A rigid plate C is used to transfer a 20-kip load P to a steel (E = 30,000 ksi) rod A and to an aluminum alloy (E = 10,000 ksi) pipe B as shown. The supports at the top of the rod and the bottom of the pipe are rigid, and the materials were unstressed before the load was applied. The cross-sectional areas of rods A and pipe B are 0.80 in2 and 3.00 in2 respectively. Determine The normal stress in rod A in ksi, The normal stress in pipe B in ksi and The displacement of plate C in inches. Note: Draw the Free Body Diagram, include the units/dimensions, use the proper formula and round-off all the answers and final answers to 5 decimal places. A10 in.PP22В20 in. Fundamental Equations of Mechanics of MaterialsAxial LoadShearNormal StressAverage direct shear stressVATavgADisplacementP(x)dxTransverse shear stress8 =A (x)EVQItPLAEShear flowVQq = Tt =δ- α ΔΤL= aTorsionStress in Thin-Walled Pressure VesselShear stress in circular shaftТрCylinderprpr2twhereSpherepr= solid cross section01 = 02 =2tJ =- (c,“ – c,")Stress Transformation Equationstubular cross section0x + 0yO, – 0,+ycos 20 +Powersin 202P = Tw = 2™fTOx – 0yAngle of twistsin 20 + Try cos 202T(x)dxPrincipal StressJ(x)GTxytan 26,TL$ = EJG(ox – 0,)/2Ox + 0y1,20,- 0,yAverage shear stress in a thin-walled tube+ Tây2TTavg21A mMaximum in-plane shear stress(0, – 0,)/2Shear Flowtan 20,TTxyq = Tavg!2A mTmax+ Txy2BendingOx + 0yO avgNormal stressМyAbsolute maximum shear stressIσ.maxUnsymmetric bendingTabsmaxfor ởmax, O min same sign2MạymaxỞ mintan aTabsmaxfor omax, Omin Opposite signstan 02Geometric Properties of Area ElementsMaterial Property RelationsPoisson's ratio-A = bhElat1, = bh!, = hblongC%3DGeneralized Hooke's Law:[0, - vo, + o.)]Rectangular area%3DExE[0, – v(o, + 0,)]-A = bhEEz[0. - v(0, + 0,)]E1Tyz, Yzx1%3DYxyTxy YyzTriangular areawhereEG =2(1 + v)A =a(9 + D)w :Relations Between w, V, Mh(2a + b\a + bdM= VdxdVbw(x),dxTrapezoidal areaElastic Curve1M-A =EId*vEI=w(x)!, = }ar*%3Dd'vEl= V(x)Semicircular areadvEldx?M(x)-A = Tr²BucklingCritical axial load1, = fart!, = fartT²EIPer(KL)²Critical stressVIJAO crCircular area(KL/r)²*Secant formulaесPỞ max1 +secAV EAžabA =Energy MethodsConservation of energyU. = U;zero slopeStrain energyN²LSemiparabolic areaU; =2AEconstant axial load"M²dxU, =bending moment%3D2EIA =U; =transverse shear%3D2GAzero slope -pLT°dxtorsional moment%3D2GJExparabolic area

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Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

A rigid plate C is used to transfer a 20-kip load P to a steel (E = 30,000 ksi) rod A and to an aluminum alloy (E = 10,000 ksi) pipe B as shown. The supports at the top of the rod and the bottom of the pipe are rigid, and the materials were unstressed before the load was applied. The cross-sectional areas of rods A and pipe B are 0.80 in² and 3.00 in² respectively. Determine The normal stress in rod A in ksi, The normal stress in pipe B in ksi and The displacement of plate C in inches.

Note: Draw the Free Body Diagram, include the units/dimensions, use the proper formula and round-off all the answers and final answers to 5 decimal places.

Fundamental Equations of Mechanics of Materials
Axial Load
Shear
Normal Stress
Average direct shear stress
V
A
Tavg
A
Displacement
P(x)dx
Transverse shear stress
8 =
A (x)E
VQ
It
PL
AE
Shear flow
VQ
q = Tt =
δ- α ΔΤL
= a
Torsion
Stress in Thin-Walled Pressure Vessel
Shear stress in circular shaft
Тр
Cylinder
pr
pr
2t
where
Sphere
pr
= solid cross section
01 = 02 =
2t
J =
- (c,“ – c,")
Stress Transformation Equations
tubular cross section
0x + 0y
O, – 0,
+
y
cos 20 +
Power
sin 20
2
P = Tw = 2™fT
Ox – 0y
Angle of twist
sin 20 + Try cos 20
2
T(x)dx
Principal Stress
J(x)G
Txy
tan 26,
TL
$ = E
JG
(ox – 0,)/2
Ox + 0y
1,2
0,- 0,
y
Average shear stress in a thin-walled tube
+ Tây
2
T
Tavg
21A m
Maximum in-plane shear stress
(0, – 0,)/2
Shear Flow
tan 20,
T
Txy
q = Tavg!
2A m
Tmax
+ Txy
2
Bending
Ox + 0y
O avg
Normal stress
Мy
Absolute maximum shear stress
I
σ.
max
Unsymmetric bending
Tabs
max
for ởmax, O min same sign
2
Mạy
max
Ở min
tan a
Tabs
max
for omax, Omin Opposite signs
tan 0
2
Geometric Properties of Area Elements
Material Property Relations
Poisson's ratio
-A = bh
Elat
1, = bh
!, = hb
long
C
%3D
Generalized Hooke's Law
:[0, - vo, + o.)]
Rectangular area
%3D
Ex
E
[0, – v(o, + 0,)]
-A = bh
E
Ez
[0. - v(0, + 0,)]
E
1
Tyz, Yzx
1
%3D
Yxy
Txy Yyz
Triangular area
where
E
G =
2(1 + v)
A =
a
(9 + D)w :
Relations Between w, V, M
h
(2a + b\
a + b
dM
= V
dx
dV
b
w(x),
dx
Trapezoidal area
Elastic Curve
1
M
-A =
EI
d*v
EI
=w(x)
!, = }ar*
%3D
d'v
El
= V(x)
Semicircular area
dv
El
dx?
M(x)
-A = Tr²
Buckling
Critical axial load
1, = fart
!, = fart
T²EI
Per
(KL)²
Critical stress
VIJA
O cr
Circular area
(KL/r)²*
Secant formula
ес
P
Ở max
1 +
sec
A
V EA
žab
A =
Energy Methods
Conservation of energy
U. = U;
zero slope
Strain energy
N²L
Semiparabolic area
U; =
2AE
constant axial load
"M²dx
U, =
bending moment
%3D
2EI
A =
U; =
transverse shear
%3D
2GA
zero slope -
pLT°dx
torsional moment
%3D
2GJ
Exparabolic area

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