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Systematic Analysis of Stringer & Panel Method stringer stringer   s    t   r    i   n   g   e   r   s    t   r    i   n   g   e   r v 1 3 4 a practical design method for structural concrete plate elements loaded in plane analysis method developed for orthogonal geometries that can optimally take advantage of two-way slab design by allocating stringer elements to beam locations, with distributed intermediate reinforcement in panel elements structural model consists of a series of horizontal and vertical stringers for the transfer of normal forces, whereas the rectangular fields between can be filled with panels for the transfer of shear  forces 1 3 4 only shear force exists in the shear  panel, and it has the same value, v, per unit length at all positions in the panel; this demand forms the basis of the two-way reinforcement a statical or equilibrium based method (lower bound) such that any distribution of internal stresses satisfying equilibrium is valid as long as sufficient member and deformation capacity exists to allow the structure to redistribute stresses as  prescribed  designer must carefully consider the  forces in the structure and choose an equilibrium system in which the load "flows" to the supports in a recognizable way v v v 2F   -    2    F  F   -    F  F   -    F -F -F  1 - 3   -    F    1 2    F    1 2    F 2F  F  - a  - 1  a  - 1  a a 1  a 1  a

Stringer and Panel Example

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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.

    ***