Chapter # 7 Design of Two-Way Slabs (Part-B) Problem # 4 Compute the positive and negative moments in the column and middle strips of the exterior panel of the Flat Plate Slab between columns B and E in figure. The slab is 8 in. thick and supports a superimposed service dead load of 25 psf and service live load of 60 psf. The edge beam is 12 in wide by 16 in. in overall depth and is cast monolithically with the slab. Problem # 4 Column and middle strips Problem # 4 - Solution Solution: 1) Compute the factored loads 8 ππ’ = 1.2 × 150 + 25 + 1.6 60 = 246 ππ π 12 Note: Live load reduction must be carried out if local building code allows it 2) Compute moments in span BE of the slab a) Compute ππ and π2 (average) and divide the slab into column and middle strips Problem # 4 - Solution 1 14 1 16 ππ = 21.0 ππ‘ − − = 19.75 ππ‘ 2 12 2 12 20 18 π2 = ππ‘ + ππ‘ = 19 ππ‘ 2 2 The column strip extends the smaller of π2 Τ4 or π1 Τ4 on each side of the column centerline. Because π1 is greater than either value of π2 , base this on π2 The column strip extends 20Τ4 = 5ππ‘ toward AD and 18Τ4 = 4.5ππ‘ toward CF from line BC as shown in figure. The total width of column strip is 9.5 ft. The half-middle strip between BE and CF has a width of 4.5 ft, and the other one is 5 ft, as shown Problem # 4 - Solution b) Compute ππ ππ’ π2 ππ2 ππ = 8 0.246 × 19 × 19.752 → ππ = = 228 πππ − ππ‘ 8 c) Divide MO into negative and positive moments The distribution of the moment to the negative and positive regions is as given by Table 13-2 (refer next slide). In the terminology of Table 13-2, this is a “slab without beams between interior supports and with edge beam.” From Table 13-2, the total moment is divided as follows: πΌππ‘πππππ πππππ‘ππ£π ππππππ‘ ππ’ = −0.70ππ = −0.70 × 228 = −160 πππ − ππ‘ πππ ππ‘ππ£π ππ’ = 0.50ππ = 0.50 × 228 = +114 πππ − ππ‘ πΈπ₯π‘πππππ πππππ‘ππ£π ππππππ‘ ππ’ = −0.30ππ = −0.30 × 228 = −68.4 πππ − ππ‘ Problem # 4 - Solution Problem # 4 - Solution d) Divide the moments between the column and middle strips (i) Interior Negative Moments This division is a function of πΌπ1 π2 Τπ1 , which equals zero, because there are no beams parallel to BE. Using Table 13-3, Interior Column − strip negative moment = 0.75 × −160 = −120 πππ − ππ‘ −120 = = −12.6 πππ − ππ‘/ππ‘ width of column strip 9.5 (π€πππ‘β ππ ππππ’ππ π π‘πππ) Interior Middle − strip negative moment = 0.25 × −160 = −40 πππ − ππ‘ Half of this goes to each half-middle strips beside column strip BE Problem # 4 - Solution Problem # 4 - Solution (ii) Positive Moments This division shall be carried out as per Table 13-4, shown below Problem # 4 - Solution Column − strip positive moment = 0.60 × 114 = 68.4 πππ − ππ‘ Middle − strip positive moment = 0.40 × 114 = 45.6 πππ − ππ‘ Half of this goes to each half-middle strip (iii) Exterior Negative Moments As per Table 13-5, the division of exterior negative moment is a function of πΌπ1 π2 Τπ1 , which is again equal to zero, because there are no beams parallel to π1 . Also: πΈππ πΆ π½π‘ = πΈππ πΌπ and πΆ=ΰ· π₯ 1 − 0.63 π¦ π₯ 3π¦ 3 The πΈππ and C refer to the attached torsional member shown in figures on slide # 13 Problem # 4 - Solution Computation of value of “C” To compute C, divide the edge beam into rectangles. The two possibilities are shown in figures opposite. Considering figure-a: πΆ=ΰ· π₯ 1 − 0.63 π¦ π₯ 3π¦ 3 (a) Attached torsion member (b) Attached torsion member 1 − 0.63 × 12Τ16 123 × 16 1 − 0.63 × 8Τ8 83 × 8 πΆ πππ. −π = + = 5367 ππ.4 3 3 1 − 0.63 × 20Τ8 203 × 8 1 − 0.63 × 12Τ8 123 × 8 πΆ πππ. −π = + = 3741 ππ.4 3 3 Problem # 4 - Solution Larger of the two governs → πΆ = 5367 ππ.4 Is is the moment of inertia of the strip of slab section perpendicular to the edge, which has b = 8 in., then 12 × 19 × 83 πΌπ = = 9728 ππ.4 12 Because ππ′ is the same in slab and beam, πΈππ = πΈππ and πΈππ πΆ π½π‘ = πΈππ πΌπ 5367 → π½π‘ = = 0.276 2 9728 Problem # 4 - Solution Interpolating from Table 13-5, we get: π½π‘ = 0: 100 πππππππ‘ ππ ππππ’ππ π π‘πππ π½π‘ = 2.5: 75 πππππππ‘ ππ ππππ’ππ π π‘πππ ⇒ πππ π½π‘ = 0.276, 97.2 πππππππ‘ ππ ππππ’ππ π π‘πππ Exterior Column − strip negative moment = 0.972 × (−68.4) = −66.5 πππ − ππ‘ = −7 πππ − ππ‘ /ππ‘ Exterior Middle − strip negative moment = 0.028 × (−68.4) = −1.92 πππ − ππ‘ Problem # 4 - Solution Flow diagram of the calculation of moments in an exterior span of a two-way flat-plate slab Class Work Problem Compute the column strip and middle moments in the direction perpendicular to the edge of the exterior bay of the flat plate slab shown in figure on next slide. The slab is 7.5 in. thick and supports a superimposed service dead load of 25 psf and service live load of 50 psf. There is NO edge beam and columns are all 18 in. square. Class Work Problem
