf it may be assumed that bf = 20 tf then both of them are known. tf to be taken to the nearest 2mm and bf to the nearest 2cm.

Now we can calculate Ix of the section and then Sx as;

47

Sx = Ix / (0.5 hw + tf)

Above is estimation of section dimensions, therefore we should do the following checks;

- Check for compactness:

1- For hot rolled section

For web hw / tw < 127/√Fy For flange c/ tf < 16.9 /√Fy see the distance c in below figure. Section is compact c

tf

2- For built up sections

For web hw / tw < 190/√Fy For flange c/ tf < 21 /√Fy Section is not compact

- Check bending stress:

Fact = Mx / Sx < Fbcx Fbcx may be taken 0.64 Fy if the section is compact, while not compact Fbcx = 0.58 Fy

- Check fatigue stress:

Fatigue stress = Mf / Sx < Fsr units (t/cm2) for the calculation of Mf , this will be explained in detail in next section. For rolled sections Fsr = 1.68 t/cm2 from code For built up sections Fsr = 1.26 for stringer and MG but is 1.12 t/cm2 for XG

48

(always from code)

- Check shear stress:

q = Qmax / (hw x tw) < 0.35 Fy

- Check deflection:

This check is done due to LL only. For MG and stringer

ᵟact = ᵟdistributed + ᵟconcentrated < ᵟall = span / 600

ᵟact = (5Wdistr. L4 / 384EI) + (ME / EI)

We have the following two cases of loading

P P 1.2m

0.3 0.9m

+

W

L

For the distributed load the value of deflection at the middle is known which is written in the first term of above equation (5Wdistr. L

4 / 384EI). For the second term of concentrated load we may use the elastic load method in order to calculate the elastic moment ME in the second term (ME / EI).

For the case of XG we have the following system of loading (see XG in previous chapter).

P2 P2 P1 P1 P3 P3 2m

W2 W1 W2

0.5 0.5 m

B

49

It may be difficult to calculate deflection for above system of loading, therefore it is more convenient to transfer to UDL (WLL) as follow:

WLL = [W1 x 3m + W2 (B – 3) + 2 (P1 + P2 + P3)] / B = … t/m

And maximum deflection is;

ᵟact = (5Wdistr. L4 / 384EI)

This should be less than ᵟall = B / 600

Calculation of Fatigue moment Mf

In general fatigue moment Mf is obtained from the equation:

Mf = Mmax - Mmin

Generally we should calculate MDL and MLL for both cases concentrated and distributed. Procedure to get Mf is shown hereafter in detail.

For simple beam:

For DL , MDL = WL2 / 8

For LL , we have the following systems;

(available for all simple beams; MG, XG and stringer)

P P 1.2m

W

0.3 0.9m

L

Above system is equal to the following;

50

P P 1.2m

0.3 0.9m

+

W

L

And from IL of maximum moment we can obtain Mmax as explained in previous chapter.

Now, as in code, we obtain the value;

Mmax = MDL + 0.7 Mconcentrate + 0.3 Mdistributed

Mmin =MDL

Mf = Mmax – Mmin = 0.7 Mconcentrated +0.3 Mdistributed

Remember this is for case of simple beam

For the case of XG having cantilever

In this case we should have cases of loading and using IL concept, see figures below.

P2 P2 P1 P1 P3 P3 2m

W2 W1 W2

0.5 0.5 m

B

51

MLL = Mdistributed + Mconcentrated

-0.5 Lc - 0.5 Lc

+

B/ 4

Influence line for BM

As explained before we can obtain Mconcentrated and Mdistributed . The reader may note that for above system of loading moment for both concentrated and distributed is positive therefore we obtain maximum moment.

In order to obtain the minimum moment due to LL (negative moment) we should apply distributed load only over the 2 cantilevers, this will gave the Mdistributed negative

Thus ; Mmax = MDL + 0.7 Mconcentrated (positive) + 0.3 Mdistributed (positive)

Mmin = MDL + 0.3 Mdistributed (negative)

Mf = Mmax – Mmin = 0.7 Mconc (positive) + 0.3 Mdistr (positive) - 0.3 Mdistr (negative)

Remember this is for XG having cantilevers

Max. max. BMD & SFD

In our previous study the two quantities Mmax & Qmax were calculated several times, this was for the maximum moment in the middle of span and maximum shear at supports. However, any other section has also its maximum moment and shear. Thus if we calculate the maximum for all sections and collect them in one drawing this is called the max. max. straining action (for moment and shear).

This concept is explained in detail hereafter;

52

P

A B

L

Let us calculate the maximum moment and shear for this simple beam. As we know that critical section is in the middle for moment, thus we should put load at the middle in order to get Mmax as;

L/2

Mmax at the middle

But for shear critical section is at support therefore we should put loads at support to obtain Qmax

P

A B

A 0 B

Maximum shear at left support and zero in all other sections

Let us now try to calculate the maximum moment and shear for many sections indicated in figure. Section 1 in the middle, sections 2 and 5 at supports, section 3 at L/4 and section 4 at 3L/4.

53

2 3 1 4 5

A B

L/2 1

Section 1;

Maximum at this section is PL/4

For shear (remember load at the section that is in the middle)

P

A B

P/2 L P/2

P/2

A B

-P/2 We get at section 1 two values for shear P/2 and –P/2

Section 2;

P

A B

P L 0

For moment it is zero (Mmax =0 for this section)

54

For shear as usual we have 2 values maximum positive = +p

And maximum negative = 0

A 0 B

Section 3;

P

A B

3P/4 L/4 P/4

3P/4 +

A B

-P/4 S.F.D

+

3PL/16 B.M.D

For moment Mmax = 3PL/16

For shear maximum positive = + 3P/4

For shear maximum negative = - P/4

55

After studying all sections (4 and 5 may be obtained by symmetry) we can put the obtained values in one drawing for moment known as max. max. moment diagram, and another one for max. max. shear diagram.

Zero zero

3PL/16 3PL/16

PL/4

Max. max. B.M.D.

P 3P/4 P/2 P/4

Zero

Zero -P/4 -P/2

-3P/4 -P

Max. max S.F.D.

Above is the ideal analysis to obtain max. max. diagrams, but in order to simplify the procedure the code suggest a simplified drawing and equation for max. max. moment and shear shown in drawing below. x origin mid span

M@X MLL

X

0.06L 0.06L

M approximate max. max. moment

M@X = [1 – (X/0.44L)2] MLL

mid span

MLL

0.12L

Approximate curve for shear is given below;

56

Q

QLL (+) Q@X origin

X (-)

X

Qpositive = (X/L)2 QLL and Qnegative = [(L – X)/L]

2 QLL

For dead load there are no cases of loading therefore the usual parabola (WL2/8) with origin in the middle as in figure below;

X origin

MDL @Xd MDL max

Xd

M max. max. moment for DL

MDL @Xd = [1 – (Xd /0.5L)2 MDL]

And for shear;

Q

QDL Qdl@x

A origin B

X

57

Max. max shear for DL

QDL@X = X QDL / 0.5L

Important note:

In general when required to draw max. max. for moment and shear the beam is divided into six sections, shown in figure below, and substitute in equations of dead loads above, and then for live loads. The reader should note that in case of shear for live load the origin is located at right support, therefore all sections below should be measured from right support. For moment also the origin is shifted 0.06L and all sections should be measured from this origin.

6 5 4 3 2 1

A B

X

X for each of above section is;

Section 1 = 0 , Section 2 = 0.06L , Section 3 = 0.125L, Section 4 = 0.25L,

Section 5 = 0.375L, Section 6 = 0.5L

Above values are for the case of dead loads. But for moment of live load, as mentioned in above note, the origin has a shift of 0.06L therefore we should reduce from above values 0.06L; for instance sec. 1 will be (-0.06L) and sec.2 (0),…..

For shear of live load we have; sec. 1 (0.5L), sec. 2 (0.56L) and so on.

The reader may refer to the drawing of max.max. moment and shear shown previously.

58

Curtailment of flange plate

To obtain a good design the flange of plate girder of any beam should follow the value of BM. Therefore it should be reduced when the moment become smaller. Reduction of moment may be known from above study of max. max. moment for both dead and live loads. This reduction of flange plate is known as, Curtailment of flange plate. In general the reduction in flange is easier done by reducing the breadth of the flange.

The code imposes the following conditions for curtailment of flange plate;

- If length of beam is less than 30m, reduction is done starting from support (left and right supports) till a distance of L/6. That is 2L/3 in the middle will remain without any reduction

- For spans greater than 30 m the reduction is done twice;

1. From support to L/9 (the smallest section)

2. From L/9 to L/4

We can note that sections of length L/2 in the middle of beam remain without any reduction, then from L/4 first reduction till L/9. Second reduction is from L/9 till support.

It is important to underline that reduction in practice should be done gradually not suddenly.

It is clear that in order to do the curtailment it is required to calculate the maximum moment at the section of curtailment. This will only be known from drawing the max. max. moment for dead and live loads.

After calculating the new moment at the position in curtailment a new section is design in which the flange is reduced. This is done as follow;

Stress = MY/I thus 0.58 Fy = M@curtailment (hw + tf) / Ix(new)

From which we obtain new Ix(new) and from the equation of Ix we have;

Ix(new) = (hw3 tw) / 12 + [(bf

3new tf) / 12]2 + bf new tf (hw/2 + tf/2)2

59

From which bf new is obtained. For simplicity and without error the term of bf3

may be neglected because it is very small.