homogeneous 10 m rigid bar has a mass of 800 kg and is suspended at both ends by a 4 m bronze cable on the left and a 3 m steel cable on the right. Determine the smallest area in mm in each cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel.
homogeneous 10 m rigid bar has a mass of 800 kg and is suspended at both ends by a 4 m bronze cable on the left and a 3 m steel cable on the right. Determine the smallest area in mm in each cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
A homogeneous 10 m rigid bar has a mass of 800 kg and is suspended at both ends by a 4 m bronze cable on the left and a 3 m steel cable on the right. Determine the smallest area in mm in each cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel.
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.
![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%2F77c85352-a22a-4ee1-9ae3-735f88247e30%2F17y0tt_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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