-
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.
***