Difficult slab analysis problems may require a finite element program. If one is not available a grid program can be used. The method is based on the paper "The Grid-Framework Method for Plates in Flexure", by Alan L. Yettram and Husain M. Husain, Jouenal of the Engineering Mechanics Division, American Society of Civil Engineers, Vol. 93, June 1963, pages 53 to 64. If we set the value of mu, Poissons Ratio, to 1/3, then the finite element system reduces to a series of beams in flexure and torsion. The moment of inertia is the usual bh³/12,and the torsion, J, is the same value but, of course, multiplied by the ratio of G/E. In the case of beams, at ultimate loads, the torsion becomes soft so the constant should be reduced.

The structure chosen to illustrate the method is shown in this sketch. It is one half of a 20 ft. square slab on columns connected by marginal beams, The sturucture is symmetrical about the North/South axis. The columns will be neglected in this analysis but they could be included for moment by stretching them out as members in both the X and Y directions. The slab is divided into 5 bands in both directions with the center bands under the 5 k load. Note that the positive directions are (with respect to the drawing) down and to the left. The thickness chosen for this demonstration are: Slab = 6 in., Beam = 12 wide by 16 in. deep. The live load was assumed as 40 psf and there is a 5 kip load at the very center. The modulus of elasticity was assumed to be 3000 ksi

The moments of inertia and the torsion factor for the slab was calculaed using the equation shown above, and for the beam the coefficients were taken from a mechanics of materials text. The valuse for torsion was reduce to 75 percent for softness at ultimate loads. The load factors used were the code, desd load 1.3 and live load 1.7.

Following is the input to the program:

4010 ' No. of members, No. o joints, Value of E
4020 DATA 22,15,422000
4025 ' Coordinates: Joint, X, Y
4030 DATA 1,20,0, 2,20,5, 3,20,10, 4,15,0, 5,15,5
4035 DATA 6,15,10, 7,10,0, 8,10,5, 9,10,10, 10,5,0
4040 DATA 11,5,5, 12,5,10, 13,0,0, 14,0,5, 15,0,10
4050 ' Member properties, No., I value, J value (zero to exit).
4060 DATA 1,0.35,.034, 2,.0521,.0065, 3,.026,.013, 0
4070 ' Member incidences
4080 DATA 1,1,2,1, 2,2,3,1, 3,4,5,2, 4,5,6,2, 5,7,8,2
4085 DATA 6,8,9,2, 7,10,11,2, 8,11,12,2, 9,13,14,1, 10,14,15,1
4090 DATA 11,10,13,3, 12,7,10,3, 13,4,7,3, 14,1,4,3, 15,11,14,2
4092 DATA 16,8,11,2, 17,5,8,2, 18,2,5,2, 19,12,15,1, 20,9,12,1
4095 DATA 21,6,9,1, 22,3,6,1
4100 ' Supports and releases, Jt., vert, Mx, My (zero to exit).
4110 DATA 1,1,0,1, 3,0,1,1, 4,1,0,1, 7,1,0,1, 10,1,0,1, 13,1,0,1, 15,0,1,1, 0
4120 ' Load numb., Descrip., No. joint loads, No. member loads
4130 DATA 1,1.3 DEAD+1.7 LIVE, 13,0
4140 ' Concentrated loads: Joint, Z, Tx, Ty
4150 DATA 1,1.16,0,0, 2,2.32,0,0, 4,2.32,0,0, 5,4.3,0,0, 6,2.32,0,0
4155 DATA 7,6.16,0,0, 8,4.3,0,0, 9,2.32,0,0, 10,2.32,0,0, 11,4.3,0,0
4160 DATA 12,2.32,0,0, 13,1.16,0,0, 14,2.32,0,0
4170 ' Uniform loads: Member, U

The slab can be designed with bands just as a typical flat slab with middle and column strips. In this case the middle strip, Member 5, that is 5 feet wide, would be designed for a moment of 33.54 kip feet, and Member 3 for a moment of 24.35 kf. The beam, Member 20, has a moment of 59.06 kf.

The use of a grid program makes it possible to solve slab problems that, unless you have a finite element program, would require a lot of guessing.

If you would like to have a copy of this example "slab-1.zip", you can download by clicking HERE. It will require a program such as WinZip to unzip it.