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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Answered: HELLO SIR, COULD YOU SOLVE THIS TABLE…

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

Problem 1.1MA

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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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