The Analysis of Rectangular solid slabs by Finite Differences I have a BAS IC program program which provides provides an elasti c analysis of a rectangula rect angularr slab/beam syst em with variable variable sect ion properties properties and any c ombination of supported nodes, t he program program also treats non-linear cases where where lift-off occurs at supports which allows foundation rafts and slabs with unrestrained corners to be analysed. This sheet is a trial to see how the analysis might transfer to Mathcad format and, in the first instance, is limited to the t he analysis analysis of rectangular slabs us ing a square mesh (the program program allows a rect angular angular mesh), Pois son's ratio is zero and beam supports are not yet included. The loads are specified as point loads at any node, by apportioning distributed loads appropriately any combination of point, wholewhole- or part-U part-UD D or hydr hydrostatic ostatic loads can be accepted. The stiffness stiff ness matrix is as sembled from the finite-difference operator patterns and then inverted, the deflexions are obtained by multiplying the inverted inverted matrix by the load vector. vector. The stress resultants are calculated from t he deflexions deflexions using the appropriate appropriate finite diff erence erence formula f ormulae. e. Finite differences have been pretty well well been superseded by finite elements for s lab design but, when the plan-form is rectangular, they still have a lot to offer by way of simplicity and accuracy. h
2
1
N-2
3
N-1
N h
N+1
N+2
N+3
2N-2
2N-1
2N
2N+1
2N+2
2N+3
3N-2
3N-1
3N
No. of longit. grid lines, N
No. of transverse grid lines, M (M-3)*N+1
(M-3) (M-3)*N *N+2 +2 (M-3) (M-3)*N+ *N+3 3
(M-2)*N-2 (M-2)*N-1 (M-2)*N
(M-2)*N+1
(M-2) (M-2)*N *N+2 +2 (M-2) (M-2)*N *N+3 +3
(M-1)*N-2 (M-1)*N-1 (M-1)*N
(M-1)*N+1
(M-1)*N+2 (M-1)*N+3
M*N-2
M*N-1
M*N
TYPICAL RECTANGULAR SLAB Enter the data:Width of slab (mm)...................................................... wid idth th 7000
width
Number of longit grid lines, N
h
Grid line spacing (mm)...................................................... h 500 Number Number of of t ransve ransverse rse grid grid lines............. ........ ........ ......... .... M 15 Depth of slab (mm)........................................................... d 200 Young Young's 's modulus modulus (kN/mm^2).............. (kN/mm^2).............. ........ ......... ........ ...... E 28
Note:- N and M not to be less than 5
1
Length of slab, (mm),
length h ( M 1) 3
distributed inertia, length
1
D
d
12
N
E
The stiffness matrix is assembled from the appropriate finite difference patterns for a rectangular slab with all edges free/free in the collapsed section below. For information on the method see Structural Analysis, Ghali & Neville, Chapman & Hall, or s ome similar text
Infill below leading diagonal (at t he same time c orrecting for the half-value leading diagonal c oefficient input above). T
A A A
Supply some supports, initially, s ay a line support on the left hand side and two point supports near the other two corners:(If the slab edges are not tied down, lift-off from the line support can be engineered by trial and error, after the first run any support with a tensile reaction is freed. After s everal runs, repeating this procedure this results in only the seven central wall nodes being supported, as follows),
Dimension the array: Support
MN
0
Set the supported nodes to a high number: 12
i 4N 1 5N 1 ( M 5) N 1
Add to leading diagonal of A matrix:
i 1 M N
Then part of the stiffness matrix looks like:-
A
2
Support 10 i
12
i N 4
Support 10
i ( M N) 4
Support 10
i
12
i
A A Support i i
i i
i
W 10
Supply a load vector, say a UDL at the rate of W kN/sq.m.
W length width
Panel_load
6
( M 1) ( N 1) 1 10 initially set the whole load matrix so that each node is loaded by Panel_load
i 1 2 M N
Load Panel_load i
then modify the edge and corner nodes:Load i 1 2 N
Load i
Load
i
i N 2 N M N
2
Load i
i
2
Load i 1 N 1 ( M 1) N 1
i
Load i
Load
2
i ( M 1) N 1 ( M 1) N 2 M N
Load i
i
2
In the BASIC version of the program the heavily banded stiffness matrix is solved using Choleski decomposition. With Mathcad it seems much simpler to invert the matrix
Solve the problem:
1
Deflexions A
Load
2
h
D
Deflns Rearrange deflexions into a rectangular array, (for plotting).
p 0 for i 1 N M p p 1
1 N
q ceil p
r p ( q 1) N D
Deflexions
q r
D
B Deflns
3
p
A table containing the deflexions at the nodes, in mm units, is as follows:-
Deflns
and a graph of deflexions appears thus:-
DEFLEXIONS
Deflns (Notice how the unrestrained rear corners pick themselves up off t he end support wall)
4
The support reactions are found by multiplying the support stiffnesses by the deflexions at the nodes and the appropriate constant, thus:
i 1 M N
Rn Deflexions Support i
i
D
i 2
h
Check that sum of reactions is equal to sum of applied loads: Sum of reactions=
Sum of Loads=
Sum_reactions
Sum_loads
The two column reactions are:-
Rn
Load
Rn
N 4
Sum_reactions (If these two are equal we have a balance of vertical forces, if not, we have an error!.)
Sum_loads
and
Rearrange reactions into a rectangular array, for plotting:
Rn
)4 (MN
Rect( Rn)
p 0 for i 1 N M p p 1
1 N
q ceil p
r p ( q 1) N
A plot of the reactions is as follows:-
D
rq
D
REACTIONS at SUPPORTED NODES
Rect(Rn)
5
Rn
p
Calculate some bending moments:
Mx
p 0 for i 1 M p p 1 q 1 for j 1 N 2 q q 1 M1
p q
M1
M N
0
M1 M1
D 2
h Table of Mx values in kN:-
M1
Mx
Bending Moment Mx
Mx 6
B
2Bp q Bp q1
p q 1
My
p 1 for i 1 M 2 p p 1 q 0 for j 1 N q q 1 M1
p q
M1
M N
B
2Bp q Bp1 q
p 1 q
0
M1 M1
D 2
h M1 My in kN. :-
My
Bending Moment My
My
7
Calculate the twisting moments Mxy
p 1 for i 2 M 1 p p 1 q 1 for j 2 N 1 q q 1 M1
p q
Bp1 q1 Bp1 q1 Bp1 q1 Bp1 q1 4
M N BM N1 BM1 N1 BM1 N B B B B M1 M 1 M 2 M 1 M 1 1 M 1 2 M1 B B B B 1 1 2 2 2 1 1 1 1 2 M1 B B B B 1 N 2 N 2 N 1 1 N 1 1 N M1
M N
B
p 1 for i 2 N 1 p p 1
2 p1 B2 p1 B1 p1 B1 p1 .5 B B B B M1 .5 M p M p1 M p 1 M 1 p 1 M 1 p 1 M1
1 p
B
p 1 for i 2 3 M 1 p p 1
p1 2 Bp1 1 Bp1 1 Bp1 2 .5 B B B B M1 .5 p N p 1 N p 1 N 1 p 1 N1 p 1 N M1
p 1
Table of Mxy, in kN:-
B
M1 M1
D 2
h M1
Mxy
8
and a plot of Mxy:-
Mxy
9
