1/19 M.Chrzanowski: Strength of Materials SM1-05: Statics 4: statically determined bar structures STATICALLY DETERMINED PLANE BAR STURCTURES (FRAMES, ARCHES)
Dec 18, 2015
1/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
STATICALLY DETERMINED PLANE BAR STURCTURES (FRAMES, ARCHES)
2/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
Formal definition: A frame is a plane (2D) set of beams connected at stiff and/or hinged joints (corners)
Joints have to be in the equilibrium!
X = 0Y = 0MK = 0
?
? !?!
!?!
Frames
STIFF JOINT HINGED JOINT
3/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
What is the difference between beams and frames?
Why do we need to make frames?
Beam FrameBeam or frame?
Hey, you!
Frames
4/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
Equilibrium equationsX = 0Y = 0MK = 0
MA2 = 0MA1 = 0
MAn = 0…………
+ For n hinged joints (if any!) at A1, A2 …An points
+ kinematic stability of a structure (c.f. Theoretical Mechanics)
Centre of instability
Examples of unstable structures
HYPER-STIFF
UNSTABLE
Frames
5/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
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
x
y
Wq’=q·Δx/Δs=q ·cosW = q·Δx
W
Δs
Δx
x
s y
q’
qs =q’·cos= q·cos2Δx/Δs = cosq
q s
s ns =q’·sin= q·sin ·cos
= q’·Δs
qsq’
ns
Frames
n s
6/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
Example: Diagrams of M, Q, N for a simple frame1,5 kN/m
2 kN
2 m
1,5 m
1 m 2 kN1 kN
2 kN
sin=0,6 cos =0,8
+-
-
+
Q [kN]
M [kNm]
0,8
31
3
0,33
1
2
1,62
1,2
0,6
2 2
0,5 m
+-
-
2N [kN]
2
Frames
Q
No
n
3kN
7/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
2
Checking the equilibrium at a joint
M [kNm]
31
3
0,25
1
+-
2
1,2
0,6
-
2N [kN]
-
+
Q [kN]
0,8
1,62
2 2
0,5 m
+-
2
3
3
3
3 2
3
3 2
2
2
Q
No
nFrames
8/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
FRAMEARCH
M
N
+ --
Stones and other brittle materials do not sustain an extension
Arches
Formal definition: An arch is a plane (2D) set of curved beams connected at stiff and/or hinged joints
9/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
X = 0Y = 0MK = 0
Mc = 0
C
Arches
10/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
x
y
2l
xC
x
yC
CC
y
Parabolic arch
Semi-circular arch
To determine reactions we only need to know position and magnitude of loads and position of the hinge and supports
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 + cx2x = r·cos
rC
rC
xC , yC r, C
h
a,b,c from: for x = 0 y = 0
for x = l y = hfor x = 2l y = 0
(Symetric arch)
nn
= arctg dy/dx
Parabolic arch
Semi-circular arch
(in polar coordinates)
Arches
11/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
Example: parabolic arch under concentrated force
x
y
l l
h
C
A B
Mc = 0
RA RBVB
HB
X = 0
MA = 0 P·l - VB ·2l =0
VA
HA
VB = P/2
HA/VA = HB/VB = l/h
Higher the ratio l/h (i.e. lower the ratio h / l) – higher the value of horizontal reaction H
HA = HB = VA· l/h
P
Y = 0 = VA
Arches
12/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
Arches
x
y
l l
h
C
A B
RA
VBVA = P/2
HA=
(P/2) ·(l/h)
P
Symmetry axis
RA RBM
Symmetry axis
RB
Δ
M = RA·Δ
ΔR
A
N Q
NA
QA
NC=HAQA = VA
13/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
VA=P/2
HA=(P/2)(l/h)
QV
QH NV
NH Bar axis
sincos AAHV HVQQQ
QN
nnN
Q
cossin AAHV HVNNN cossin2 hlPN
A
CAt C: =0 2PQC
hlPNC 2
At A: ǂ0
For (l/h)>>1 – shallow arch
For (l/h)<<1 – steep arch
0sin lhtg 1cos
0cos12sin112sin12 PtgPhlPQA
hlPN A 02
0cos,1sin2
sincos2 hlPQ
000202 PhlPQA
2012 PPN A
hl
AN
14/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
N Q
|NA|>P/2
NC=HA =- (P/2) (l/h)
+
- --
+
QA<P/2
QA = VA=P/2
Anti-symmetric axis
P/2
P/2
Symmetry axis
Q
No
n
15/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
P
r
r r
P/2
P/2
P/2
P/2
M 0,3r
0,2Pr
P/2
P/2
P/2
P/2
P/2
P/2
P/2
Q
P/2
P/2
P/2
P/2
P/2 +
- P/2
P/2+
-P/2
0,7P
N
Q
N
M
o
n
tension
+-+
Example: semi-circular arch under horizontal force
16/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
r
P
r r
r
P
r r
r
P
r
r r
r
P
r
r r
Quantitative comparison of frame, quasi-arch and arch
17/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
r r
P
P
r
P/2P/2
M
Q N
+-
+
Pr
Pr/2
Pr/2
P/2
P
P
Q
N
M
o
n
„Frame”
18/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
P
P/2r r
P
r
P/2
M
Q
tension
Pr/20,55 Pr
P/2P+ -
N
P/2
P/2
P
-
+~1,1P
Quasi-arch
19/19M.Chrzanowski: Strength of Materials
SM1-05: Statics 4: statically determined bar structures
Comparison
M
Q
N
0,2Pr
+ P
P/2
Pr/20,55Pr
Pr/2Pr
Pr/2
-P/2
P
++
-P/2
P
-+ P
P/2
1,1P
P/2
+-
P/2
P/2+
P/2
+
-
0,7P
P/2
P/2P/2