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No of Test - P Length (MM) I (MM) E (N/MM ) Mass (KG) Newto N (N) L (MM) L (MM )

1) Calculations were shown to determine the Young's modulus of an unknown material through a bending test. The average Young's modulus calculated was 5353.04 N/mm2. 2) Reactions and internal forces were calculated for a truss under a loading condition. The reaction forces were determined to be 12.26 N and 24.53 N. 3) Deflections of each member of the truss were calculated using the internal force and geometry of the members. The total deflection of the truss was calculated to be 0.2592 mm.

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Muhamad Syafik
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27 views4 pages

No of Test - P Length (MM) I (MM) E (N/MM ) Mass (KG) Newto N (N) L (MM) L (MM )

1) Calculations were shown to determine the Young's modulus of an unknown material through a bending test. The average Young's modulus calculated was 5353.04 N/mm2. 2) Reactions and internal forces were calculated for a truss under a loading condition. The reaction forces were determined to be 12.26 N and 24.53 N. 3) Deflections of each member of the truss were calculated using the internal force and geometry of the members. The total deflection of the truss was calculated to be 0.2592 mm.

Uploaded by

Muhamad Syafik
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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CALCULATIONS

Determined Young Modulus :No


of
test
.
1
2
3
4
5

Mass
(Kg)
0.2

Length

I
(mm4)

(mm)
Newto
n (N)

1.962

L
(mm)
160

L
(mm)
409600
0

7.73
7.54
7.62
8.15
8.31

3.98
Average

PL
48 EI

E
(N/mm)

E=

5441.96
5579.09
5520.52
5161.51
5062.13
5353.04

P L
48 I

P = load applied
L = length of the satay stick
= deflection of the satay stick
I = moment of inertia.
Where,

Moment of Inertia, I
D2
I=
64
2

I=

(3)
64

I = 3.98 mm4
Therefore, the average E obtained is 5353.04 N/mm.

Determined the reaction of the truss :10

30

40

12.26

30

12.26

10

12.26 N

12.26

MB = 0

-(12.26)(0.3m) (12.26)(70) + RH (1.0) = 0


RH = 12.26N

fy = 0

- RB + 24.53 + 24.53 + 24.53 = 0


RB = 24.53N

fx = 0

Determined the deflection of the truss :=

F ' L
AE

where,
F' = internal force of each member
E = one unit load at point E
L = length of each member
A = area of each member
E = Youngs Modulus
= deflection
(Using Alternative Method and Unit Load Method)

Member
AC
AB
BD
BC
CE
CD
DF
DE
EF
EG
GI
GF
FI
FH

F' (N)
0
0
0
-12.26
-8.67
8.67
6.13
-6.13
17.34
-18.39
-18.39
0
0
18.39

E (N)
0
0
-1
0
0
0
-1
0
0
0
0
0
0
-1

L (mm)
141.42
100
100
100
141.42
141.42
200
200
282.84
200
200
200
282.84
200

A (mm)
7.069
7.069
7.069
7.069
7.069
7.069
7.069
7.069
7.069
7.069
7.069
7.069
7.069
7.069

E (N/mm)
5353.04
5353.04
5353.04
5353.04
5353.04
5353.04
5353.04
5353.04
5353.04
5353.04
5353.04
5353.04
5353.04
5353.04

(mm)
0
0
0
0
0
0
-0.0324
0
0
0
0
0
0
-0.0972

HJ
HI
IJ
IK
KN
KJ
JN
JL
LM
LO
LN
NO
OP
OM
MP

18.39
0
0
-18.39
-18.39
0
17.34
6.13
0
8.67
-6.13
-8.67
0
-12.26
0

-1
0
0
0
0
0
0
-1
-1
0
0
0
0
0
0

200
7.069
5353.04
200
7.069
5353.04
282.84
7.069
5353.04
200
7.069
5353.04
200
7.069
5353.04
200
7.069
5353.04
282.84
7.069
5353.04
200
7.069
5353.04
100
7.069
5353.04
141.42
7.069
5353.04
200
7.069
5353.04
141.42
7.069
5353.04
141.42
7.069
5353.04
100
7.069
5353.04
100
7.069
5353.04
Therefore the total deflection is

-0.0972
0
0
0
0
0
0
-0.0324
0
0
0
0
0
0
0
0.2592m
m

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