HELLO SIR, COULD YOUSOLVE THIS TABLE TOPICABOUT STRENGTH OFMATERIALS LABORATORY?BENDING TESTSECOND STAGEAL ESRAAExample: Copper MaterialDistance between anchor= 100 mmL = 1mh = 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)11000.9810.09810.13961.2385.463*10514.008*1010 1.04*10322001.9620.19620.38328.94710.926*105 11.06 *101033000.5344000.75001.076.6001.2977001.4988001.79.9001.891010002.16

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Engineering

Mechanical Engineering

HELLO SIR, COULD YOU SOLVE THIS TABLE TOPIC ABOUT STRENGTH OF MATERIALS LABORATORY ?BENDING TEST

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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Concept explainers

Thermal Properties of Materials

In mechanical engineering, the thermal property is represented as the physical quantity that is relevant to the application of heat. The thermal properties are evaluated by variations in any specific material due to low heat or high heat. The quantities that influence the thermal property of the materials are mass, density, temperature, and many others.

Question

HELLO SIR, COULD YOU
SOLVE THIS TABLE TOPIC
ABOUT STRENGTH OF
MATERIALS LABORATORY
?BENDING TEST
SECOND STAGE
AL ESRAA
Example: Copper Material
Distance between anchor= 100 mm
L = 1m
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)
1
100
0.981
0.0981
0.13
961.238
5.463*105
14.008*1010 1.04*103
2
200
1.962
0.1962
0.38
328.947
10.926*105 11.06 *1010
3
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

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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…
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