4 Output Description
5.3. Building Construction Plate
This example is used for the explanation of punching checks at single columns according to DIN1045−1. For this purpose a building construction plate sup-ported on 5 columns and external walls is supposed to be calculated for the ultimate limit state. The figure shows the deformations and edge support re-actions of the load case dead weight (the walls are supported elastically).
Building construction plate
In the following figure the element mesh of the columns is enlarges repre-sented. The four innermost small QUAD elements of the inside columns are identically here above the column. At the edge columns only the upper half of the support area with two small QUAD’s was defined for the sake of simplicity − the lower half of the support area is not represented exactly and is not to be recognized in the picture. The real column dimension is quadratic at the edge as well as at the inside columns. Thus the edge columns stand in a flush way at the plate edge. The support nodes are marked with a small cross.
Element mesh of the plate and columns
The materials were defined with the program AQUA and the system was gen-erated with the program GENF. The load input occurred with the program SOFILOAD. A dead weight load as well as two field−by−field imposed load caseswere calculated. In the program MAXIMA the extreme bending mo-ments and shear forces were saved in the load cases 2101 to 2110 as well as the maximum support reactions in the load cases 2155 and 2156 for the ulti-mate limit state:
PROG MAXIMA
HEAD Superposition ULS − Ultimate Limit State ECHO FULL NO ; ECHO TABS YES
COMB 1 desi BASE 2100
ACT G GAMU 1.35 1.0 PSI0 1.0 1.0 1.0 LC 1 G
ACT Q_B GAMU 1.50 0.0 PSI0 0.7 0.5 0.3 $ Category B offices LC 2,3 Q
SUPP 1 EXTR mami ETYP qua* TYPE m titl ’ULS_QUAD’
SUPP 1 EXTR mami ETYP qua* TYPE vx,vy titl ’ULS_QUAD’
SUPP 1 EXTR mami ETYP NODE TYPE pz titl ’ULS_NODE’
END
For the design these load cases have to be indicated now. With CTRL ULTI it was defined additionally, that the internal forces to be used are already the ultimate internal forces. The load safety factors were considered already in these internal forces. In addition the column dimensions 40⋅40 cm were input globally with the record PUNC in this example:
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PROG BEMESS
HEAD Design inclusive Punching Checks ECHO FULL NO ; ECHO PARA,PUNC FULL
CTRL ULTI $ The internal forces and moments contain
$ already the load load safety factors.
LC AUTO $ selects all necessary load cases of the MAXIMA
$ ULS−Superposition
$ please check the selected load cases in the result
$ file PUNC D 0.40 B 0.40
GEOM − 30 10 30 10 $ mm DIRE 0 0
END
Due to the input ECHO PUNC FULL a detailed output of the punching checks occurs at first. At the following output of an edge column the effective per-imeter u is determined to 56 % from u0. With 6.91 cm2/m logitudinal reinfor-cement the punching check can be filled without shear reinforreinfor-cement. The de-signations of the calculated values correspond to the dede-signations in DIN 1045−1.
Punching Design (DIN1045−1)
Node number = 230 X= 5.800 [m] Y= 0.000 [m]
Max. shear force V−ULS= 133.5 [kN] integrated of QUADs Column size b= 0.400 [m] d= 0.400 [m]
Plate thickness h−slab= 0.200 [m] depth 0.165 [m]
1. perimeter at 1.5*d= 0.247 [m] utot= 3.155 [m] ucrit= 1.753 [m]
(u= 56 % of utot due to openings, edges or walls−> edge column)
Min.reinforc. as−upper= 4.89 [cm2/m] (Min.design−moment−> edge column) Min.reinforc as−lower= 2.40 [cm2/m] (Min.design−moment−> edge column) Tension reinfor. as >= 6.77 [cm2/m] mue= 0.41 [o/o] VRdct 106.7 [kN/m]
V−Ed = 1.40*V/ucrit = 106.7 [kN/m] <= 106.7 [kN/m] =Vrdct No punching shear reinforcement necessary.
At the end of all detailed checks a summary follows, that is printed also at ECHO PUNC YES (default, if no ECHO input was done):
Punching Design (DIN1045−1)
484 L 11.600 8.000 60.9 0.271 0.908 36 0.57 − 4.60 − 632 W 17.400 −0.200 8.7 0.271 0.629 25 0.13 − 0.05 − Typ I=inner column, E=edge column, C=corner column, F=foundation,
W=end of wall, L=wall corner, G=end_of_girder
ucrit =effective length of 1. perimeter, reduced due to openings and edges %u0 =reduktionfactor due to openings and free edges = u0/u0−tot in % AssSum=shear reinforcemend − total sum of all nperi perimeters
ast = min. required tension reinforcement in the punching zone nperi =up to this perimeter, shear reinforcement is required
Minimim design moments and collaps reinforcement are taken into account.
At punching−nodes the bending moments have been reduced (rounded)
At column−nodes the slab−thickness for bending−design has been increased with 1:3 starting at the column−edge (not at wall punching nodes)
In a second design calculation the column dimensions were varied and an en-larged column head is input at a column. The reinforcement ratio of the longi-tudinal reinforcement was limited to 0.8 % with RO_V in the record PUNC.
PROG BEMESS
HEAD Smaller and larger columns − with shear reinforcement ECHO FULL NO ; ECHO PARA,PUNC,REIN FULL
CTRL ULTI $ The internal forces and moments contain
$ already the load load safety factors.
LC AUTO $ selects all necessary load cases of the MAXIMA
$ ULS−Superposition
$ please check the selected load cases in the result
$ file CTRL LCR 2
PUNC X Y D B HEAD DHEA REIN=0.8 5.800 .000 0.20 0.20
5.800 4.000 0.20 0.20 0.60 0.40 5.800 8.000 0.20 0.20
11.600 .000 0.60 0.60 11.600 4.000 0.20 0.20 GEOM − 30 10 30 10 $ mm DIRE 0 0
END
At the inside column X=11.60, Y=4.00 (in the following section pictures at the right−hand side below) the support reaction V can not be included now any-more without shear reinforcement. BEMESS determines itself a longitudinal reinforcement of 13.2 cm2/m, with which the punching check with 5.76 cm2 shear reinforcement can occur:
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Node number = 476 X= 11.60 [m] Y= 4.000 [m]
Max. shear force V−ULS= 334.4 [kN] integrated of QUADs Column size b= 0.200 [m] d= 0.200 [m]
Plate thickness h−slab= 0.200 [m] depth 0.165 [m]
1. perimeter at 1.5*d= 0.247 [m] utot= 2.355 [m] ucrit= 2.355 [m]
Min.reinforc. as−upper= 6.20 [cm2/m] (Min.design−moment−> inner column) Min.reinforc as−lower= 5.97 [cm2/m] (collapsreinforcement V−ULS/fyk) Tension reinfor. as >= 13.20 [cm2/m] mue= 0.80 [o/o] VRdct 133.3 [kN/m]
mue necessary to satisfy von vRD,max acc. DIN 1045−1 equation 107!
V−Ed = 1.05*V/ucrit = 149.1 [kN/m] > 133.3 [kN/m] =Vrdct 1. design cut of shear reinforcement at point 0.5d −> u= 1.318 [m]
Shear reinforcem. Ass= (V−Ed*ucrit/u−VRdc)*u/fyd/kappa−s (kappa−s=0.70) Shear reinforcem. Ass= 5.76 [cm2] ass= 35.32 [cm2/m2]
to be provided in the 1. perimeter up to columnedge + 0.144 [m]
2. perimeter Ass= 1.74 [cm2] ass= 10.66 [cm2/m2] til 0.268 [m]
Ass= (V−Ed(u)−VRdc)*u*sw/d/fyd/kappa−s Second perimeter only necessary due to DIN 1045−1 13.3.3(7)
The collaps design acc. DIN 1045−1 13.3.2(12) increased the lower reinforcement.
Punching Design (DIN1045−1) CONCLUSION
NodeNo Typ X Y V−ULS d−col ucrit =%u0 v−max AssSum ast nperi No [m] [m] [kN] [m] [m] [o/o] [MPa] [cm2] [cm2/m]
1 W 0.000 −0.200 7.6 0.271 0.629 25 0.11 − 0.03 − 230 E 5.800 0.000 133.5 0.226 1.308 56 0.87 3.90 13.20 2 238 I 5.800 4.000 333.4 0.577 3.369 100 0.63 − 6.26 − 247 I 5.800 8.000 318.8 0.226 2.355 100 0.86 6.96 13.20 2 468 E 11.600 0.000 127.9 0.678 1.891 56 0.57 − 4.73 − 476 I 11.600 4.000 334.4 0.226 2.355 100 0.90 7.50 13.20 2 484 L 11.600 8.000 60.9 0.271 0.908 36 0.57 − 4.60 − 632 W 17.400 −0.200 8.7 0.271 0.629 25 0.13 − 0.05 −
Cuts about the columns show now the computed necessary reinforcement:
Upper longitudinal reinforcement design case 1 in cut direction at 40⋅40 cm columns
Upper longitudinal reinforcement design case 2 with varied column dimen-sions
Shear reinforcement design case 2 with varied column dimensions in cm2/m2 Notes:
Upper edge column at the left−hand side (in figure Upper longitudinal rein-forcement in cut direction at 40⋅40 cm columns" the upper point at the left−
hand side = node no. 230): In the 1st design case (mentioned in the following LCD 1) the punching check with the column dimension 40⋅40 cm requires a necessary upper longitudinal reinforcement of 7.06 cm2/m. In the LCD 2 the round cut is smaller due to the smaller column 20⋅20 cm. The raised shear stress requires a shear reinforcement of 2.9 cm2 (32.38 cm2/m2) in the first
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round cut with a raised maximum longitudinal reinforcement of 13.20 cm2/m (Figure Upper longitudinal reinforcement with varied column dimensions").
Lower inside column at left−hand side (node no. 238): In the 2nd design case LCD 2 an effective enlarged column head of 60 cm was defined. Through that the necessary longitudinal reinforcement from punching reduces to 0.38% = 6.26 cm2/m. However, the bending design supplies a higher reinforcement of 12.71 cm2/m due to the smaller column dimension of 20⋅20 cm, because the support moment reduction is smaller and the design thickness is smaller in the central node than in LCD 1:
LCD 1: design thickness = 20cm+0.5dS / 3= 20cm + 0.5⋅45.2 / 3 = 27.5cm LCD 2: design thickness = 20cm+0.5dS / 3= 20cm + 0.5⋅22.6 / 3 = 23.8cm Lying at the certain side, the thickness increase is assumed not from the en-larged column head but only from the column with 1/3.
Design thickness in central node
Upper edge column at the right−hand side (node no. 468): In the 2nd design case a column with 60⋅60 cm was defined. Here the punching check succeeds now without problems. It is determined a necessary longitudinal reinforce-ment of 0.29% = 4.73cm2/m. However, the bending design requires a higher reinforcement of 6.27 cm2/m.
Lower inside column at the right−hand side (node no. 476): In LCD 2 the shear force can be included only at the column 20⋅20 cm with a shear reinforcement of 5.8 cm2 (see detailed output). The necessary maximum longitudinal rein-forcement is 13.20 cm2/m (limitation at 0.8%).
The lower inside column at the right−hand side (node no. 476) requires a punching shear reinforcement of 5.8 cm2 (35.5 cm2/m2) in the first round cut and 1.7 cm2 in the second round cut.This corresponds to a distributed shear reinforcement of 10.66 cm2/m2 distributes to the area of the punching cone from the column outer edge to 1.5⋅d