Force in a Statically Indeterminate Cantilever Truss
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Transcript of 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
BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2
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
BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2
Figure 3: Forces in the Truss due to the load on the Redundant members
4.0 PROCEDURE
3
BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2
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
BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2
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
BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2
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
BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2
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
BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2
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
BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2
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
BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2
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
BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2
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
BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2
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
BFC 31701 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS GROUP 2
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
BFC 3051 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS
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)
BFC 3051 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS
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
BFC 3051 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS
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.
BFC 3051 FORCE IN A STATICALLY INDETERMINATE CANTILEVER 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
BFC 3051 FORCE IN A STATICALLY INDETERMINATE CANTILEVER TRUSS
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