Home Work:1. Draw shear force and bending moment diagrams for the beam shown in figure.30kN2mВL10KN/mC10mHello sir, a pictureunderneath, 1 solvedthis, is it true or not,please, if it is in error,I wánt the samesource and my sourcelawsegment AB:-70VAB = 74- 10X%3DMAB= 74 x- 10X (*)RA 74MAB- 74X-5x30 KNEMA3 KN1830 *2- Re*1o +10* 1o *5SIment Bc:-RARex to = 60 + 5 002m.8mR.VBC= 74-30 – 1º (X)RA = 74Rc- 560Rc- 56 KNVBc- 44-1o XC =S.F-DEFyMBc = 74 X -30 (X-2)– 10 (X) (X)MB c - 74X- 30X+60 - 5 xMBc-44X-5X+60) segment Be:-segment AB:X= 0mVBC2.4m%3D65.6 kN.MRA-30-100 + 56 -02RA - +74 KNX = 2 MM.D= 24 KNMB C= 128 KNX 10 m: M Max =65.6194-10X-• → X= X-9.4TAB -74 KNo knX =4.4 m-56 KMMmax - 44( 4.4) – 5CY.45+60Mmax - 113.6-88+ 6.VBC=MAB=X 2 mMBC = 0 KN.MM Maxa 165.6 (N-m/VAB - 54 KNMAB128 KNm)

First calculate the reactions at the supports,

Answered: 1. Draw shear force and bending moment…

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Mechanical Engineering

1. Draw shear force and bending moment diagrams for the beam shown in figure. 30kN 2m B 10KN/m C 10m

Elements Of Electromagnetics

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Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Concept explainers

Combined Loading

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial forces, torsion, and so on, then such body is said to be under combined loading. A very practical example of combined loading is the crankshaft of an internal combustion engine (IC engine). The crankshaft of the IC engine is under high rpm, which is caused due to the torsional forces. Due to the presence of piston, counterbalances, and internal imbalances, the crankshaft experiences lateral forces and due to the rise in temperature, the crankshaft undergoes thermal expansion, which pulls the crankshaft along the axial direction and lateral direction. The presence of unbalanced loads along the periphery of the crankshaft also induces a bending moment.

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)

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