HELLO SIR, COULD YOUSOLVE THIS TABLE TOPICABOUT STRENGTH OFMATERIALS LABORATORY?BENDING TESTSECOND STAGEExample: Copper MaterialDistance between anchor= 100 mmL= Imh = 6.5mm, b 25.5 mmI= bh/12 = 583,6 mmY = h/2 = 6.5/2MassF (N)Moment02 (N/m)=E (N/m?)1/RNo.R (m)(gm)(N.m)(mm)(m")5.463*10 14.008*1010 1.04*1010.926* 10 11.06 *10101000.9810.09810.13961.23822001.9620.19620,38328.9473000.5344000,75001.076.6001.2977001.498.8001.799001.891010002.165. CALCULATIONRWhere:M- MomentI- Area Moment of Inertia (kg/m)-x (m)12o Stress (MPA)ye= Thickness h/2 (mm)E- Young's ModulusR- 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 ofthe bar material). By knowing the magnitude of the effective moment (M) and themoment 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 approximationaccording to the following figure:RR-8L/2From the above triangle and according to Pythagoras law:R' -A+ (R-8) .2LE+R -2 x 8xR+8 .3RWhile & is a small number, then; we can neglect (8) from other side and theequation will be come:2xRx8 = L.4LR =8x8Now as we can calculate (I, M, R) then we can also calculate Young's Modulus forthe material.By return to Equation (1) we find that:Ex yR.7The value of (6) resulting from the two equations (6 and 7) proves the validity ofthe equation (1).

Answered: Image /qna-images/answer/e8807e07-d024-4733-a2de-079d7451daca.jpg

SOLVE THIS TABLE TOPIC ABOUT STRENGTH OF MATERIALS LABORATORY ?BENDING TEST SECOND STAGE Example: Copper Material Distance between anchor= 100 mm L= Im h=6.5mm, b 25.5 mm I= bh/12 = 583.6 mm Y = h/2 = 6.5/2

SOLVE THIS TABLE TOPIC ABOUT STRENGTH OF MATERIALS LABORATORY ?BENDING TEST SECOND STAGE Example: Copper Material Distance between anchor= 100 mm L= Im h=6.5mm, b 25.5 mm I= bh/12 = 583.6 mm Y = h/2 = 6.5/2

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Combined Loading

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial forces, torsion, and so on, then such body is said to be under combined loading. A very practical example of combined loading is the crankshaft of an internal combustion engine (IC engine). The crankshaft of the IC engine is under high rpm, which is caused due to the torsional forces. Due to the presence of piston, counterbalances, and internal imbalances, the crankshaft experiences lateral forces and due to the rise in temperature, the crankshaft undergoes thermal expansion, which pulls the crankshaft along the axial direction and lateral direction. The presence of unbalanced loads along the periphery of the crankshaft also induces a bending moment.

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= Im
h = 6.5mm, b 25.5 mm
I= bh/12 = 583,6 mm
Y = h/2 = 6.5/2
Mass
F (N)
Moment
02 (N/m)=
E (N/m?)
1/R
No.
R (m)
(gm)
(N.m)
(mm)
(m")
5.463*10 14.008*1010 1.04*10
10.926* 10 11.06 *1010
100
0.981
0.0981
0.13
961.238
2
200
1.962
0.1962
0,38
328.947
300
0.53
4
400
0,7
500
1.07
6.
600
1.29
7
700
1.49
8.
800
1.7
9
900
1.89
10
1000
2.16
5. CALCULATION
R
Where:
M- Moment
I- Area Moment of Inertia (kg/m)-x (m)
12
o Stress (MPA)
ye= 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' -
A
+ (R-8) .2
L
E+R -2 x 8xR+8 .3
R
While & is a small number, then; we can neglect (8) from other side and the
equation will be come:
2xRx8 = L
.4
L
R =
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:
Ex y
R
.7
The value of (6) resulting from the two equations (6 and 7) proves the validity of
the equation (1).

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