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I have never used 3D elements but I am confident in saying that any combination of solid elements and line elements should have a boundary condition that allows compatibility equations to be solved based on the boundary condition specified. Is your model stable? Solid 3D elements without such boundary condition should behave like unreinforced concrete, where there is a significant increase in deflection once cracking moment is achieved because of loss of member moment of inertia.

1). Tie beams provide stability to structure (depending upon framing) and carry load from slab to foundation. The load distribution for tie beams depends on what level tie-beams have been provided. 2) You can if you are providing tie-beams. 3) If you have designed the slab to be supported on tie-beams instead of grade, then they will transfer the tributary load from the slab to the columns and foundation. 4) Slab on grade lays mostly at grade level. Raft is generally provided below frost zone.

Stirrup spacing, among others, depends upon seismic requirements (see chapter 21 of ACI 318-08). Moreover, spacing of stirrups is closer near the support, as compared to middle portion of the beam.

Shear reinforcement in SAP2000 is given in the units of 'length^2/unit length'. For example, if the displayed unit of length is in inches, shear reinforcement is in 'in^2/in.' units. To convert shear reinforcement value ( given by SAP2000 in 'inch' units) into stirrup spacing, multiply the given value by 12 to convert it in into normally used units of 'in^2/ft.' Now, divide the product of 'cross-sectional area of one stirrup & 12' by the required area of shear reinforcement, calculated earlier. The result of division will be the stirrup spacing in inches. For example, if the shear reinforcement value (given by SAP2000 in 'inch' units) is 0.025, required As =0.025x12=0.3 in^2/ft. Cross-sectional area of one #3, 2-legged stirrup is 0.11 x 2 = 0.22 in^2. Then, required spacing of #3 stirrup will be = 0.22 x 12 / 0.3 = 8.8 in. Alternately, you may get the same result by dividing the 'cross-sectional area of stirrup' with the 'required shear reinforcement value, given by SAP2000 in inch units, directly. [ 0.22 / .025 = 8.8 in.] Regards.

You can calculate centre of rigidity by calculating stiffness of each column i.e 6EI/L^2 and multiplying by the distance from the origin. Adding all these values and then divide by sum of all column stiffness. Do the same for both directions. Remember to see the column orientation while calculating inertia.

keep spreading good work..... beavo