Example: Copper Material Distance between anchor= 100 mm L = Im h = 6.5mm, b = 25.5 mm I bh³/12=583.6 mm Y=h/2-6.5/2 Mass Moment 8 (gm) F (N) (N.m) 100 0.981 0.0981 200 1.962 0.1962 300 400 500 600 700 800 900 1000 No. 1 2 3 4 5 6 7 8 9 10 (mm) 0.13 0.38 0.53 0.7 1.07 1.29 1.49 1.7 1.89 2.16 R (m) 02 (N/m²) 01 5.463 10 328.947 10.926 10 961.238 E (N/m²) 14.008 10¹0 11.06 1010 1/R (m¹) 1.04 10
Example: Copper Material Distance between anchor= 100 mm L = Im h = 6.5mm, b = 25.5 mm I bh³/12=583.6 mm Y=h/2-6.5/2 Mass Moment 8 (gm) F (N) (N.m) 100 0.981 0.0981 200 1.962 0.1962 300 400 500 600 700 800 900 1000 No. 1 2 3 4 5 6 7 8 9 10 (mm) 0.13 0.38 0.53 0.7 1.07 1.29 1.49 1.7 1.89 2.16 R (m) 02 (N/m²) 01 5.463 10 328.947 10.926 10 961.238 E (N/m²) 14.008 10¹0 11.06 1010 1/R (m¹) 1.04 10
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
![HELLO SIR, COULD YOU
SOLVE THIS TABLE TOPIC
ABOUT STRENGTH OF
MATERIALS LABORATORY
?BENDING TEST
SECOND STAGE
Example: Copper Material
Distance between anchor= 100 mm
L=1m
h = 6.5mm, b = 25.5 mm
I=bh³/12=583.6 mm*
Y=h/2= 6.5/2
Mass
1/R
F (N)
R (m)
02 (N/m²)
01
E (N/m²)
(gm)
(m¹)
100
0.981
961.238 5.463 105
328.947 10.926*10
14.00810¹0 1.04 10³
11.06*10¹0
200
1.962
300
400
COO
500
600
7
700
8
800
9
900
10
1000
5. CALCULATION
M
E
==
R
Where:
M- Moment
I= Area Moment of Inertia (kg/m³)=bxh (m³)
12
o Stress (MPA)
y c Thickness-h/2 (mm)
E-Young's Modulus
R- Radius of Curvature (mm)
If an arm is subjected to a bending moment, we can use Equation (1) to calculate
(bending stress, bending amount, bend radius of the bar, modulus of elasticity of
the bar material). By knowing the magnitude of the effective moment (M) and the
moment of inertia of the rod section (I), and in the given case in the experiment.
the radius of the rod curve (R) can be calculated with an acceptable approximation
according to the following figure:
R.
R-8
L/2
From the above triangle and according to Pythagoras law:
R =
- (2)² + (R-SF..
.................2
-2x8xR+8².................. 3
1²
R² =
4
While 8 is a small number, then; we can neglect (82) from other side and the
equation will be come:
L²
2xRx8=
4
R=L²
8x8
Now as we can calculate (I, M, R) then we can also calculate Young's Modulus for
the material.
By return to Equation (1) we find that:
Exy
R
Mxy
I
σ. =
.7
The value of (6) resulting from the two equations (6 and 7) proves the validity of
the equation (1).
No.
1
2
3
4
5
6
Moment 8
(N.m) (mm)
0.0981
0.13
0.38
0.1962
0.53
0.7
1.07
1.29
1.49
1.7
1.89
2.16](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F526088ef-969f-40cf-b1e7-a9ca6278b9a3%2F94255d29-caf6-4f0a-b00d-d95f07900e4d%2Fujlugx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:HELLO SIR, COULD YOU
SOLVE THIS TABLE TOPIC
ABOUT STRENGTH OF
MATERIALS LABORATORY
?BENDING TEST
SECOND STAGE
Example: Copper Material
Distance between anchor= 100 mm
L=1m
h = 6.5mm, b = 25.5 mm
I=bh³/12=583.6 mm*
Y=h/2= 6.5/2
Mass
1/R
F (N)
R (m)
02 (N/m²)
01
E (N/m²)
(gm)
(m¹)
100
0.981
961.238 5.463 105
328.947 10.926*10
14.00810¹0 1.04 10³
11.06*10¹0
200
1.962
300
400
COO
500
600
7
700
8
800
9
900
10
1000
5. CALCULATION
M
E
==
R
Where:
M- Moment
I= Area Moment of Inertia (kg/m³)=bxh (m³)
12
o Stress (MPA)
y c Thickness-h/2 (mm)
E-Young's Modulus
R- Radius of Curvature (mm)
If an arm is subjected to a bending moment, we can use Equation (1) to calculate
(bending stress, bending amount, bend radius of the bar, modulus of elasticity of
the bar material). By knowing the magnitude of the effective moment (M) and the
moment of inertia of the rod section (I), and in the given case in the experiment.
the radius of the rod curve (R) can be calculated with an acceptable approximation
according to the following figure:
R.
R-8
L/2
From the above triangle and according to Pythagoras law:
R =
- (2)² + (R-SF..
.................2
-2x8xR+8².................. 3
1²
R² =
4
While 8 is a small number, then; we can neglect (82) from other side and the
equation will be come:
L²
2xRx8=
4
R=L²
8x8
Now as we can calculate (I, M, R) then we can also calculate Young's Modulus for
the material.
By return to Equation (1) we find that:
Exy
R
Mxy
I
σ. =
.7
The value of (6) resulting from the two equations (6 and 7) proves the validity of
the equation (1).
No.
1
2
3
4
5
6
Moment 8
(N.m) (mm)
0.0981
0.13
0.38
0.1962
0.53
0.7
1.07
1.29
1.49
1.7
1.89
2.16
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