Transcript Title
SM1-05: Statics 4: statically determined bar structures
STATICALLY DETERMINED PLANE
BAR STURCTURES
(FRAMES, ARCHES)
M.Chrzanowski: Strength of Materials
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SM1-05: Statics 4: statically determined bar structures
Frames
Formal definition:
A frame is a plane (2D) set of beams connected at stiff and/or hinged joints (corners)
HINGED JOINT
STIFF JOINT
Joints have to be in
the equilibrium!
?
X = 0
Y = 0
MK = 0
?
!?!
!?!
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SM1-05: Statics 4: statically determined bar structures
Frames
What is the difference between beams and frames?
Why do we need to make frames?
Hey, you!
Beam
M.Chrzanowski: Strength of Materials
Beam or frame?
Frame
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SM1-05: Statics 4: statically determined bar structures
Frames
Equilibrium equations
X = 0
Y = 0
MK = 0
+
+
MA1 = 0
MA2 = 0
…………
MAn = 0
For n hinged joints (if any!)
at A1, A2 …An points
kinematic stability of a structure (c.f. Theoretical Mechanics)
Examples of unstable structures
Centre of
instability
M.Chrzanowski: Strength of Materials
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SM1-05: Statics 4: statically determined bar structures
Frames
After we determine reactions and check stability we can deal with a frame as a set
of individual beams, applying all techniques which have been demonstrated for
beams. But, besides of diagrams of M and Q we have to make diagrams of N, too.
Some problems can be encountered with sloping members
W
q
W = q·Δx = q’·Δs
qs =q’·cos= q·cos2
Δx/Δs = cos
ns =q’·sin= q·sin ·cos
y
s
y
W
q’
q’
q’=q·Δx/Δs=q ·cos
x
s
qs
ns
x
Δx
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SM1-05: Statics 4: statically determined bar structures
Frames
Example: Diagrams of M, Q, N for a simple frame
3
3
2 kN
1
sin=0,6
cos =0,8
1,5 kN/m
3kN
1,5 m
1 kN
2 kN
1m
2m
n
2 kN
M [kNm]
N
o
2
2
Q [kN]
-
2
2
Q
-
+
-
+
+
N [kN]
2
0,5 m
2
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SM1-05: Statics 4: statically determined bar structures
n
Frames
N
o
Checking the equilibrium at a joint
Q
3
3
1
3
2
3
M [kNm]
Q [kN]
2
2
-
2
2
3
-
+
+
M.Chrzanowski: Strength of Materials
+
-
2m
0,5
3
2
-
3
2
2
N [kN] 3
2
2
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SM1-05: Statics 4: statically determined bar structures
Arches
Formal definition:
An arch is a plane (2D) set of curved beams connected at stiff and/or hinged joints
+
-
-
N
M
FRAME
ARCH
Stones and other brittle materials do not sustain an extension
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SM1-05: Statics 4: statically determined bar structures
Arches
C
X = 0
Y = 0
MK = 0
Mc = 0
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SM1-05: Statics 4: statically determined bar structures
Parabolic y
arch
Arches
y
C
n
rC
xC n C
C
yC h
Semi-circular
arch
r
x
2l
x
To determine reactions we only need to know position and magnitude of loads
and position of the hinge and supports
r, C
xC , yC
But to determine the cross-sectional forces we do need the equation describing
shape of the arch: including coordinates of any point and its tangent.
y = a + bx + cx2
a,b,c from:
for x = 0
for x = 2l
(Symetric arch)
for x = l
y=0
y=0
y=h
x = r·cos
(in polar coordinates)
= arctg dy/dx
M.Chrzanowski: Strength of Materials
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SM1-05: Statics 4: statically determined bar structures
Arches
Example: parabolic arch
under concentrated force
P
C
y
h
A
HA
l
RA
MA = 0
Mc = 0
B
VA
P·l - VB ·2l =0
HA/VA = HB/VB = l/h
x
l
VB
HB
RB
VB = P/2 = VA
X = 0
Y = 0
HA = HB = VA· l/h
Higher the ratio l/h (i.e. lower the ratio h / l)
– higher the value of horizontal reaction H
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SM1-05: Statics 4: statically determined bar structures
Arches
P
C
Symmetry
axis
y
Symmetry
axis
Δ
M = RA·Δ
h
HA=
(P/2) ·(l/h) A
l
RA
VB
N
N
A
M.Chrzanowski: Strength
of Materials
RB
x
l
VA = P/2
NC=HA
M
RA
B
ΔRA
RB
QA = VA
QA
Q
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SM1-05: Statics 4: statically determined bar structures
Q QV QH VA cos H A sin
Q P 2cos l hsin
HA=(P/2)(l/h)
N NV N H VA sin H A cos
NH
Bar axis
Q
QH
C
QV
At C: =0
QC P 2
NC P 2l h
At A: ǂ0
For (l/h)<<1 – steep arch
NV
A
VA=P/2
N
N P 2sin l hcos
2 sin 1, cos 0
QA P 20 l h P 20 0 0
N A P 21 0 P 2
For (l/h)>>1 – shallow arch
nn
tg sin h l 0 l h
cos 1
N QA P 21 l hsin P 21 1 tg sin P 21 cos 0
Q
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N A P 20 l h N A
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SM1-05: Statics 4: statically determined bar structures
n
N
o
NC=HA =- (P/2) (l/h)
Q
QA = VA=P/2
Anti-symmetric
axis
Symmetry
axis
QA<P/2
|NA|>P/2
+
-
N
M.Chrzanowski: Strength of Materials
P/2
P/2
+
-
Q
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SM1-05: Statics 4: statically determined bar structures
tension
P
Example: semi-circular arch
under horizontal force
r
P/2
P/2
Q
P/2
P/2
P/2
0,7P
n
+
P/2
N
N
P/2
P/2
0,2Pr
o
r
Q M
r
P/2
M
P/2
P/2
0,3r
P/2
+
P/2
P/2
-
M.Chrzanowski: Strength of Materials
P/2
-
+
P/2
P/2
-
P/2
+
P/2
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SM1-05: Statics 4: statically determined bar structures
Quantitative comparison of
frame, quasi-arch and arch
P
r
r
r
P
r
P
r
r
P
r
r
r
M.Chrzanowski: Strength of Materials
r
r
r
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SM1-05: Statics 4: statically determined bar structures
„Frame”
P
M
r
Pr/2
P
Pr
Pr/2
r
N
r
o
n
Q
P/2
M
P/2
N
+
Q
+
-
P
P/2
P
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SM1-05: Statics 4: statically determined bar structures
Quasi-arch
tension
P
r
M
P
0,55 Pr
P/2
P/2
r
Pr/2
r
N
Q
P/2
+
P
~1,1P
M.Chrzanowski: Strength of Materials
P/2
P
+
P/2
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SM1-05: Statics 4: statically determined bar structures
Comparison
Pr/2
M
0,55Pr
0,2Pr
Pr/2
Pr
Pr/2
P
+
Q
-
P/2
P
P/2
P/2
+
+
-
1,1P
P/2
+
P
M.Chrzanowski: Strength of Materials
+
P
P/2
+
0,7P
P/2
N
-
P/2
+
P/2
P/2
P/2
-
P/2
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SM1-05: Statics 4: statically determined bar structures
stop
M.Chrzanowski: Strength of Materials
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