40 kips/ft 40 kip A 1.5 + 1.5 ft
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Show detailed solution refer to the table of equations.

Transcribed Image Text:For the simply supported beam with shown loading, determine the deflection at
midspan. (EI = 750 kips · ft²)
40 kips/ft
40 kips/ft
В
A
1.5 ft_,
1.5 ft
-4 ft-

Transcribed Image Text:TABLES OF EQUATIONS
Normal Stress
Bearing Stress
td
Shear Stress (Single)
Shear Stress (Double)
P
2A
Tangential Stress (Cylindrical)
Longitudinal Stress (Cylindrical)
PD
PD
2t
4t
Stress on Spherical Vessels
Hooke's Law
PD
O, =
4t
o = Ee
Deformation
Axial Strain
PL
8 =
AE
Generalized Hooke's Law
Oy
V下
Ex
E
Dilatation
dy
Ey =
E
e = e, +€y +6
E
E, =
Shear Strain
Shear Modulus
E
G =
2(1+ v)
Thermal Deformation
Thermal Strain
ốr = a(AT)L
ET = aAT
Shear Stress due to Torsion
T'c
TmaxT
Shear Stress at any Point
T=Tmar
Polar Moment of Inertia of Circle
Minimum Shear Stress (Hollow)
Tmin=
-Tmax
Polar Moment of Inertia of Hollow Circle
Angle of Twist
TL
JG
Shear Stress due to Torsion (Non-Circular Tubes)
T
Power
P = Ta
2tA
Angle of Twist (Non-Circular Tubes)
TL ( ds
Bending Stress
Maximum Bending Stress
Mc
Om =T
Elastic Section Modulus
Deflection of a Beam
Second Moment-Area Theorem
d'y
dr
El
= M
tC/p = (area bet. C and D)X,
Area of General Spandrel
bh
A =
Centroid of General Spandrel
b
c=+2
n+1
Deflection in Simply Supported Beams
Deflection in Cantilever Beams
8g = (area of M/El diagram)Xg
tCIA
Maximum and Minimum Stress in Beams with
Combined Axial and Lateral Loads in Beams
Combined Axial and Lateral Loads
P My
AT
P. Mc
Gmazmin =7+T
Transformation of Plane Stress
Ox + dy, O dy
2
cos 20 + Tyy Sin 20
Principal Stresses
2
cos 20 -Ty sin 20
2
d, + dy
+ Tầy
Omax.min D
2
6-dy sin 20 + ry cos 20
sin 20 + Txy Cos 20
2
Maximum Shear Stress and Corresponding Angle
Angle of Principal Plane
Tmax =
+ Ty
tan 20, =
tan 28, = --0,
Coordinates of Center of Mohr's Circle
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