rep truss-fida '10

8/7/2019 REP TRUSS-FIDA '10

1/19

1.0 OBJECTIVE

1.1 To observe the effect of redundant member in a structure and understand the

method of analyzing type of this structure.

2.0 LEARNING OUTCOME

2.1 Application of engineering knowledge in practical application.

2.1 To enhance technical competency in structure engineering thr

laboratory application.

3.0 THEORY

3.1 In a statically indeterminate truss, static equilibrium alone cannot be used to

calculated member force. If we were to try, we would find that there would

be too many unknowns and we would not be able to complete the

calculations.

3.2 Instead we will use a method known as the flexibility method, which uses

an idea know as strain energy.

3.3 The mathematical approach to the flexibility method will be found in the

most appropriate text books.


2/19

Figure 1: Idealized Statically Indetermined cantilever Truss

Basically the flexibility method uses the idea that energy stored in the

frame would be the same for a given load whether or not the redundant

member whether or not.

In other word, the external energy is same internal energy.

In practice, the loads in the frame are calculated in its released from

(that is, without the redundant member) and then calculated with a unit load in

place of the redundant member. The value of both is combined to calculate the

force in the redundant member and remaining members.

The redundant member load in given by:

=

ln

fnlP

2

The remaining member force is then given by:

Member force = Pn + f

Where,

P = Redundant member load (N)

L = length of members (as ratio of the shortest)

n = load in each member due to unit load in place of redundant

member (N)

F = Force in each member when the frame is release (N)


3/19

Figure 2 shows the force in the frame due to the load

of 250 N. You should be able to calculate these values from Experiment:

Force in a statically determinate truss

Figure 2: Force in the Released Truss

Figure 3 shows the loads in the member due to the unit load being

applied to the frame.

The redundant member is effectively part of the structure as the

idealized in Figure 2.

Figure 3: Forces in the Truss due to the load on the redundant members

Method of Joints

The method that had been used for data analysis in this experiment is Method of

Joints


4/19

Method of Joints suitable to use in calculating the entire member forces for a

truss.

This method entails the use of a free body diagram of joints with the equilibrium

equations

Fx = 0 and Fy = 0.

Calculation only can be started for joint where the numbers of unknowns are two

or less.

4.0 PROCEDURE

1. The thumbwheel on the redundant member up to the boss was wind and

hand tightens it. Do not use any tools to tighten the thumbwheel.

2. The pre-load of 100N downward was applied, re-zero the load cell and

carefully zero the digital indicator.

3. A load of 250N was applied carefully and checked the frame was stable and

secure.

4. The load to zero (leaving the 100N preload) was returning. Rechecked and re-

zero the digital indicator. Never apply loads greater than those specified on

the equipment.


5/19

5. Loads in the increment shown in table 1 was applied, the strain readings and

the digital indicator readings was recorded.

6. Subtracted the initial (zero) strain reading (be careful with your signs) and

completed table 2.

7. Calculated the equipment member force at 250 N and enter them into table 3.

8. Plotted a graph of Load vs Deflection from Table 1 on the same axis as Load

vs deflection when the redundant removed.

9. The calculation for redundant truss is made much simpler and easier if the

tabular method is used to sum up all of the Fnl and n2l terms.

10.Refer to table 4 and enter in the values and carefully calculated the other terms

as required.

11. Enter your result in to Table 3.5.0 RESULT

Member strains ()

Load

(N) 1 2 3 4 5 6 7 8

Digital

Indicator

reading (mm)

0 178 224 -44 -67 139 24 58 40 0.00350 190 219 -54 -81 143 17 70 47 0.027

100 204 214 -63 -96 147 9 83 54 0.053

150 215 209 -71 -109 151 3 95 60 0.075

200 230 204 -81 -123 155 -5 108 67 0.098

250 243 199 -90 -137 158 -12 121 74 0.118

Table 1: Strain Reading and Frame Deflection

Member strains ()

Load

(N)

1 2 3 4 5 6 7 8


6/19

0 0 0 0 0 0 0 0 0

50 12 -5 -10 -14 4 -7 12 7

100 26 -10 -19 -29 8 -15 25 14

150 37 -15 -27 -42 12 -21 37 20

200 52 -20 -37 -56 16 -29 50 27

250 65 -25 -46 -70 19 -36 63 34

Table 2 : True Strain Reading

Member Experimental Force (N) Theoretical Force (N)

1 392.44 374.942 -150.94 -125.06

3 -277.73 -250.00

4 -422.63 -375.06

5 114.71 124.94

6 -217.35 -176.927 380.36 354.00

8 205.28 177.08

Table 3: Measured and Theoretical in the Redundant Cantilever Truss

Member Length F n Fnl n2l Pn Pn + f

1 1 -250 -0.707 -176.75 0.50 -124.94 374.94

2 1 250 -0.707 176.75 0.50 -124.94 -125.06

3 1 250 0 0.00 0.00 0.00 250.00

4 1 -500 -0.707 353.50 0.50 -124.94 -375.06


7/19

5 1 0 -0.707 0.00 0.50 -124.94 124.94

6 1.414 0 1.0 0.00 1.414 176.92 -176.92

7 1.414 354 0 0.00 0.00 0.00 354.00

8 1.414 354 1.0 500.56 1.414 176.92 177.08

Total 854.06 4.828

P = Total FnlTotal n2l

Table 4: table for calculating the Forces in the Redundant Truss

6.0 DATA ANALYZE AND CALCULATION

6.1 The Formula

If to analysis we can use the formula Youngs Modulus relationship, the method

is calculate the equivalent member force. Complete the experimental force in

Table 3. (Ignore member 6 at this stage)

E =

Where;

E = Youngs Modulus (N/m2)

= stress in the member (N/m2)

= Displayed strain

=

Where;

F = Force in the member (N)

A = Cross section in the member (m2)

6.2 Experimental Force

Calculation For True Strain Reading ;


8/19

True Strain Reading (load 50, 100,150,200,250)

= Strain reading (load 50, 100,150,200,250) Strain reading (load0)

Example for True Strain Reading (member 1)

True Strain Reading (load 50) = 190 - 178= 12

True Strain Reading (load 100) =204 - 178= 26



True Strain Reading (load 250) = 243 178

= 65

Example for True Strain Reading (load = 250 N)

True Strain Reading (member 1) = 243 178

= 65

True Strain Reading (member 2) = 199 - 224

= -25

True Strain Reading (member 3) = - 90 (-44)

= - 46

True Strain Reading (member 4) = - 137 (-67)

= - 70


= 19True Strain Reading (member 6) = - 12- 24

= -36


= 63


9/19

True Strain Reading (member 8) = 74 (40)

= 34

Calculation for Cross Section Area of the member;

Diameter, D = 6.06 mm

From equation;4

A

2D=

4

)05.6( 2=

75.28= mm2

Calculation for Experimental Force (N); Load = 250 N

Given,

E steel = 2.10 x 105 N/mm2

Calculation for Member 1;

F = A E

= 28.75 x (2.10 x 105) x (65 x 10-6)= 392.44 N


F = A E

= 28.75 x (2.10 x 105) x (-25 x 10-6)

= -150.94 N


F = A E

= 28.75 x (2.10 x 105) x (-46 x 10-6)= -277.73 N


F = A E

= 28.75 x (2.10 x 105) x (-70 x 10-6)

= -422.63 N


10/19


F = A E

= 28.75 x (2.10 x 105) x (19 x 10-6)

= 114.71 N


F = A E

= 28.75 x (2.10 x 105) x (-36 x 10-6)

= -217.35 N


F = A E = 28.75 x (2.10 x 105) x (63 x 10-6)

= 380.36 N


F = A E

= 28.75 x (2.10 x 105) x (34 x 10-6)

= 205.28 N

6.3 Calculation the Forces in the Redundant Truss

Sample calculation to get the value n

1.414 = 1

1.0 = x

x = . 1 .

1.414x = 0.707

Calculation of the value Fnl in the table 4


11/19

nl

Member F n L F x n x l

1 250 -0.707 1 (250)(-0.707)x 1 = -176.752 -250 -0.707 1 (250)(-0.707)x 1 = 176.753 250 0 1 250x0x1=04 -500 -0.707 1 (-500)(-0.707)x 1= 353.50

5 0 -0.707 1 (0)(-.0.707)x1 = 06 0 1.0 1.414 (0)(1)(1.414) = 07 354 0 1.414 (354)(0)(1.414) = 08 354 1.0 1.414 (354)(1)(1.414) = 500.56

Total Fnl = 854.06

Calculation the value of

Member n n l nl

1 -0.707 0.50 1 0.50x 1 =0.50

2 -0.707 0.50 1 0.50x 1 =0.50

3 0 0 1 0x1 = 0

4 -0.707 0.50 1 0.50x1 = 0.50

5 -0.707 0.50 1 0.50x1 =0

6 1.0 1 1.414 1x1.414 =1.414

7 0 0 1.414 0x1.414 =08 1.0 1 1.414 1x1.414 =1.414

Calculation the value of Pn

P = Total FnlTotal nl

P = -(854.06)(4.828)

P = -176.92

Member P n P x n

1 -176.92 -0.707 124.942 -176.92 -0.707 124.94


12/19

P n

+ F

PAB

500

250

PAE

PAC

A

3 -176.92 0 0.004 -176.92 -0.707 124.945 -176.92 -0.707 124.946 -176.92 1.0 -176.927 -176.92 0 0.00

8 -176.92 1.0 -176.92

Calculation the value of

Member Pn F Pn + F = Theoretical Force

1 124.94 250 374.942 124.94 -250 -125.063 0.00 -250 -250.004 124.94 -500 -375.065 124.94 0 124.946 -176.92 0 -176.927 0.00 354 354.008 -176.92 354 177.08

7.4 Theoretical Force

Calculation for Theoretical Force (N);

At Point A ;


13/19

500

PBA

PBC

sin = (1/2)

cos = (1/2)

FX = 0

FAX + FAE + FAC (cos ) = 0

(-500) + FAE + FAC (1/2) = 0

FAE + FAC (1/2) = 500 N .................(1)

FY = 0

FAY - FAB - FAC (cos ) = 0(250) FAB FAC(1/2) = 0

FAB + FAC (1/2) = 250 N .................(2)

At Point B ;

FX = 0

500 + FBC = 0

FBC = - 500 N (Compression Force)

FY = 0

FBA = 0

From equation (2) ;


14/19

PEA

PEC

PED

PCE

PCA

PCB

PCD

C

E

FAB + FAC (1/2) = 250 N

0 + FAC (1/2) = 250 N

FAC = 250 (2)

FAC = 353.55 N (Tensioned Force)

From equation (1) ;

FAE + FAC (1/2) = 500 N

FAE = 500 [ 353.55 (1 / 2) ]

FAE = 500 (250)

FAE = 250 N (Tensioned Force)

At Point C;

FX = 0

FCB + (FCD) FCA (1/2) = 0

(500) + FCD [353.55 (1/2)] = 0

500 + FCD - 250 = 0

FCD = - 250 N (Compression Force)

FY = 0

FCE + FCA (1/2) = 0

FCE = 353.55 (1/2)

FCE = 250 N (Compression Force)

At Point E;


15/19

FX = 0

FEA FDE (1/2) = 0

(250) + FDE (1/2) =0

FDE = 250 (2)

FDE = 353.55 (Tensioned Force)


16/19

8.0 DISCUSSION

1. From table 3, compare your answer to the experimental values. Comment on the

accuracy of your result.

From the result, the value of forces obtains are just small different and we can said

that there are almost equal event from the Experimental and Theoretical (calculation)

are almost equivalent. This indicate that the Experimental Force obtain are accurate

and correct. Difference between this values may be happen because of these errors

during the experiment conducted:

No follow the prescribed procedure.

Inconsistent when take a reading recorded at changing meter tool

inconsistent, caused by some factor such as vibrant wind influence and

table.

Inaccurate screw adjustment where found when taking reading in

digital Reading force value imposed changing.

2. Compare all of the member forces and the deflection to those from statically

determinate frame. Comment on them in terms of economy and safety of the

structure.


17/19

Based on one force burdened levied at a truss structure analysis made to show every

member in that truss structure will respond with force burdened imposed whether that

member compression strength or tension force. For example in this experiment, member

of 2, 3, 4 and 6 are in compression while the members of 1,5,7 and 8 are in tensioned.

From this analysis it shows that security and cost saving are needed in each member in

the truss. For example at member 1 the needs material that capable to hold tension force

because the member only in a tension force. This case done to other member based on

the force that imposed.

3. What problems could you for see if you were to use a redundant frame in a real

life application. ( Hint: look at the zero values for the strain reading once you have

included the redundant member by winding up thumnut).

Usage of the redundant member has several advantages such as can reduce deflection

in that structure. Apart from that it can increase strength in structure. Possible problem

would arise when using excess framework are upgrading again construction cost. Apart

from that with member increase in this structure and would complicate the connection in

every member. Indirectly this also increase construction cost.

Application

In the construction industry there are few products which need to be specifically

designed and tailor made for each development - the roof truss is one of them. A wooden

truss is a very strong and stable frame made of wood and held together with metal

connector plates. It is of an ancient design. A truss adds support and shape to a roof.

Some others describe it as an open web type of design that is appropriate for supporting

roofs, decks, or floors. Rooms are now possible inside the attic due to the Attic Type

Roof Truss.

Here are some examples.

1. Large roof

2. Curved Truss


18/19

Truss bridge for a single track railway, converted to pedestrian use and pipeline support

8.0 CONCLUSION

From the experiment, we have observed the effect of redundant members in a

structure and we learnt the method analyzing types of structures. We distinguished

between statically determinate trusses, the axial forces in the statically determinate

trusses can be calculated considering the equilibrium of forces only but the calculation of

the axial forces in the statically indeterminate trusses require in addition to statics,

consideration of deformation compatibility. In this experiment, we also noticed that the

framework comply with the pin joint theory even though the joint are not truly pin joint.

Therefore, Method Of Joint is appropriate to determine the internal forces for eachmembers.

Compared the statically indeterminate structures with statically determinate

structure, it shows the following behavior:

1) The structure is not in stress when subjected to movement of support.

2) Changes in temperature of members caused the structures to deform, but didnt

stress the structure.

3) Incorrect length of members caused only deformation, but didnt stress the

members.
http://en.wikipedia.org/wiki/Truss_bridgehttp://en.wikipedia.org/wiki/File:RRTrussBridgeSideView.jpghttp://en.wikipedia.org/wiki/Truss_bridge


19/19

9.0 REFERENCE

1. http://en.wikipedia.org/wiki/Trusses

2. http://web.mit.edu/emech/dontindex-build/full-text/emechbk_5.pdf
http://en.wikipedia.org/wiki/Trusseshttp://en.wikipedia.org/wiki/Trusseshttp://web.mit.edu/emech/dontindex-build/full-text/emechbk_5.pdfhttp://en.wikipedia.org/wiki/Trusseshttp://web.mit.edu/emech/dontindex-build/full-text/emechbk_5.pdf

rep truss-fida '10

Documents