For the beam with shown loading, show that the equation of the elastic curve is Wo 7w,L3 + x3 y = EI 120L 36 360 В A L-

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Answer the problem with detailed step by step solution refer to the table of equations when answering the problem.
For the beam with shown loading, show that the equation of the elastic curve is
1
Wo
7w,L3
y = El
x5
+
120L
36
360
В
A
L-
Transcribed Image Text:For the beam with shown loading, show that the equation of the elastic curve is 1 Wo 7w,L3 y = El x5 + 120L 36 360 В A L-
TABLE OF EQUATIONS
Normal Stress
Bearing Stress
P
A
td
Shear Stress (Single)
Shear Stress (Double)
P
= 1
2A
Tangential Stress (Cylindrical)
PD Pr
O = =T
Longitudinal Stress (Cylindrical)
PD Pr
O = 2t
4t
Stress on Spherical Vessels
Hooke's Law
PD Pr
Os =
4t
O = Ee
2t
Deformation
Axial Strain
PL
ΔΙ.
E =
AE
1-1-,
Generalized Hooke's Law
dy
E
Ex =
Dilatation
dy
Ey =
E
e = €x + €y + €z
E
dy
E, = -v
E
E
Shear Strain
Shear Modulus
E
G =
2(1+ v)
y =
Thermal Strain
Thermal Deformation
8, = a(AT)L
ET = aAT
Shear Stress due to Torsion
Shear Stress at any Point
Tc
Tmax =
T=-Tmax
Polar Moment of Inertia of Circle
Minimum Shear Stress (Hollow)
Tmin =-Tmax
C2
Angle of Twist
Polar Moment of Inertia of Hollow Circle
TL
JG
Shear Stress due to Torsion (Non-Circular Tubes)
Power
T
P = T
2tA
Angle of Twist (Non-Circular Tubes)
TL ( ds
Bending Stress
Ox = --om
4A?G
Maximum Bending Stress
Mc
Om =T
Elastic Section Modulus
Deflection of a Beam
d?y
Edr = M
Second Moment-Area Theorem
tc/p = (area bet.C and D)x
Area of General Spandrel
bh
A =
n+1
Centroid of General Spandrel
b
n+2
Deflection in Simply Supported Beams
8p = 0Ạx - tạ/A
tc/A
Deflection in Cantilever Beams
8R = (area of M/El diagram)Xg
Maximum and Minimum Stress in Beams with
Combined Axial and Lateral Loads in Beams
Combined Axial and Lateral Loads
P My
A I
P. Mc
Omax,min = t-
O =--
Transformation of Plane Stress
cos 20 +Txy sin 20
2
Principal Stresses
O + dy Ox - 0y
2
Ox-dy sin 20 + Tyy cos 20
2
cos 20 - Tzy sin 20
Omar, min=
2
Maximum Shear Stress and Corresponding Angle
Angle of Principal Plane
+ Tây
Tmax =
21.xy_
O- dy
2Txy
tan 20
tan 20,
%3D
Coordinates of Center of Mohr's Circle
(*.0)
(0x+ dy
2
Transcribed Image Text:TABLE OF EQUATIONS Normal Stress Bearing Stress P A td Shear Stress (Single) Shear Stress (Double) P = 1 2A Tangential Stress (Cylindrical) PD Pr O = =T Longitudinal Stress (Cylindrical) PD Pr O = 2t 4t Stress on Spherical Vessels Hooke's Law PD Pr Os = 4t O = Ee 2t Deformation Axial Strain PL ΔΙ. E = AE 1-1-, Generalized Hooke's Law dy E Ex = Dilatation dy Ey = E e = €x + €y + €z E dy E, = -v E E Shear Strain Shear Modulus E G = 2(1+ v) y = Thermal Strain Thermal Deformation 8, = a(AT)L ET = aAT Shear Stress due to Torsion Shear Stress at any Point Tc Tmax = T=-Tmax Polar Moment of Inertia of Circle Minimum Shear Stress (Hollow) Tmin =-Tmax C2 Angle of Twist Polar Moment of Inertia of Hollow Circle TL JG Shear Stress due to Torsion (Non-Circular Tubes) Power T P = T 2tA Angle of Twist (Non-Circular Tubes) TL ( ds Bending Stress Ox = --om 4A?G Maximum Bending Stress Mc Om =T Elastic Section Modulus Deflection of a Beam d?y Edr = M Second Moment-Area Theorem tc/p = (area bet.C and D)x Area of General Spandrel bh A = n+1 Centroid of General Spandrel b n+2 Deflection in Simply Supported Beams 8p = 0Ạx - tạ/A tc/A Deflection in Cantilever Beams 8R = (area of M/El diagram)Xg Maximum and Minimum Stress in Beams with Combined Axial and Lateral Loads in Beams Combined Axial and Lateral Loads P My A I P. Mc Omax,min = t- O =-- Transformation of Plane Stress cos 20 +Txy sin 20 2 Principal Stresses O + dy Ox - 0y 2 Ox-dy sin 20 + Tyy cos 20 2 cos 20 - Tzy sin 20 Omar, min= 2 Maximum Shear Stress and Corresponding Angle Angle of Principal Plane + Tây Tmax = 21.xy_ O- dy 2Txy tan 20 tan 20, %3D Coordinates of Center of Mohr's Circle (*.0) (0x+ dy 2
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