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
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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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
Transcribed Image Text: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
Transcribed Image Text: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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