Draw the shear force diagrams. For each of the diagram, use the full span of the beam and not the cut position load scenario Acul3.92 N-AY140mm* D'YTomRAshear force.SFul-3.92+5.9493.SFut-1.8293 NRRA300mmload scenario B.cutRA196N 3.92220mm 1260min300mmload scenario DcutIRB240mm43/N 392400mm300mmRBRBAs length of beam 440(140+300)cut 4 point B at end & collapseTaking moment about BEMB=0-3.92 (440) + RA (300) = 0RA= 5.7493 N.Z Fy = 0.RA+RB - 3.92.RB 182932ReactionsΣΜΑΞΟ-RB (440) + 3-92 (260) +1.96 (220)=6RB 3.2963 N.shear forceSFcut = RB -SFcut=3.2963 N'ReacronsEMARB (446) + 392 (400) + 4.91 (240)=0RB = 6-2418 N.- shear force.SFcutSFcut= 6.2418- 3.92.2.32. N

Note:- As per our guidelines we are supposed to answer only one question. Kindly repost other…

load scenario A cut 3.92 N -Y 140mm * D'Y RA 300mm shear force. SFul-3.92+5.9493. SFeut-1.8293 N IRB → load sse ri R. As length of beam 440 (140+300) cut 4 point B at end & collapse Taking moment about B EMB=0 -3.92 (440) + RA (300) = 0 RA= 5.7493 N. Z Fy = 0. RA+RB- 3.92. A RB 18293 =

load scenario A cut 3.92 N -Y 140mm * D'Y RA 300mm shear force. SFul-3.92+5.9493. SFeut-1.8293 N IRB → load sse ri R. As length of beam 440 (140+300) cut 4 point B at end & collapse Taking moment about B EMB=0 -3.92 (440) + RA (300) = 0 RA= 5.7493 N. Z Fy = 0. RA+RB- 3.92. A RB 18293 =

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

Slope and Deflection

The slope in a deflected beam is described as an angle in radians made by the tangent of the section with the original axis of the beam. The deflection at any point on the axis of the beam is defined as the lateral displacement of the point before and after loading.

Question

Draw the shear force diagrams. For each of the diagram, use the full span of the beam and not the cut position

load scenario A
cul
3.92 N
-AY
140mm
* D'Y
Tom
RA
shear force.
SFul-3.92+5.9493.
SFut-1.8293 N
RRA
300mm
load scenario B.
cut
RA
196N 3.92
220mm 1
260min
300mm
load scenario D
cut
IRB
240mm
43/N 392
400mm
300mm
RB
RB
As length of beam 440
(140+300)
cut 4 point B at end & collapse
Taking moment about B
EMB=0
-3.92 (440) + RA (300) = 0
RA= 5.7493 N.
Z Fy = 0.
RA+RB - 3.92.
RB 18293
2
Reactions
ΣΜΑΞΟ
-RB (440) + 3-92 (260) +1.96 (220)=6
RB 3.2963 N.
shear force
SFcut = RB -
SFcut
=
3.2963 N'
Reacrons
EMA
RB (446) + 392 (400) + 4.91 (240)=0
RB = 6-2418 N.
- shear force.
SFcut
SFcut
= 6.2418- 3.92.
2.32. N

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