force in a statically indeterminate cantilever truss

BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2

1.0 OBJECTIVE

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

method of analysing type of this structure.

2.0 LEARNING OUTCOME

2.1 Aplication of engineering knowledge in practical aplication.

2.2 To enchance technical competency in structure engineering through laboratory

aplication.

3.0 THEORY

3.1 In a statically indeterminated 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 “unknows” and we would not be able to complete the calculations

3.2 Instead we will use a method know as the flexibility meethod, 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.

Figure 1 : Idealised Statically Indetermined cantilever Truss

Basically the flexibility method usues the idea that energy stored in the frame would

be the same for a given load wheather or not the redundant member whether or not.

In other word, the external energy = internal energy.

In practise, 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

1

1

2

34

5 7

8

F

6


redundant member. The value fo both are combined to calculate the force in the

redundant member and remaining members.

The redundant member load in given by:

P =

The remaining member force are 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)

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 idealised in Figure 2

2

-250N

250N

250N-500N

0 354N354N

F=250N

0

0

1

1


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

4.0 PROCEDURE

3


1. Wind the thumbwheel on the ‘redundant’ member up to the boss and hand – tighten

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

2. Apply the pre-load of 100N downward, re-zero the load cell and carefully zero the

digital indicator.

3. Carefully apply a load of 250N and check the frame is stable and secure.

4. Return the load to zero (leaving the 100N preload). Recheck and re-zero the digital

indicator. Never apply loads greater than those specified on the equipment.

5. Apply loads in the increment shown in table 1, recording the strain readings and the

digital indicator readings.

4


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

table 2.

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

8. Plot 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 (με)

5


Load

(N)

1 2 3 4 5 6 7 8 Digital

Indicator

reading (mm)

0 124 204 -33

50 132 195 -42

100 140 185

150 149 176

200 157 168

250 166 158

Table 1: Strain Reading and Frame Deflection

Member strains (με)

Load

(N)

1 2 3 4 5 6 7 8

0 0 0 0 0 0 0 0 0

50 13 5 -8 -13 5 -7 13 7

100 26 -9 -17 -28 10 -14 27 15

150 41 -13 -24 -38 16 -19 41 24

200 53 -16 -32 -49 21 -24 53 31

250 65 -21 -40 -62 25 -31 65 37

Table 2 : True Strain Reading

6


Member Experimental Force (N) Theoretical Force (N)

1 385.91 375.12 -124.68 -124.93 -237.481 -2504 -368.09 -374.95 148.43 125.16 -184.05 -176.97 385.91 3548 219.67 177.1

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 0.5 125.1 375.12 1 -250 -0.707 176.75 0.5 125.1 -124.93 1 -250 0 0 0 0 -2504 1 -500 -0.707 353.5 0.5 125.1 -374.95 1 0 -0.707 0 0.5 125.1 125.16 1.414 0 1 0 1.414 -176.9 -176.97 1.414 354 0 0 0 0 3548 1.414 354 1 500.56 1.414 -176.9 177.1

Total854.06 4.828

P = -Total Fnl

Total n2l

= -854.06

4.828

= -176.9 N

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

6.0 CALCULATION

6.1 Calculation of real force, F

7

8

6


m = 8 m = 2j – 3j = 5 8 > 2(5) – 3r = 3 8 > 10 – 3

8 > 7

So, the structure is statically internal indeterminate.

CALCULATION FOR THEORETICAL FORCE (N)

RAY

RAX A 1 E

5 2 7 1m

RBX B 4 C 3 D 250N 1m 1m

∑MA = 0 250 ( 2 ) - RBX = 0

500 - RBX = 0 - RBX = -500 RBX = 500N

JOINT METHOD CALCULATION

8

24cm 24cm250N

24cm

B C

DEA

FBA

5

500

4FBC


MEMBER 4 MEMBER 5

∑Fx = 0 ∑Fy = 0 500 + FBC = 0 FBA = 0 FBC = -500N FBA = 0

MEMBER 3 MEMBER 7

∑Fx = 0 ∑Fy = 0 -FDE (1/1.414) – FDC = 0 -250 + FDE(1/1.414) = 0 - FDC - 354 (1/1.414) = 0 FDE = 354N

FDC = -250N

9

3

D

7

FDE

FDC

250

FCEFCA

FCD

3C 4

FCB

FCA

2 8


MEMBER 2

∑Fy = 0 250 + FCE = 0 FCE = -250N

MEMBER 8

∑Fy = 0 - 250 + FCA (sin 45°) = 0 FCA = 250N Sin 45°

= 354 N

10

C

4

8

FCB

FCA

2 8

FABFAC

FAEA500N

FAC Cos 45

FCA Sin 45

FCA Cos 45


MEMBER 1

∑Fx = 0 500 –FAC (Cos 45°) - FAE = 0 500 - 354 (Cos 45°) = FAE

FAE = 250N

CALCULATION FOR n

POINT A

11

FAC Sin 45

B C

DEA

1

1

FAB

A

1


Σfy = 0

FAB + (1 / 33.941) x 24 = 0

FAB = - 0.707N

Σfx = 0

FAE + (1 / 33.941) x 24 –= 0

FAE = - 0.707N

POINT B

Σfy =0

FAB – FBE (24 / 33.941) = 00.707 - FBE (0.707) = 0FBE = 1

Σfx =0

FBC + FBE (24 / 33.941) = 0 FBC + FBE (0.707) = 0FBC = - 0.707

POINT C

12

FAE

FBC

FBEFAB

FBC


Σfx= 0

FBC + (1 / 33.941) x 24 + FCD (24

/33.941) = 0

0.707 + (- 0.707) + FCD (0.707) = 0

FCD = 0

Σfy = 0

-(1 / 33.941) x 24 – FCE – FCD = 0

- 0.707 - FCE – 0 = 0

FCE = - 0.707

POINT D = Zero Bar

POINT E

Σfy = 0

(1 / 33.941) x 24 – FCE = 0

0.707 – 0.707 = 0

0 = 0 (CHECKING)

Σfx = 0

FAE + FED – (1 / 33.941) x 24 = 0

0.707 + FED - 0.707 = 0

FED = 0

13

1

FEC

FCD

FEC

FDEFAE

FBE

BFC 3051 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS

GROUP 4

6.3 Calculation of Internal Forces AC

P value, internal forces AC = P = -

= -

= - 176.9 N (compression )

6.4 Example of calculation for member 1 (Pn + f)

Given;

Length, L = 1cm

Force, F = 250N

Load in each member due to unit load in place of redundant

member,

n = - 0.707N

Area, A = πd²/4 = π(6)²/4 = 28.274 mm² = 0.283cm²

Fnl = (250)(- 0.707)(1) = -176.75 N.cm

n2l = (0.707) 2(1) = 0.5 cm

Pn = (-176.9)(-0.707) = 125.1

Pn + f = 125 + 250

= 375 N = Theoretical Force (N)

854.06

4.828

Σ F’nl / AE

Σ n2 l / AE


GROUP 4

6.5 Calculation of Experimental Force (N)

ε250N (True strain reading for load 250N)

Strain for load 250N – 0N

1 2 3 4 5 6 7 8

0N 153 243 -27 -48 125 27 22 8

250N 218 222 -67 -110 150 -4 87 45

250N 65 -21 -40 -62 25 -31 65 37

Given;

Area, A = πd²/4 = π (6²)/4 = 28.27mm²

Modulus young, Esteel = 2.10 x 105 N/mm2

AE = 28.27 x 2.10 x 105 = 5.937

Member 1,

F = AEε

= (5.937 x 106) (65 x 10-6)

=385.91 N

Member 2,

F = AEε

= (5.937 x 106) (-21 x 10-6)

= -124.68 N

Member 3,

F = AEε

= (5.937 x 106) (-40 x 10-6)

= -237.481 N

Member 4,

F = AEε

= (5.937 x 106) (-62 x 10-6)

=-368.09 N

Member 5,

F = AEε

= (5.937 x 106) (25 x 10-6)

= 148.43 N

Member 6,

F = AEε

= (5.937 x 106) (-31 x 10-6)


GROUP 4

= -184.05 N

Member 7,

F = AEε

= (5.937 x 106) (65 x 10-6)

= 385.91 N

Member 8,

F = AEε

= (5.937 x 106) (37 x 10-6)

= 219.67 N


GROUP 4

6.0 DISCUSSION AND CONCLUSION

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

the accuracy of your result

From the table 3, the experimental and theoretical forces are not accurate.

We can see that there are huge difference value between experiment and theory.

It is mean that, the accuracy of the result is not exact but for the compression

and tension member, we can conclude that the following tension and

compression is same with the value of the force is different. The experiment

value is different compared to the theoretical value. Only member 2 for

experiment value is near with theoretical value.

It’s probably because of the error while setting the apparatus of the

experiment laboratory. Since the equipment is under the air conditioner, the

factor of wind and human mistakes will be taken in this experiment, so we can

assume that the apparatus are not calibrated.

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.

For the determinate truss it has extra member and for the

indeterminate truss it has extra member. From that, for the determinate

truss it safety for the structure however for the indeterminate truss it more

safety for the structure. In the economy, the indeterminate truss will cost

more expensive than determinate truss because the indeterminate truss has

more member than the determinate truss.


GROUP 4

3. What problem could you for seen if you were to use a redundunt frame in

a “real life’ aplicatioin. (Hint: look at the zero value for the strain reading

once you have included the redundant member by winding up thumnut).

Redundant member is use to make beauty for the truss. When they are

redundant in truss it will rising the cost because there are redundant member in

that truss.

7.0 CONCLUSION

From what we have learn and get form the experiment we can say that, the

statically indeterminate structure can be classified if the equilibrium equations

were not adequate to calculate the external reactions of all the internal forces.

However there are pro and contra in this statically indeterminate analysis. The

advantages of statically indeterminate are:

The maximum stresses and deflections are smaller than statically

determinate counterpart.

Can support loads with support loading on thinner members with

increased stability.

Have a tendency to redistribute its load to redundant supports in

the case of faulty design or overloading occurs.

The disadvantages of this truss are: Redundant structures can induce problems such as differential

displacement

High cost of statically indeterminate structure compare to

determinate structure.

REFERENCES


GROUP 4

1. Yusof Ahmad(2001), “Mekanik Bahan dan Struktur”: Universiti Teknologi Malaysia,Skudai, Johor Darul Takzim

2. Bambang Prihartanto(2008), “Structural Analysis”:Universiti Tun Hussein Malaysia.

3. “Structural Use Of Steelwork In Building”:British Standard.

4. Plane Frame Example – http:// www.amazon.com

http://www.amazon.com/

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