1. Draw shear force and bending moment diagrams for the beam shown in figure. 30kN 2m B 10KN/m C 10m
1. Draw shear force and bending moment diagrams for the beam shown in figure. 30kN 2m B 10KN/m C 10m
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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Question
![Home Work:
1. Draw shear force and bending moment diagrams for the beam shown in figure.
30kN
2m
В
L10KN/m
C
10m
Hello sir, a picture
underneath, 1 solved
this, is it true or not,
please, if it is in error,
I wánt the same
source and my source
law
segment AB:-
70
VAB = 74- 10X
%3D
MAB= 74 x- 10X (*)
RA 74
MAB- 74X-5x
30 KN
EMA
3 KN
18
30 *2- Re*1o +
10* 1o *5
SIment Bc:-
RA
Rex to = 60 + 5 00
2m.
8m
R.
VBC= 74-30 – 1º (X)
RA = 74
Rc- 560
Rc- 56 KN
VBc- 44-1o X
C =
S.F-D
EFy
MBc = 74 X -30 (X-2)– 10 (X) (X)
MB c - 74X- 30X+60 - 5 x
MBc-44X-5X+60) segment Be:-
segment AB:X= 0mVBC
2.4m
%3D
65.6 kN.M
RA-30-100 + 56 -0
2
RA - +74 KN
X = 2 M
M.D
= 24 KN
MB C= 128 KN
X 10 m
: M Max =65.6194-10X-• → X= X-9.4
TAB -74 KN
o kn
X =4.4 m
-56 KM
Mmax - 44( 4.4) – 5CY.45+60
Mmax - 113.6-88+ 6.
VBC=
MAB=
X 2 m
MBC = 0 KN.M
M Maxa 165.6 (N-m/
VAB - 54 KN
MAB
128 KNm)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2a89ebd-de96-4375-bbbf-a6b4128fe997%2Fc5a2a027-d6e7-496c-8f05-085f845c3e81%2F7deyup8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Home Work:
1. Draw shear force and bending moment diagrams for the beam shown in figure.
30kN
2m
В
L10KN/m
C
10m
Hello sir, a picture
underneath, 1 solved
this, is it true or not,
please, if it is in error,
I wánt the same
source and my source
law
segment AB:-
70
VAB = 74- 10X
%3D
MAB= 74 x- 10X (*)
RA 74
MAB- 74X-5x
30 KN
EMA
3 KN
18
30 *2- Re*1o +
10* 1o *5
SIment Bc:-
RA
Rex to = 60 + 5 00
2m.
8m
R.
VBC= 74-30 – 1º (X)
RA = 74
Rc- 560
Rc- 56 KN
VBc- 44-1o X
C =
S.F-D
EFy
MBc = 74 X -30 (X-2)– 10 (X) (X)
MB c - 74X- 30X+60 - 5 x
MBc-44X-5X+60) segment Be:-
segment AB:X= 0mVBC
2.4m
%3D
65.6 kN.M
RA-30-100 + 56 -0
2
RA - +74 KN
X = 2 M
M.D
= 24 KN
MB C= 128 KN
X 10 m
: M Max =65.6194-10X-• → X= X-9.4
TAB -74 KN
o kn
X =4.4 m
-56 KM
Mmax - 44( 4.4) – 5CY.45+60
Mmax - 113.6-88+ 6.
VBC=
MAB=
X 2 m
MBC = 0 KN.M
M Maxa 165.6 (N-m/
VAB - 54 KN
MAB
128 KNm)
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