For the beam with shown loading, determine the deflection at C when E = 200 GPa,I = 26.9 × 106 mm*.

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, determine the deflection at C when E =
200 GPa,I = 26.9 × 106 mm*.
60 kN
C
В
A
1 т
1 m
Transcribed Image Text:For the beam with shown loading, determine the deflection at C when E = 200 GPa,I = 26.9 × 106 mm*. 60 kN C В A 1 т 1 m
TABLE OF EQUATIONS
Normal Stress
Bearing Stress
P
A
td
Shear Stress (Single)
Shear Stress (Double)
P
P
A
= 1
2A
Tangential Stress (Cylindrical)
PD Pr
O = =-
Longitudinal Stress (Cylindrical)
PD Pr
2t t
4t
2t
Stress on Spherical Vessels
Hooke's Law
PD Pr
O = Ee
4t
2t
Deformation
Axial Strain
PL
ΔΙ.
8 =
AE
E =
Generalized Hooke's Law
dy
E
Ex =
Dilatation
dy
Ey =
E
e = €x + €y + €z
E
E, = -v
E
E
Shear Strain
Shear Modulus
E
G =
2(1+ v)
y =
Thermal Strain
Thermal Deformation
8, = a(AT)L
Er = 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
IG
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
Second Moment-Area Theorem
d?y
ElT = M
tc/p = (area bet.C and D)
dx
Area of General Spandrel
bh
A =
n+1
Centroid of General Spandrel
b
n+2
Deflection in Simply Supported Beams
8 = 0,x - 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
Р Му
A I
Combined Axial and Lateral Loads
P. Mc
Omax,min = t-
O =--
Transformation of Plane Stress
cos 20 +Txy sin 20
Principal Stresses
2
Oz + dy Ox - 0y
2
dy =
cos 20 - Tzy sin 20
Omax, min=
2
7.
O- dy cin 20 + Tyy Cos 20
y =-
2
Maximum Shear Stress and Corresponding Angle
Angle of Principal Plane
2Txy
+ rảy
Tmax =
tan 20
tan 20, =
Coordinates of Center of Mohr's Circle
(0x+ dy
2.0)
2
Transcribed Image Text:TABLE OF EQUATIONS Normal Stress Bearing Stress P A td Shear Stress (Single) Shear Stress (Double) P P A = 1 2A Tangential Stress (Cylindrical) PD Pr O = =- Longitudinal Stress (Cylindrical) PD Pr 2t t 4t 2t Stress on Spherical Vessels Hooke's Law PD Pr O = Ee 4t 2t Deformation Axial Strain PL ΔΙ. 8 = AE E = Generalized Hooke's Law dy E Ex = Dilatation dy Ey = E e = €x + €y + €z E E, = -v E E Shear Strain Shear Modulus E G = 2(1+ v) y = Thermal Strain Thermal Deformation 8, = a(AT)L Er = 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 IG 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 Second Moment-Area Theorem d?y ElT = M tc/p = (area bet.C and D) dx Area of General Spandrel bh A = n+1 Centroid of General Spandrel b n+2 Deflection in Simply Supported Beams 8 = 0,x - 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 Р Му A I Combined Axial and Lateral Loads P. Mc Omax,min = t- O =-- Transformation of Plane Stress cos 20 +Txy sin 20 Principal Stresses 2 Oz + dy Ox - 0y 2 dy = cos 20 - Tzy sin 20 Omax, min= 2 7. O- dy cin 20 + Tyy Cos 20 y =- 2 Maximum Shear Stress and Corresponding Angle Angle of Principal Plane 2Txy + rảy Tmax = tan 20 tan 20, = Coordinates of Center of Mohr's Circle (0x+ dy 2.0) 2
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