statically determinate trusses

7/28/2019 Statically Determinate Trusses

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Analysis of Statically Determinate Trusses

Example 1:

The following roof structure is used for a factory building. The trusses are spaced at 4 m spacing and thepurlin are spaced at 1,8 m. The roof covering is of galvanized sheeting with a mass of 10 kg/m

2. There is an

external imposed load of 0,3 kN/m2

to allow for servicing, rain, hail etc. Assume the trusses and purlin tohave a self-weight of 18 kg/m

2. Determine:

a) The point loads at each of the nodesb) The reactionsc) The forces in the members AB, AF, BF, FC, FG and GH

Trusses Page 1 of 17 7/23/2003


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Definition of sin and cos .

Length

projectionY =sin

Length

projectionX =cos

Equilibrium of Forces

For equilibrium we can write the following:

0= Y

0= X

0= M

This allows us to solve 3 unknowns and only 3.

Solution to Questions:

a) Load from the sheeting:

A one metre length when seen vertically from the top has a length of 1/cos which in this case is equal to1/cos 20 = 1,064 m. The load is then equal 10 kg/m

2x gravity acceleration x 1,064 m. Assume gravity

acceleration to be = 10 m/s2. The load in when seen vertically from the top = 106,4 N/m

2= 0,1064 kN/m

2.

Total vertical load =Sheeting 0,106 kN/m

2

Trusses 0,180 kN/m2

Additional imposed 0,300 kN/m2

Total 0,586 kN/m2

Load at each node = load intensity times the area carried by the node= 0,586 kN/m

2x 4 m x 1,8 m

= 4,22 kN

b) Reactions as a result of the symmetrical structure and loading are equal to 4 x 4,22 kN

Reactions = 16,88 kN



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Member X-projection, x Y- projection, y Length

AB 1,800 0,655 1,9155

AF 2,038 0 2,038

BF 0,238 0,655 0,6969

BC 1,800 0,655 1,9155

FC 1,562 1,310 2,0386

FG 2,038 0 2,038

GH 6,248 0 6,248

For node AAssume that all members are in tension



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= 0Y

011,288,16 =+++ AFAB YY

0sinsin11,288,16 =+++ AFAB FF

011,288,16 =

+

++Length

projectionYF

Length

projectionYF AFAB

0038,1

0

9155,1

655,011,288,16 =+++ AFAB FF

+ 14,77 + 0,34195 FAB = 0FAB = - 43,194 kN

= 0X

0=+ AFAB XX

0=+AFAFAF

ABABAB

L

xXL

xF

0038,2

038,2

9155,1

800,1194,43 =+ AFX

FAF = + 40,590 kN

Equilibrium of node B

= 0Y 022,4 =+ BFBCBA YYY

022,4 =+BF

BFBF

BC

BCBC

BA

BABA

L

yF

L

yF

L

yF

0

6969,0

655,0

9155,1

655,0

9155,1

655,0194,4322,4 =++ BFBC FF

0,34195 FBC 0,93988 FBF = - 10,550 (1)

= 0X 0=++ BFBCBA XXX

06969,0

238,0

9155,1

80,1194,43

9155,1

80,1=++ BFBC FF

0,9397 FBC + 0,3415 FBF = - 40,5895 (2)

FBC = - 41,753 KNFBF = - 3,968 kN



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Alternative method is to use the moment equilibrium of a portion of the structure. This enables us to solvethree unknown forces. It is usual to choose a section where two of the forces cut in a point so that theirmoment is equal to zero, thereby we can solve one of the unknowns immediately.

= 0CM 16,88 x 3,6 2,11 x 3,6 4,22 x 1,8 FFG x 1,310 = 0FFG= 34,791 kN

= 0Y 022,411,288,16 =++ FCBC YY

0036,2

310,1

9155,1

655,055,10 =++ FCBC FF

0,34195 FBC + 0,64342 FFC = - 10,55 (3)

= 0X 0=++ FGFCBC XXX

0791,34038,2

038,2

036,2

562,1

9155,1

8,1=++ FCBC FF

0,9397 FBC + 0,76719 FFC = - 34,791 (4)

FBC = - 41,753 kNFFC = 5,793 kN

Parallel Chord Trusses

Parallel chord trusses may be treated in the same way as beams. The top andbottom chords resist the bending moment and the diagonal and vertical membersresist the shear forces.

Examples of parallel chord trusses



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The second truss must have a very small slope otherwise one underestimates the forces in the top andbottom chords.

Example:

Determine the member forces in the following trusses. The roof sheeting has a unit mass of 12 kg/m2, the

ceiling 15 kg/m2, the trusses 15 kg/m

2and there is a live load of 0,4 kN/m

2. The loads are to be increased

with load factors of 1,2 on the permanent (dead load) and 1,6 on the imposed (live load).

The dead load is equal to:

Roof sheeting: 12 kg/m2

x 10 m/s2

= 120 N/m2

= 0,12 kN/m2

Ceiling: 15 kg/m2

x 10 m/s2

= 150 N/m2

= 0,15 kN/m2

Trusses and purlin 15 kg/m2

x 10 m/s2

= 150 N/m2

= 0,15 kN/m2

Total dead load = 0,42 kN/m2

Total live load = 0,40 kN/m2

Factored Load 1,2 x DL + 1,6 x LL = 1,2 x 0,42 + 1,6 x 0,40 = 1,144 kN/m2

Load at each node intensity x tributary area = 1,144 x 2 x 3,5 = 8,008 kN

Draw the Shear force and bending moment diagrammes:



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8/17

Solve the forces in the members through using sections.

Section 1:

In the panel between A and C the vertical shear force of 17,333 kN must be carried by the diagonal memberBC. Therefore:

YBC = 17,333 kN kNFL

yBC

BC

BC 333,17= kNFBC 684,332,1

332,2333,17 ==



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The bending moment from the external forces must be in equilibrium with theinternal forces in the top and bottom members. The bending moment from theexternal forces is shown in the bending moment diagramme.

Take moments about C.

FBD x 1,2 + 34,666 = 0

kNFBD 888,282,1

666,34=

=

All the sections may be tackled in a similar fashion.

Section 2:

To solve for the force in member CE take moments about D.

- FCE x 1,2 + 34,666 = 0 FCE = + 28,888 kN

The vertical force in member CD must be equal to the vertical component of theforce in BC. The vertical component of the force in BC is equal to the shear betweenA and C. FCD = 17,333 kN.



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Portal Frame Analysis

Determine the bending moment diagramme of the following structure.

Plywood unit weight = 7 kN/m3

Portal unit weight = 7 kN/m3

Tiles have a mass of 55 kg/m2

Purlin have a unit weight of 5 N/m3

Imposed or live load = 0,3 kN/m

2

We would like to convert these loads into a uniformly distributed load on the portal frame. Remember that thetiles, plywood and the portal itself are at an angle of 20,556. We would like to have the loading horizontal.

Dead Loads

Tiles = 55 x 10 /1000 = 0,55 kN/m2

on the slope = 0,55/cos 20,556 = 0,587 kN/m2

horizontalPlywood = 0,022 x 7 = 0,154 kN/m

2on slope = 0,154/cos 20,556 = 0,164 kN/m

2

Portal = 0,1 x 0,5 x 7 = 0,350 kN/m on the slope = 0,35/cos 20,556 = 0,374 kN/m

Purlin = 0,075 x 0,225 x 5 = 0,084 kN/m length of purlin

Tributary length = the spacing of the portals = 4 m.

Total dead load: Tiles = 0,587 x 4 = 2,348 kN/mPlywood = 0,164 x 4 = 0,656 kN/mPortal = 0,374 = 0,374 kN/mPurlin = 0,084 x 4 / 1,1= 0,305 kN/m

Total = 3,683 kN/m

Live load Imposed = 0,3 x 4 = 1,2 kN/m

Factored Loads 1,2 x DL + 1,6 x LL1,2 x 3,683 + 1,6 x 1,2 = 6,340 kN/m



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Line sketch of portal frame:

= 0AM 0162

1634,6

2

= EY

YE = 50,72 kN

In a similar fashionYA = 50,72 kN

= 0CM 02

834,678

2

= AA XY

02

834,67872,50

2

= AX

XA = 28,983 kN



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Virtual Work Used to Calculate Deflection of Trusses.

Example 1:

The following timber truss has loads as shown. The sections have an area of 2700 mm2 and a modulus ofelasticity, E = 7800 MPa. Determine the vertical deflection of the node G and of the node F.

Solution to the forces in the members.

Member L P0 pv P0pvLAB 2,207 -7,238 -1,183 18,898

BC 2,207 -4,825 -1,183 12,598

CD 2,207 -4,825 -1,183 12,598

DE 2,207 -7,238 -1,183 18,898

AH 2,000 +6,559 +1,072 14,062

HG 2,000 +6,559 +1,072 14,062

GF 2,000 +6,559 +1,072 14,062

FE 2,000 +6,559 +1,072 14,062

BH 0,933 0 0 0

DF 0,933 0 0 0

CG 1,866 +2,040 +1,000 3,807

BG 2,207 -2,413 0 0GD 2,207 -2,413 0 0

P0pvL (kN.m) 123,047

External virtual work = internal virtual work

1,0 x =

EA

LpP v0

= mmx

x

78002700

10047, 6123= 5,84 mm



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For the deflection at F:

Member L P0 pv P0pvL

AB 2,207 -7,238 -0,591 9,441

BC 2,207 -4,825 -0,591 6,293

CD 2,207 -4,825 -0,591 6,293

DE 2,207 -7,238 -1,774 28,338

AH 2,000 +6,559 +0,536 7,031

HG 2,000 +6,559 +0,536 7,031

GF 2,000 +6,559 +1,608 21,094

FE 2,000 +6,559 +1,608 21,094

BH 0,933 0 0 0

DF 0,933 0 +1,000 0

CG 1,866 +2,040 +0,500 1,903

BG 2,207 -2,413 0 0

GD 2,207 -2,413 -1,183 6,300

P0pvL (kN.m) 114,818

External virtual work = internal virtual work

1,0 x =

EA

LpP v0

= mmx

x

78002700

10818,6

114= 5,45 mm



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Virtual work applied to bending members:

Example 1:

Determine the deflection in the middle of the following beam.

External Virtual work = Internal virtual work

1,0 x = dsIE

mL v 0

0M

= dsIE

mL v

0

0M 22

22

2

0xwxLw

M

= xmv = 5,0

1,0 x = dxIE

xwxLwL

0

32

22 2

=2

0

43

86

L

xwxLw

EI

1

=

12848

44 LwLw

EI

1

=

EI

L4

384

w5

Instead of integrating the bending moment equations one can use standard integration equations.



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Apply the table to the previous problem:



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[ ]0 1 2 20

2 26

L

vM m hds a a c EI

= +

h =2

La1 =

4

La2 =

8

2Lwc2 =

32

2Lw3

= +

2 2

0

0

32 212 4 8 32

L

vM m L L w L w LdsEI

IE

Lw

=

4

384

51

Example 2:

Determine the deflection in the middle of the following beam. E = 200 GPa and I = 43,60 x 10-6

m4:

External Virtual work = Internal virtual work

EI * 1,0 x = M dsmL

v 0

0

= 213

aah

+ [ ]21122121 226

bbbabaaah

+++ + [ ]21122121 226

bbbabaaah

+++ + 213

aah

= 505,13

3 + [ ]4525,2225,250455,1505,12

6

5,+++

1

+ [ ]405,125,1454025,24525,226

5,1+++ + 405,1

3

3

= 388,125 kN.m

=EI

125,388 =66 1060,4310200

125,388 xx

= 0,0445 m = 44,5 mm



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Example 3:The following beam has been strengthened and stiffened by welding 10 mm thick plates over the central halfof the member. Determine the deflection in the middle of the beam and compare this to the deflection of the

un-stiffened beam. I of un-stiffened = 44,3 x 106 mm4. With the stiffening, the I increases to 81,8 x 106 mm4.E = 200 GPa

1,0 x = dsIE

mML v

0

0

= dsIE

mL v

20

0M 2

= [ ] ( )[ ] 211221211221 /262/262 EIbacbbaah

EIcaah

+++++

= [ ] ( )[ ] )8,81200/(35,1375,8429035,675,16

32)103,4410200/(375,3925,675,1

6

32 66 +++++ xx

= 2,476 x 10-2

+ 6,911 x 10-2

m= 9,387 x 10

-2m

= 93, 9 mm

Un-stiffened member:

45

384

w L

EI

=

4

3 6

5 5 12000152,4

384 200 10 44,3 10mm

x x

= =

T P 17 f 17 7/23/2003

statically determinate trusses

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