The figure shown is a cross section of a concrete column that is loaded uniformly in compression. Determine: A. The average compressive stress in kips/in2 in the concrete if the load is equal to 3200 kips. B. Determine the coordinates of the point where the resultant load must act in order to produce uniform normal stress in the column. 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
The figure shown is a cross section of a concrete column that is loaded uniformly in compression. Determine: A. The average compressive stress in kips/in2 in the concrete if the load is equal to 3200 kips. B. Determine the coordinates of the point where the resultant load must act in order to produce uniform normal stress in the column. 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
Related questions
Question
The figure shown is a cross section of a concrete column that is loaded uniformly in compression. Determine:
A. The average compressive stress in kips/in2 in the concrete if the load is equal to 3200 kips.
B. Determine the coordinates of the point where the resultant load must act in order to produce uniform normal stress in the column.
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%2F2074eebc-8363-4924-91f2-79e5c4eea24f%2Fwvr1i8m_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
Transcribed Image Text:24 in 20 in.→
– 24
20 in.
16'in.
8 in.
--
8 in.
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