A simply supported beam is made of A-36 steel and is subjected to the loadingshown below. Let P = 20.2 kN, I = 145.7 × 10⁰ mmª, and E = 200 GPa. You are required to usethe Method of Superposition to solve thisproblem.(a) Find the deflection at point C of the beam(in mm).(b) Find the slope at end A of the beam (indegrees).5mP4 kN/m5 mB EQUATIONS FOR DEFLECTIONS OF SIMPLY SUPPORTED BEAMSSIMPLY SUPPORTED BEAMSLENGTH, LAyAAMYdydxdydydxMaxL/2dydxL/2F'YM₂Max9bU/2 FdydxY Masdy(3).L/2dxdxBdiadydxCANTILEVER BEAMSLENGTH, Ldydx/uwindydxBBAnYst*7%LENNETУмMo(dyУмиSLOPEdx/Maxdy -Fab(L+b)(dx6EI L(dy)dxdy(dx)dydx/dydx/dydx=dydx=Fab(L+a)6EI L=--dy3q, L'dx), 128EIdy 7q,1(dx) 384E1,(dydx-FL²16EIdyML6EI,==9.L²dx 45EI,MLЗЕІ,aMax9.1.²24EI,MaSLOPEMax-79 L'360EIX dy =9,L¹dx/stas6EL,dy -FL²dxSEI,Max-FL²2EI,MLEI,9.L²48EI-Fxy= (3L²-4x²)48EIУмиEQUATIONS FOR DEFLECTIONS OF CANTILEVER BEAMSy=Умакy=y=Y 6EL,Lу"DEFLECTIONEQUATIONS-Mxy=- -(x²-3Lx+21²)6EI,L-ML-²√243EI atx=0.4226L-9x(x²-2Lx² +1²)24EI,УмитУман|y=-Fbx6EI,LУмаy=-FL²48EL,y=(16x-24Lx² +9L')384EI-9.L384E1,+17L²x-L¹) L/2≤x≤L-Fba (1-b²-a¹)360EI,L(L-b²-x²)0≤x≤aУманy=-0≤x≤L/2-Sq₂L¹384EI,(8x³-24Lx¹-6.563x10¹q LEI,-9x(3x-10L²x² +7L')-Fx².6EI-6.52 x 10 ³q,L*EI,Mechanics of Materials - 4550≤x≤L/2DEFLECTIONEQUATIONS-FL²ЗЕІ,-(3L-x)-9₁x²24EI,Y =at x= 0.4598L-Fx²y=- (2-x)6EI,y=-9₁x²x²24EIу-FL²3x-1-1)24EIy =at x 0.5193-9,L¹SEI(x²-4Lx+6L²)Mx²2EI₂-9²²² (₁ML²Уми "2EI,x²-21x4+0≤x≤L/2LL/2≤x≤L0≤x≤L/2

The applied force is The moment of inertia of beam is The elasticity of beam is

A simply supported beam is made of A-36 steel and is subjected to the loading shown below. Let P = 20.2 kN, I = 145.7 x 106 mm, and E= 200 GPa. You are required to use the Method of Superposition to solve this problem. (a) Find the deflection at point C of the beam (in mm). (b) Find the slope at end A of the beam (in degrees). 5 m P 4 kN/m Imtion 5m

A simply supported beam is made of A-36 steel and is subjected to the loading shown below. Let P = 20.2 kN, I = 145.7 x 106 mm, and E= 200 GPa. You are required to use the Method of Superposition to solve this problem. (a) Find the deflection at point C of the beam (in mm). (b) Find the slope at end A of the beam (in degrees). 5 m P 4 kN/m Imtion 5m

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

A simply supported beam is made of A-36 steel and is subjected to the loading
shown below. Let P = 20.2 kN, I = 145.7 × 10⁰ mmª, and E = 200 GPa. You are required to use
the Method of Superposition to solve this
problem.
(a) Find the deflection at point C of the beam
(in mm).
(b) Find the slope at end A of the beam (in
degrees).
5m
P
4 kN/m
5 m
B

EQUATIONS FOR DEFLECTIONS OF SIMPLY SUPPORTED BEAMS
SIMPLY SUPPORTED BEAMS
LENGTH, L
A
y
A
A
M
Y
dy
dx
dy
dy
dx
Max
L/2
dy
dx
L/2
F
'YM₂
Max
9
b
U/2 F
dy
dx
Y Mas
dy
(3).
L/2
dx
dx
B
dia
dy
dx
CANTILEVER BEAMS
LENGTH, L
dy
dx/u
win
dy
dx
B
B
An
Yst
*7
%
LENNET
Ум
Mo
(dy
Уми
SLOPE
dx/Max
dy -Fab(L+b)
(dx
6EI L
(dy)
dx
dy
(dx)
dy
dx/
dy
dx/
dy
dx
=
dy
dx
=
Fab(L+a)
6EI L
=--
dy
3q, L'
dx), 128EI
dy 7q,1
(dx) 384E1,
(dy
dx
-FL²
16EI
dy
ML
6EI,
=
=
9.L²
dx 45EI,
ML
ЗЕІ,
a
Max
9.1.²
24EI,
Ma
SLOPE
Max
-79 L'
360EI
X dy =9,L¹
dx/stas
6EL,
dy -FL²
dx
SEI,
Max
-FL²
2EI,
ML
EI,
9.L²
48EI
-Fx
y= (3L²-4x²)
48EI
Уми
EQUATIONS FOR DEFLECTIONS OF CANTILEVER BEAMS
y=
Умак
y=
y=
Y 6EL,L
у"
DEFLECTION
EQUATIONS
-Mx
y=- -(x²-3Lx+21²)
6EI,L
-ML-²
√243EI atx=0.4226L
-9x(x²-2Lx² +1²)
24EI,
Умит
Уман
|y=
-Fbx
6EI,L
Ума
y=
-FL²
48EL,
y=(16x-24Lx² +9L')
384EI
-9.L
384E1,
+17L²x-L¹) L/2≤x≤L
-Fba (1-b²-a¹)
360EI,L
(L-b²-x²)
0≤x≤a
Уман
y=-
0≤x≤L/2
-Sq₂L¹
384EI,
(8x³-24Lx¹
-6.563x10¹q L
EI,
-9x(3x-10L²x² +7L')
-Fx².
6EI
-6.52 x 10 ³q,L*
EI,
Mechanics of Materials - 455
0≤x≤L/2
DEFLECTION
EQUATIONS
-FL²
ЗЕІ,
-(3L-x)
-9₁x²
24EI,
Y =
at x= 0.4598L
-Fx²
y=
- (2-x)
6EI,
y=-9₁x²x²
24EI
у
-FL²
3x-1-1)
24EI
y =
at x 0.5193
-9,L¹
SEI
(x²-4Lx+6L²)
Mx²
2EI₂
-9²²² (₁
ML²
Уми "2EI,
x²-21x4
+
0≤x≤L/2
L
L/2≤x≤L
0≤x≤L/2

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Step 2: Calculation of deflection of beam at C and slope at A for distributed load

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Step 3: Calculation of deflection of beam at C and slope at A for point load at C

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Step 4: Calculation of net slope at A and net deflection at C

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