A box has a weight of 150 pounds, is resting on a tailgate of a pickup weighing 60 pounds. The tailgate is supported by two cables (loads are indicated in the figure and only one is showing on the figure). If the cross-sectional area of the cables is both 0.017 in2 , Find the tensile force and normal stress in each cable in lb/in2. Note: Draw the Free Body Diagram, include the units/dimensions, use the proper formula and round-off all the answers and final answers to 3 decimal places
A box has a weight of 150 pounds, is resting on a tailgate of a pickup weighing 60 pounds. The tailgate is supported by two cables (loads are indicated in the figure and only one is showing on the figure). If the cross-sectional area of the cables is both 0.017 in2 , Find the tensile force and normal stress in each cable in lb/in2. Note: Draw the Free Body Diagram, include the units/dimensions, use the proper formula and round-off all the answers and final answers to 3 decimal places
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
A box has a weight of 150 pounds, is resting on a tailgate of a pickup weighing 60 pounds. The tailgate is supported by two cables (loads are indicated in the figure and only one is showing on the figure). If the cross-sectional area of the cables is both 0.017 in2 , Find the tensile force and normal stress in each cable in lb/in2.
Note: Draw the Free Body Diagram, include the units/dimensions, use the proper formula and round-off all the answers and final answers to 3 decimal places.
Transcribed Image Text:Wc= 150 lb
d. = 18 in.
Cable
H= 12 in.
Crate
Tail gate
Truck
d7 = 14 in.
Wj= 60 lb
L = 16 in.
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F97214b4e-52b7-4faa-8f78-ab75f2ef67af%2Fde6f83bd-a9c2-4843-9a43-fa6ce01f8616%2F3czw2jp_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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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