solid mechanics 2. The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 14mm² and is made from 6061-T6 aluminum. Determine the vertical deflection of the bar at Dwhen the distributed load is applied as shown in Figure 2.300 N/m1.5 m-2 mFigure 2-2 mD Average normal stress in an axial memberFƠNg AAverage direct shear stressVTavgAvAverage bearing stressσbAbAverage normal strain in an axial memberAL8ElongLLAhEla =hAxial deformationFLA,E,8 =AdorFLorAELIST OF FORMULABasic definition-Relationship between E, G, and vEG = 2(1 + v)Definition of allowable stresso failureFSoro allowFactor of safetyo failureFSor FSo actualTallowTfailureTactualHooke's law (one-dimensional)o = EeandT = GyPoisson's ratioElatElongfailureFS

Answered: Image /qna-images/answer/ebf0c45b-5edb-4e64-a99e-0eaf70e0ddc1.jpg

2. The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 14 mm² and is made from 6061-T6 aluminum. Determine the vertical deflection of the bar at D when the distributed load is applied as shown in Figure 2. 300 N/m 1.5m -2m- Figure 2 B -2 m. D

2. The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 14 mm² and is made from 6061-T6 aluminum. Determine the vertical deflection of the bar at D when the distributed load is applied as shown in Figure 2. 300 N/m 1.5m -2m- Figure 2 B -2 m. D

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

Mechanical Drawings

A mechanical drawing is described as a technical drawing created by engineers following standard layout conventions, terminology, interpretation, and many other factors. It is also known as an engineering drawing.

Question

solid mechanics

2. The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 14
mm² and is made from 6061-T6 aluminum. Determine the vertical deflection of the bar at D
when the distributed load is applied as shown in Figure 2.
300 N/m
1.5 m
-2 m
Figure 2
-2 m
D

Average normal stress in an axial member
F
ƠNg A
Average direct shear stress
V
Tavg
Av
Average bearing stress
σb
Ab
Average normal strain in an axial member
AL
8
Elong
L
L
Ah
Ela =
h
Axial deformation
FL
A,E,
8 =
Ad
or
FL
or
AE
LIST OF FORMULA
Basic definition
-
Relationship between E, G, and v
E
G = 2(1 + v)
Definition of allowable stress
o failure
FS
or
o allow
Factor of safety
o failure
FS
or FS
o actual
Tallow
Tfailure
Tactual
Hooke's law (one-dimensional)
o = Ee
and
T = Gy
Poisson's ratio
Elat
Elong
failure
FS

Branch of science that deals with the stationary and moving bodies under the influence of forces.

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