Direct Design Method for Prestressed Concrete Slabs

DIRECT DESIGN METHOD FOR PRESTRESSED CONCRETE SLABS

Chen-Hwa Wang Associate Professor of Civil Engineering and Mechanics The Catholic University of America Washington, D.C.

Although several methods such as with maximum bending moment moment-balancing( 1 , load-balanc- may be considered as the control ing( 2 ) and structural membrane section. However, a continuous slab theory (31 can be applied to design has several maximum moment peaks of slabs, none of these is a direct in the middle portion of spans and design method. At present, the load- over supports. Therefore, selection balancing method is widely applied of a control section which has the because it is simpler than the others. largest maximum moment may not It is a trial and analysis approach necessarily provide the optimum de- for obtaining the expected stresses sign. Conventional reinforcing steel and behavior in a slab by the proper or discontinuous tendons for resist- determination of slab thickness, bal- ing partial moment stress at several anced load, eccentricity, and applied sections with moments higher than prestressing. Since this approach is at the control section becomes more time consuming and difficult to a practical and economical. Hence, new design engineer in practice, the selection of the control section this paper presents a straightfor- really depends on the type of struc- ward method and procedures for de- ture and engineering judgment sign of slabs with consideration of whether or not to use mild steel re- the selection of control sections and inforcement or local prestressing in distribution of prestressing. Illustra- the design. In any case, the selection tive examples are also included for of a control section in each direction practical application. of a two-way slab should be considered independently. CONTROL SECTIONS Control sections, which are the DIRECT DESIGN basis of the direct design method, The direct design method is de- should be first selected considering veloped based on the load-balancing maximum moment and the economy analysis for obtaining the desired of the entire slab; the behavior of minimum and/or allowable maxi- the slab under design load can then mum compressive stresses at the ex- be predicted. treme fibers of the control sections. For a simple span, the section It is governed by the appropriate

62 PCI Journal A direct method for design of slabs, instead of the usual process of trial and analysis, is introduced considering desired behavior and necessary distribution of prestressing in the slabs. Direct design equations are developed, procedure is outlined, and illustrative problems are included.

computation of the balanced load control sections, using continuous and the amount and eccentricity of draped tendons. However, the ideal- the applied prestressing force de- ized configuration as shown in Fig. pending on the load and boundary 1 is used in the direct design method conditions of the slab. When the for simplicity. The effect of pre- dead load is significantly smaller stressing due to the portion of ten- than the computed balanced load, don that is concave downward is concrete stresses should be investi- minor and may be neglected or com- gated for possible overstress at the puted( 4 > depending on the curva- initial stages of the member under ture of the tendon and the qualified prestressing and dead load. If the professional judgment.

allowable stress at that stage is ex- DESIGN EQUATIONS ceeded, a thicker slab or conven- The direct design equations for tional reinforcement should be pro- slabs under uniformly distributed vided to prevent tensile cracking. loads are presented below and their Prestressing is uniformly provided derivations, based on the load-bal- across the whole width of slab and ancing analysis, are given in the is proportional to the moment in the Appendix.

rt __ _^--- rt ^ a^rs4.gl ^endo^ !ft

Fig. 1. Idealized tendon configuration June 1968 63

In general, the slab thickness is where B is the moment coefficient, determined by its moment stress, M = BWL2, depending on the slab stiffness, or soundproof and fire rat- boundary condition, W is the total ing requirements, depending on the design load applied, psf or plf, and type of structure. However, if a F is measured in kips/ft. or kips. fully employed control section is ex- Similarly, for a two-way slab, Fig. pected, the slab thickness, t, should 2, the balanced load and the effec- be computed as tive prestressing in the direction "a" are t ` L k(1.8fWl + 4f) ( 1) Wa _ 48kaBaW x 1- 1 + 48(k¢B¢ + kbBb) where t = minimum slab thickness, 2 in. L = span length, ft. Cl f W¢ / (5) W1 = balanced load, psf 8kb 1 + 48kbBb (6) where C k = ratio of the transformed Lb 6B¢LQ eccentricity to slab thickness as shown in Fig. 1, 72B¢WL2 and F= k=1—r1-0.5(r2+r3) t[1 + 48(k¢B¢ + kbBb)] x f = design minimum com- — fC¢t2 pressive stress, psi (7) fl= ultimate concrete strength, psi The formulas for balanced load and When the desired minimum com- the effective prestressing in the di- pressive stress is zero, f = 0, the rection "b" can be obtained by sim- slab thickness, in Eq. 1 can be sim- ply changing the subscripts and plified to superscripts from "a" to "b" and "b" to "a" in Eqs. 5 to 7. t = 0.745 L kc (2) Loads on beams supporting two- way slabs are non-uniformly dis- Thicker slabs can be used for any tributed along their lengths. For other reason, but its maximum com- practical simplicity, as accepted in pressive stress in the control section reinforced concrete slab-beam de- is always lower than the allowable sign, a uniform load could be as- one and is inversely proportional to sumed as the square of the thicker slab when f=0. W = W^L¢ + W2L¢ (8a) According to direct design, the 2 3 balanced load, W1, and the effective for load on short beam prestressing, F, in a one-way slab or the beam supports for one-way ,, W = W1L + W2L¢ X 2slabs can be computed as 3

t C 3 — (L¢/Lb)2 1 _ 48kBW + 8k f (8b) ) (L (3) 2 Wl — 1 + 48kB for load on long beam F, = 72BW L2 + 12 f t2 (4) t(1 + 48kB) where residual load W2 = W — W1. 64 PCI Journal Therefore, Eqs. 3 and 4 could also rigidity, or other factors, and then be applied to the direct design of compute the total design load W, beams. psf or plf. From the above equations, the 2. Determine the control section desired minimum compressive stress based upon the moments and the in the control sections, and in any economy of the entire slab. other sections with lower moment 3. Compute an acceptable maximum stress, can be automatically satisfied. k-value based on the minimum The maximum allowable compres- clear coverage for the tendons sive stress can be obtained only when and the assumed slab thickness. the slab thickness satisfies Eqs. 1 Maximum sag is preferred except or 2. For any section with higher at the free ends where the center moment stress than the control sec- of gravity of the concrete should tion additional reinforcement or lo- coincide with the center of grav- cal prestressing must be provided. ity of the steel. 4. Compute the balanced load, W1, DESIGN PROCEDURES for comparison with the dead 1. Assume the dead load, i.e., thick- load of slab and for checking the ness, of slab based on its stress, possibility of reverse stress.

Fig. 2. Idealized configuration ma two-way slab June 1968 65

5. Compute the minimum thickness the purpose of getting a watertight of slab by Eq. 1 and compare it roof slab. Compute the required pre- with the assumed slab thickness stressing for f = 4000 psi. to determine whether or not it Since the slab thickness is given, needs to be revised. the total design load of slab is 6. Repeat steps 1 to 5 if slab thick- W=94+75=169 psf. Assuming ness is revised. the section over the middle support 7. Compute effective prestressing, as the control section, its maximum F, and number of tendons and moment is WLL/8 and B = 1/8. spacings. Computing k: r1 =1.47/7.5 = 0.188 8. Compute the necessary reinforce- lowest h/4 in. at % L ment or local prestressing, based from the end support on reinforced or prestressed con- r2= 1.25/7.5=0.166 crete theory, whenever it is re- minimum cover over quired. support APPLICATION r8 =0.5 at free end The use of the direct design equa- k = 1 - 0.188 - 0.5 (0.166 + 0.50) tions will be illustrated by means of = 0.479 the following example problems. Since the desired minimum compressive stress is f = 100 psi, from Eq. 3 Example 1. A one-way roof slab has the balanced load is a plan as shown in Fig. 3. The 7½-in, concrete slab weighs 94 psf. W =48X0.479XAX169+8X0.479X100X 1 1+ 48X0.479X1/$ and carries a live load of 75 psf. ( la Minimum coverage for the cables 1 7-5 / is 1-in, measured to the cgs and 30 _ 132 psf the minimum desired compressive and the slab thickness is computed stress in the concrete is 100 psi for by Eq. 1

I! V

i+ ;I

L=30

Fig. 3. Slab dimensions for Example 1 66 PCI Journal

0.132 Therefore, t=30 0.479(1.8x4+4x0.1) f1 —f2 =100 psi = 30 x 0.19 same as requested. =5.7 in. <7.5in. f1 +fz=1000 psi less than the allowable com- The reversing stress is insignificant pressive stress, 0.45f ' = 1800 2. = 169 — 132 as the residual load W psi, as expected. = 37 psf is so close to the load 132 — 94 = 38 psf. Example 2. An 8-in, slab supported Then, from Eq. 4 the required effec- on four walls, Fig. 4, is to be post- tive prestressing is tensioned in two directions. Design live load is 100 psf. Compute the F _ 72x'/sXO.169x302+12x0.10(7.5)2 amount of prestress in two direc- 7.5(1+48x1/8x0.479) tions for f = 4000 psi. Since the slab thickness is given, 1370+67.5 = 49.5 k/ft. of width 29 the total design load of slab is W = 100 + 100 = 200 psf. Consider the and no reinforcing steel or local pre- section with maximum moment at stressing is needed. If one is inter- midspan as the control section. No ested in checking the results, f I and rebar or local prestressing is needed. f 2 can be computed from Eq. A10 Let r1=1.25/8=0.156 and Al2 as minimum cover _ 49,500 r2=ra=0.5 fl 12x7.5=550 psi at free end Thus ka=kb =1-0.156 f^ -- 0.50 = 0.444

6X 1/8 X169X302 -48X X 0.479 X 100 (7.5)2 7.52(1+48X1//X0.479) From the ACI Building Code (ACI = 450 psf 318-63), B¢ = 0.061 and B° = 0.019

I tI ( ^

I I t^ ifl?

L6 = 1O

Fig. 4. Slab dimensions for Example 2 June 1968 67 for the span ratio m = 30/40 = 0.75. 36,600 + The balanced load in direction "a" 12x8 is computed from Eq. 5 and with 0.061x74.2(30)2x12x6 f=0 12(8)2 48x0.061x0.444x200 = 0 psi minimum desired Wl 1+48x0.444(0.061+0.019) stress 260 = 764 psi maximum compressive —_ 95.9 psf = 2 .705 stress fb= Lb -4- and the balanced load in direction /1 — 2 "b" is _ 20,200 + Wb = 48x0.019x0.444x200 ^12x8— 1 1+48x0.444x0.08 0.019x74.2(40)2x12x6 = 81 12(8)2 2.705 = 29.9 psf = 0 psi minimum desired stress There is no reversing stress since = 420 psi maximum compressive W2 = 200 — 125.8 = 74.2 psf is less stress than the live load. If the slab section was reduced to Computing slab thickness, t, by 5.2 in., the maximum compressive Eq. 2 stress would be 0.096 t=0.745 x 30 4 x 0.444 f=(+) (52)2 = 22.4 x 0.232 = 764 x 2.36 = 1800 psi = 5.2 in. < 8.0 in. Example 3. A flat plate has a plan It. is a partially employed section as shown in Fig. 5. The 7½-in, con- and the maximum allowable com- crete slab weighs 94 psf and carries pressive stress will not be achieved a live load of 100 psf; f = 4000 psi. under design loads. 'Minimum coverage for the cables is The required effective prestress in 11/1-in, measured to the cgs. Allowing the direction "a" is given by Eq. 7 no tension in the concrete, determine the location for the cables and Fa 72 x 0.061 x 0.200 (30)2 compute the required prestressing 8x2.705 for the interior column strip of the = 36.6 k/ft. of width slab in the long direction. An empirical method as shown in and in the direction "b" is the ACI Building Code, Sect. 2104,. _72 x 0.019 x 0.200 (40)2 is applied. Since all the values of FO 8x2.705 negative moments are approximately the same, no rebar or local prestress- =20.2 k/ft. of width ing is intended to be used. Let the Checking for the concrete stress from section at the first interior column Eqs. A21, A23 and A22, A24, the be the control section and the panel actual concrete stresses are which gives the largest k-value is the base. ! ¢—f±f21 For the interior panel: 68 PCI Journal r1=r2=r3=1.25/7.5=0.167 B= 0.18 x 1.09 x 0.922 x 0.5 and k=l-0.167-0.167=0.666 = 0.0905 This is greater than the k-value giv- M = 0.0905 x 0.194 x 252 en for the exterior panel. = 10.98 ft-k/ft. of width k=1-0.167- From Eqs. 3 and 4 the balanced 0.5(0.5+0.167)= 0.5 load is The moment in the flat plate is _ 48 x 0.666 x 0.0905>< 194 M=CMo =BWL2 Wl 1+48 x0.666x0.0905 Where C is the moment in flat plate =144psf panels in percentages of Mo, the nu- and the required prestressing is merical sum of assumed positive and average negative moments at _ 72 x 10.98 the critical design sections of a flat F 7.5 (1 + 48 x 0.666 x 0.0905) plate panel, and = 27.1 k/ft. of width Checking stresses from Eqs. A10 B = 0.18F 1 1— 3L ) r C. and Al2, the stress due to prestressing is Assuming the- diameter of column 27,100 _ is 18 in., c = 1.5, L = 25 ft., and 7.5 ` 302 psi f l 12 x F = 1.09. For the column strip C = 0.05 and the stress due to residual load is

.II

J@25 75 -1 Fig. 5. Flat plate dimensions for Example 3

June 1968 69 6 x 0.0905 x 50 x 625 mid-span of the exterior panel, meas- f2 = 7.52 ured to the center line, is 0.182 x = 302 psi 7.5=1.37in. Therefore, minimum stress at the control section is REFERENCES fl — f2 = 0 as desired 1. Brotchie, J. F. and Russell, J. J., "Flat For the exterior panel: Since a Plate Structures", ACI Proceedings continuous tendon is used and the Vol. 61, No. 8, Aug. 1964, 2, Lin, T. Y., "Load-Balancing Method effective prestressing is F = 27.1 k for Design and Analysis of Prestressed from r2 = 0.5 and r3 = 0.166, locate Concrete Structures", ACI Proceedings, the tendon in the mid-span of ex- Vol. 60, No. 6, June 1963, and Vol. terior panel. 60, No. 12, Dec. 1963. 3. Saether, K., "The Structural Membrane Since C = 0.28 in the exterior panel Theory Applied to the Design of Pre- and B = 0.18x1.09x0.922x0.28 stressed Flat Slabs", PCI Journal, Vol. = 0.056, from Eq. Al2 the value k 8, No. 5, Oct. 1963. can be computed as 4. Koons, R. L. and Schlegel, G. J., "Pre- stressed Continuous Structures", PCI 1+48 x 0.0506k=6 ><0.0506 Journal, Vol. 8, No. 4, Aug. 1963. x 194 x 252 = 36,850 5. Saether, K., "The Structural Mem- k=0.485 brane", ACI Proceedings, Vol. 57, Jan. Therefore, 1961, pp. 827-850. r, =1—k-0.5(r2+r3) 6. Rozvany, G. I. N. and Hampson, A. J. K., "Optimum Design of Pre- = 1 - 0.485 - 0.333 = 0.182 stressed Plates", ACI Proceedings, Vol. The coverage for the cable in the 60, No. 8, Aug. 1963.

APPENDIX

Derivations of the direct design f = ultimate strength of con- equations for slabs is given. crete 1. Slab Thickness Thus, from Eqs. Al and A2 Based upon allowable and desir- f1 =0.225fc+0.5f A3 able stresses, the minimum slab Due to the variation of the location thickness can be computed. From of draped tendons, minimum cover- the load-balancing analysis age for the tendons may not always be at midspan. For the purpose of fl= f2 + f Al and simplicity in a practical design, the f1+f2^0.45f A2 stress due to prestressing is computed by the linear transformation where /1 = concrete stress due to and load-balancing theories as effective prestressing f2 = concrete stress due to W1L2 residual load, W2, which f2 8t 2 [1 — rl —/z(r2 + r3)] is equal to the total = W,L2 load, W, minus the bal- A4 8t2k anced load, W1 f = desired minimum com- where W1 denotes the balanced pressive stress in con- load, rl, r2 and ra are shown in Fig. 1 crete and k=1—r1 -0.5(r2 +r3). From

70 PCI Journal

Eqs. A3 and A4, the minimum slab _ — 8kf(t/L)2 All thickness, t, in inches, for the maxi- W 1 + 48kB mum allowable and the minimum desirable stresses can be calculated. and the stress due to this residual Thus, load is - 6B(WL2 — 8k f t2) _ W, Al2 t L A5 f^t2 (1 48kB) k(1.8f+4f) + 3. Two-Way Slabs If the desired minimum compressive Let W and W be the balanced stress is zero, f — 0, then load of a two-way slab due to the effective prestress, F and Fr, in the t = 0.745 L A6 direction "a" and "b", respectively. The residual load, therefore, is 2. One-Way Slabs and Beams Wz=W—(W;+W^) A13 For a rectangular section of slab Following the same approach and or beam, the moment stress due to derivation as for the one-way slab, the residual load is and f2 = 6M/t2 the balanced load in direction "a" M = BW.,L where is the moment 2, B is computed by Eq. Al as coefficient depending upon the load- ing and boundary conditions. There- W 2 _ 6B W,L« + f 1L 1 fore, 8k,,t2t, 6BW.,L2 and f2— t2 W 48 kB"W2 + 8k f (t/Lq)2 A14 = 6B (W — W)L2 A7 t^ Similarly, the balanced load in the direction "b" is Substituting Eqs. A4 and A7 into Al, the balanced load is computed W = 48 kbBhW2 + 8k1f (t/Lb)2 A15 as From Eq. A13, the residual load, W,,, is 48kBW + 8k f 2 _ (J—)L ka k W — 8ft2 + Wl 1 + 48 kB A8 La LG Then, from Eq. A4 the stress due W2 — + 48(kaBa + kBh) A16 to prestressing is Substituting Eq. Alb in Eqs. A14 and A15, _ 6BWL2 + f t' A9 f^ t (1 + 48kB) 48kcBaW W¢ =_____ 1 + 48(k"Ba + kbBb) Hence, the total effective prestressing per unit strip is F = 12tf1 — fc— t- ) A17 (l W = 72BWL2 + 12 f t2 A10 where t(1 + 48kB) 8kL 1 + 48kbBv C^ _ — A18^ Therefore, the residual load Wz is L 6BtLQ

June 1968 71

Wb _ 48kbBbW x l — fW J l A23 1 1+ 48(kaBa + kbBb) C fCbt2 \ and in direction "b" is — W 1 A19 (l b _ 6BbW Lb Cb = 8k¢ 1 + 48/kaBa fl t2 [1 + 48(kaBa + kbBb)] X A20 L¢ 6BbLb b Cl— fW 1 A24 Hence, the moment stress in direc- is tion "a" due to W2 Finally, the effective prestress per 6Ba Lz unit strip of slab in direction "a" f2= Wvt " A21 72BaW LQ F= and the moment stress in direction t [1 + 48(k¢B- + kbBb)] X "b" due to W2 is fCat 1 A25 b _ 6BbW2 Lv C1— W) A22 f2 — tL and in direction "b"

The stress in direction "a" due to Fb = 72BbW Lb prestressing is t [1 + 48(kaBa + kbBb)] X _ 6BaW L¢ 1 i _112 X W A26 tl t2 [1 + 48(kaB" + kbBb)] Cb

Discussion of this paper is invited. Please forward your discussion to PCI Headquarters by September 1 to permit publication in the December 1968 issue of the PCI JOURNAL.

72 PCI Journal