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example stringer
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Systematic Analysis of Stringer & Panel Method
stringer
stringer
strin
ger
strin
ger
v
N1 N2
N3 N4
?a practical design method for structuralconcrete plate elements loaded in plane
?analysis method developed fororthogonal geometries that can optimallytake advantage of two-way slab designby allocating stringer elements to beamlocations, with distributed intermediatereinforcement in panel elements
?structural model consists of a series ofhorizontal and vertical stringers for thetransfer of normal forces, whereas therectangular fields between can be filledwith panels for the transfer of shearforces
N2
N1
N3N4
?only shear force exists in the shearpanel, and it has the same value, v, perunit length at all positions in the panel;this demand forms the basis of thetwo-way reinforcement
?a statical or equilibrium based method(lower bound) such that any distributionof internal stresses satisfying equilibriumis valid as long as sufficient member anddeformation capacity exists to allow thestructure to redistribute stresses asprescribed
?designer must carefully consider theforces in the structure and choose anequilibrium system in which the load"flows" to the supports in a recognizableway
v
v v
2F
-2F
F
- F
F
-F
-F-F
12F
F
-32F
-F1 2F
F
1 2F
2F
F F
- Fa - 12 Fa
- 12 Fa
Fa
12 Fa
12
Fa
F47
00 m
m
1500
500
2700
1500500 2400 700 1900
7500 mm
F
2200
mm
2000
mm
2000 mm 2500 mm 2500 mm
Consider the following example of a reinforced concrete wall w/ opening subject to a point load...
A stringer and panel model of the structure is chosen with the configuration shown below...
????????????????????????????
V3 V4 V5
V1 V2
F x 2500 - Ra x 7000 = 0
Ra = 25007000F = 0.3571F
Ra Rb
Rb = 0.6429F
Assuming the following arbitrary sign convention...
V(+) (+) Panel Shear Forcesx
y
(+) External Forces & Boundaries
(+) Stringer Shear Flow
(+) Panel Shear Forces
e.g.
(+)F
(-)v(+)v
x
y
Stringer Elements (Global Coordinates) Panel Elements (Local Coordinates)
Such that positive shear flow in a panel element will result in the following shear flow in the stringers in theglobal coordinate system...
V(+)
a
b
? = (+)Vab
? S
hear
Flo
w =
(+)V
? =
(+)
V
? = (+)Vab
? =
(+)
V
? = (+)Vab
? = (-)Vab
? =
(-)
V
? =
(+)
V
? =
(+)
V
? = (+)Vab
? = (+)Vab
For our structural model, this gives the following shear distribution, with which the equations of equilibriumin the global coordinate system may now be formulated...
V3 V4
V1
V5
S1 S2 S3 S4
S5
F
S6
S7
Ra Rb
V2
? =V2 x 25002000? =V1 x
25002000
? =V2 x 25002000? =V1 x
25002000
? =
V2
? =
V2
? =
V 1
? =
V 1
? =V5 x 25002200
? =V5 x 25002200
? =
V5
? =
V5
? =
V4
? =
V4
? =V4 x 25002200
? =V4 x 25002200
? =
V3
? =
V3
? =V3 x 20002200
? =V3 x 20002200
F
V4
Ra Rb
V3 V5
V1 V2
This system of equations has no unique solution. Any arbitrary distribution of shear forces maybe chosen to satisfy equilibrium by eliminating internal redundancies (i.e. "arbitrarily" definingpanel shear forces) to make the structure statically determinate. For example, taking V1 = -0.1gives...
Equilibrium in y
Grid/Stringer Line 1, S1: V3 + Ra = 0 ; V3 = -Ra = -0.3571F (1)
S2: V1 - V3 + V4 = 0 (2)
S3: - V1 + V2 - V4 + V5 - F = 0 (3)S4: - V2 - V5 + Rb = 0 (4)
Equilibrium in x
S5: - V3 x 20002200 - V4 x 25002200 - V5 x 25002200 = 0 (5)
S6: - V1 x 25002000 - V2 x 25002000 + V3 x 20002200 + V4 x 25002200 + V5 x 25002200 = 0 (6)
S7: V1 x 25002000 + V2 x 25002000 = 0 ; V2 = - V1 (7)
From (2): V1 - V3 + V4 = 0
V1 + 0.3571F + V4 = 0 ; V4 = -0.3571F - V1
From (3): - V1 + V2 - V4 + V5 - F = 0
- V1 - V1 - (- 0.3571F - V1) + V5 - F = 0 ; V5 = 0.6429F + V1
V1 = - 0.1000
V2 = 0.1000
V3 = - 0.3571
V4 = - 0.2571
V5 = 0.5429
And the following distribution of stresses...
V3 V4
V1
V5
S1 S2 S3 S4
S5
F
S6
S7
Ra Rb
V2
? = 0.13? = 0.13
? = 0.13? = 0.13
? =
0.1
0
? =
0.1
0
? =
0.1
0
? =
0.1
0
? = 0.61
? = 0.61
? =
0.5
4
? =
0.5
4
? =
0.2
6
? =
0.2
6
? =0.29
? = 0.29?
= 0
.36
? =
0.3
6
? = 0.32
? = 0.32
V3 V4
V1
V5
S1 S2 S3 S4
S5
F
S6
S7
Ra Rb
V2
? = 0.31? = 0.31
? = 0.31? = 0.31
? =
0.2
5
? =
0.2
5
? =
0.2
5
? =
0.2
5
? = 0.44
? = 0.44
? =
0.3
9
? =
0.3
9
? =
0.1
1
? =
0.1
1
? =0.12
? = 0.12?
= 0
.36
? =
0.3
6
? = 0.32
? = 0.32
And the stringer stress distribution...
Ra Rb
F
- 0.3
6- 0
.36
0.13
- 0.32 - 0.46
- 0.3
9- 0
.64
0.25
-0.5
1
-1.0
00.32
This procedure may be applied to the design of structural slab systems. Take the followingstructure...
A C
3
1
2
B
4
D E
ELEV.
7 m 7 m
STAIRWELL
7 m7 m
6 m
6 m
6 m
ELEV.
Floor Plan
0.32
F12
F12
F12
F12
F12
F12
F12
F12
F12
F12
F12
F12
F2
F2
F2
F2
F48
F24
F48
F24
F12
F12
F12
F24
F24
F12
F12
F12
F24
F48
F48
F24
F24
F24
F24
F24
Floor Panel Internal Forces
Stringer and Panel Model
V7 V8 V9 V10
V5 V6
V1 V2 V3 V4
TYP.F48
F48
F48
F48
Equations of equilibrium in matrix form...
1
-1
0
0
1
0
0
- F2 + F48 +
F24 +
F24 +
F48
=
F
- 7676
0
1
-1
0
0
0
0
- 7676
0
0
1
-1
0
0
0
- 7676
0
0
0
1
-1
0
0
- 7676
0
1
-1
0
0
0
- 7676
0
0
0
1
-1
0
0
- 7676
0
1
-1
0
0
0
- 7676
0
0
0
1
-1
0
0
- 7676
0
0
0
0
1
-1
0
- 7676
0
0
0
0
0
1
-1
- 7676
0
0
V1V2V3V4V5
V
V6V7V8V9
V10
A
Example Solutions...
F24 + F12 +
F12 +
F24
0
- F2 + F48 +
F24 +
F24 +
F48
0
F24 + F12 +
F12 +
F24
0
0
F24 + F12 +
F12 +
F24
V1 = - 0.1875F
V5 = - 0.0417F
V2 = - 0.0417F
V3 = 0.0417F
V4 = 0.1875F
V7 = - 0.1875F
V6 = 0.0417F
V9 = 0.0417F
V8 = - 0.0417F
V10 = 0.1875F
- 0.21 - 0.27 - 0.21
0.21 0.27 0.21
- 0.21 - 0.21 - 0.21
0.21 0.21 0.21
External, panel, and boundary forces notshown in each example solution
- 0.2
5
0.13
0.02
- 0.0
2
- 0.2
5
0.13
0.02
- 0.0
2
Stringer Stress Diagram
= 0.125 F
? =
- 0.
1875
F
F24
F48
= 0.0 F at boundary
- 0.0
4
- 0.0
4
- 0.0
4
0.10
- 0.1
0
- 0.1
0
0.04
0.04
0.04
0.100.04
- 0.0
4
***
In Summary...
Devise a stringer and panel configuration such that load can "flow" to the supports in arecognizable way.
Formulate equations of equilibrium in global coordinate system.
Eliminate a number of internal redundancies (i.e. "arbitrarily defining" panel shear forces)to make the structure statically determinate OR construct equivalent matrices from theequilibrium equations and compute.
Graphically verify that equilibrium is satisfied at all points in the structure.
(1)
V2 = - 0.104F
V5 = - 0.042F
(2)
(3)
(4)
V1 = 0.000F
V9 = 0.104F
V3 = - 0.021F
V7 = - 0.038F
V4 = 0.125F
V8 = 0.021F
V6 = 0.042F
V10 = 0.250F
- 0.4
6
0.06
0.02
- 0.0
2
- 0.0
4
0.3
5 0
.23
0.1
0
- 0.1
0
- 0.0
2
0.0
4
0.1
9
- 0.3
1
0.0
2
0.0
4
- 0.
04 -
0.04
0.0
4
0.0
4
- 0.
04 0
.04
- 0.0
4
- 0.44 - 0.41- 0.29
0.44
0.37 0.29
- 0.15- 0.07
0.150.12
Stringer Stress Diagram
Requiring that the bottom left panel carries no shear (i.e. defining V1 = 0) shifts the stress distribution carried by thepanel and stringer elements. This solution remains valid and several others (e.g. such as an axially symmetric solutionobtained by setting V1 = 0 & V2 = 0) as the system contains enough redundancy of stress distribution to satisfy theequilibrium equations.
***