Figure 1: Representative strip. = = 3.70 m. min. per unit length of the selected strip: Own weight of slab = = 0.

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1 Example (8.1): Using the ACI Code approximate structural analysis, design for a warehouse, a continuous one-way solid slab supported on beams 4.0 m apart as shown in Figure 1. Assume that the beam webs are 30 cm wide. The dead load is 300 kg/m in addition to own weight of the slab, and the live load is 300 kg/m. Use 0 kg / cm, f c f y 400 kg / cm. Solution: 1- Select a representative 1 m wide slab strip: The selected representative strip is shown in Figure 1. - Select slab thickness: Figure 1: Representative strip The clear span length l n m one-end continuous spans, Slab thickness is taken as 17 cm. h l / 4 400/ cm min 3- Calculate the factored load w u per unit length of the selected strip: Own weight of slab ton/m ( ) ( 0.30) 1.3 ton / m w u 1.0 a strip 1 m wide, w u 1.3 ton / m 4- Evaluate the maximum factored bending moments in the strip: The results are shown in the following table. Points at which moments and shear are calculated are shown in Figure.

2 A B C D E Moment coefficient - 1/4 1/14-1/10-1 /11 1/16 Factored moment in t.m Figure : Points at which moments and shear are evaluated - Check slab thickness for beam shear: Effective depth d cm, assuming φ 1 mm bars Vu max wu ln ( 1.3)( 3.70).87 ton Φ V c / ton >.87 tons ( ) ( )( ) O.K. i.e., slab thickness is adequate in terms of resisting beam shear. 6- Design flexural and shrinkage reinforcement: Steel reinforcement ratios are then calculated, and be checked against minimum and maximum code specified limits, where 0.8 f c ρ 1 fy M Φ b d f Minimum reinforcement (100) (17) 3.06 cm /m c u The reinforcement for positive and negative moments are shown in the following table. Factored moment in t.m Reinforcement ratio Steel reinforcement cm Minimum reinforcement cm Bar size mm φ 10 φ 10 φ 10 φ 10 φ 10 Bar spacing cm Calculate the area of shrinkage reinforcement: Area of shrinkage reinforcement (100) (17) 3.06 cm /m

3 Select reinforcement bars: Main and secondary reinforcement bars are also shown in the table shrinkage reinforcement use φ 10 cm, or 4 φ 10 mm/m. Check bar spacing against code specified values main reinforcement, spacing between bars is not to exceed the larger of 3 (17) 1 cm and 4 cm, which is already satisfied as shown in the table. shrinkage reinforcement, spacing between bars is not to exceed the larger of (17) 8 cm and 4 cm, which is also satisfied. 7- Prepare neat sketches showing the reinforcement and slab thickness: Figure 3 shows reinforcement details and required slab thickness. (a) (b) Figure 3: (a) Section A-A; (b) reinforcement details

4 Example (8.): Design the slab shown in Example (8.1) using any available structural analysis software. Solution: 1- Select a representative 1 m wide slab strip: The selected representative strip is shown in Figure 1. - elect slab thickness: Same as in Example (8.1), the thickness is taken as 17 cm. 3- Calculate the factored load w u per unit length of the selected strip: a strip 1 m wide, w u 1.3 ton / m 4- Evaluate the maximum factored shear forces and bending moments in the strip: Shear force and bending moment diagrams are shown in Figure 4. Figure 4: Shearing force and bending moment diagrams - Check slab thickness for beam shear: Effective depth d cm, assuming φ 1 mm bars. V u max 3.7 ton Φ ( 0.3) 0 ( 100)( 14.40) / ton 3.7 tons O.K. V c 0.7 > i.e., slab thickness is adequate in terms of resisting beam shear. 6- Design flexural and shrinkage reinforcement:

5 Steel reinforcement ratios are calculated and checked against minimum and maximum code specified limits. M u.8 t. m ρ ( 0) ( ) ( 100)( 14.4) ( 0) s ( 100)( 14.4) 4.3 cm A /m, use φ 10 1 cm. M u 1.87 t. m ρ ( 0) ( 1.87) 0.9 ( 100)( 14.4) ( 0) s ( 100)( 14.4) 3.46 cm A /m, use φ 10 0 cm. M u 1.66 t. m ρ ( 0) ( 1.66) 0.9 ( 100)( 14.4) ( 0) 0.00 s ( 100)( 14.4) 3.17 cm A 0.00 /m, use φ 10 0 cm. M u 1.68 t. m ρ ( 0) ( 1.68) ( 100)( 14.4) ( 0) 0.00 s ( 100)( 14.4) 3.17 cm A 0.00 /m, use φ 10 0 cm. M u 0.93 t. m ρ ( 0) ( 0.93) 0.9 ( 100)( 14.4) ( 0) s ( 100)( 14.4) 1.78 cm As, min A < s ( 100)( 17.0) 3.06 cm A /m, use φ 10 cm. M u 0.74 t. m A s < A s,min

6 s ( 100)( 17.0) 3.06 cm A /m, use φ 10 cm. Calculate the area of shrinkage reinforcement: Area of shrinkage reinforcement (100) (17) 3.06 cm /m, use φ 10 cm. Select reinforcement bars: It is already done in step 6. Check bar spacing: Same as in Example (8.1). 7- Prepare neat sketches showing the reinforcement and slab thickness: Figure shows details of the required reinforcement. Figure : Section A-A Example (8.3): Design a one-way ribbed slab to cover a 4 m 10 m panel, shown in Figure 6. The covering materials weigh 10 kg/m, concrete hollow blocks are 40 cm cm 0 cm in dimension, each 0 kg in weight, equivalent partition load is equal to 7 kg/m, and the live load is 00 kg/m. Use 0 kg / cm, f c f y 400 kg / cm, and 3 γ plaster 100 kg / m. Solution:

7 Figure 6: Ribbed slab 1. The direction of ribs is chosen: Ribs are arranged in the short direction as shown in Figure 6.. The overall slab thickness h is determined: Minimum slab thickness, h 400/16 cm min Topping slab thickness 0 cm Let width of web be equal to 10 cm Design of topping slab: ( ) ( 0.0) 0.74 ton / m w u 1.0 a strip 1 m wide, w u 0.74 ton / m ( 0.40) / t. m Mu wu l / t ( 0.011)( 10) ( 100) 3 Mu 3 1. cm < cm Φ b f c Area of shrinkage reinforcement ( 100)( ) 0.90 cm m A s / Use 4 φ 6 mm / m in both directions. 3. The factored load on each of the ribs is to be computed: Factored load per rib is shown below, supported by Figure 7.

8 Figure 7: Factored load calculation Total volume (hatched) m 3 Volume of one hollow block m 3 Net concrete volume m 3 Weight of concrete ton Weight of concrete /m 0.081/(0.) (0.) 0. ton/m Weight of hollow blocks /m 0/(0.) (0.) (1000) 0.16 ton/m Weight of plastering coat /m 0.0 (100/1000) 0.04 ton/m Total dead load ton/m ( 0.6) ( 0.0) 1.10 ton / m w u 1.0 w u / m 1.10 t/m w u / rib 1.10 (0/100) 0. t/m 4. Critical shear forces and bending moments are determined: Maximum factored shear force 0. (4/) 1.10 ton Maximum factored bending moment 0. (4) (4)/ ton. m. Check rib strength for beam shear: Effective depth d cm, assuming φ 1 mm reinforcing bars and φ 6 mm stirrups. ( 0.7)( 0.3) 0 ( 10)( 1.8) / ton 1.1Φ Vc 1.1 > 1.10 t Though shear reinforcement is not required, to be used to carry the bottom flexural reinforcement. 4 φ 6 mm U-stirrups per meter run are 6. Design flexural reinforcement: Since the maximum factored moment creates compression in the flange, the section of maximum positive moment is to be designed as a T-section, shown in Figure 8. Assuming that a < cm, compressive force in concrete is

9 C 0. 8 ( f ) ( b ) ( a) c e C 0. 8 (0) (0) ( a) / a ton M u Φ M n Φ ( C) ( d a / ) ( 10.6 a)( 1.8 a / ) / a 43.6 a Figure 8: Section size at maximum positive moment Solving this equation gives a 0.4 cm, which means that the assumption made before is valid and the other root is too large to be considered. From equilibrium, C T Area of flexural reinforcement ( 0.4) ( 1000 ) 10.6 A s 1.37 cm 400 Use φ 10 mm, one is straight and the other is bent-up in each rib at its bottom side. 7. Neat sketches showing arrangement of ribs and details of the reinforcement are to be prepared: Figure 9 shows required rib reinforcement and concrete dimensions. (a)

10 (b) Figure 9: (a) Reinforcement on plan; (b) Reinforcement details at section A-A Loads on main beam: The load on the main beams which run perpendicular to the ribs include: a. Load from the slab (4/)(1.10).0 t/m b. Weight of partition loads applied directly on the beam, if any. c. Part of the beam own weight is not included in slab design calculations (projection above or below the slab). Notice that loads in (b) and (c) need to be factorized by multiplying each of them by a factor of 1..

11 Example (8.4): Design the ribbed slab shown in Figure 10. The covering materials weigh 00 kg/m, concrete hollow blocks are 40 cm cm 17 cm in dimension, each 17 kg in weight, and the live load is 30 kg/m. Use 300 kg / cm, f c f y 400 kg / cm, and 3 γ plaster 100kg / m. Figure 10: Plan of ribbed slab Solution: 1- The direction of ribs is chosen: Ribs are arranged in the short direction as shown in Figure 10. The overall slab thickness h is determined: h 400/ cm, for one end continuous panels min h 10/ cm, for cantilevered panels min Use an overall slab thickness of.0 cm. Topping slab thickness 17 cm Let width of web be equal to 10 cm Area of shrinkage reinforcement A s ( 100) ( ) 0.9 cm / m

12 Use 4 φ 6 mm / m in both directions. - The factored load on each of the ribs is to be computed: Factored load per rib is shown below supported by Figure 11. Figure 11: Factored load calculation Total volume (hatched) m 3 Volume of one hollow block m 3 Net concrete volume m 3 Weight of concrete ton Weight of concrete /m 0.06 / (0.) (0.) 0.1 ton/ m Weight of hollow blocks /m 17/(0.) (0.) (1000) ton/ m Weight of plastering coat /m 0.0 (100/1000) 0.04 ton/ m Total dead load ton/ m ( 0.88) ( 0.3) 1.7 ton / m w u 1.0 a strip 1 m wide, w u 1.7 ton / m w u / rib 1.7( 0 / 100 ) 0.64 t / m 3- Critical shear forces and bending moments are determined: Analyzing the strip shown in Figure 1 using the BENARI structural software, one gets the shown shear force and bending moment diagrams.

13 Figure 1: Shearing force and bending moment diagrams 4- Check rib width for beam shear: Effective depth d cm, assuming φ 1 mm reinforcing bars and φ 6 mm stirrups. ( 0.7)( 0.3) 300 ( 10)( 18.8) / ton 1.1Φ Vc 1.1 Since critical shear section can be taken at distance d from faces of beam, the rib shear resistance will be considered adequate and the assumed web width will be kept unchanged. Though shear reinforcement is not required, 4 φ 6 mm U-stirrups per meter run are to be used to carry the bottom flexural reinforcement. - Design flexural reinforcement: a. Positive moment reinforcement: Since factored positive moments create compression in the flange, the sections at their corresponding locations are to be designed as T-sections. Since the positive moment values are relatively small, the largest of the three values will be considered here and same reinforcement will be used at the two other locations.

14 Figure 13: Section at maximum positive moment Assuming that a < cm, compressive force in concrete, shown in Figure 13, is given by C 0. 8 ( f ) ( b ) ( a ) c e C 0.8 (300) (0) (a) / a ton M u Φ M n Φ ( C) ( d a / ) ( 1.7 a) ( 18.8 a / ) / a 37.6 a Solving this equation gives a 0.37 cm, which means that the assumption made before is valid and the other root is too large to be considered. From equilibrium, C T Area of flexural reinforcement ( 0.37) ( 1000 ) 1.7 A s 1.14 cm 400 Use side. φ 10 mm, one is straight and the other is bent-up in each rib at its bottom b. Negative moment reinforcement: Since factored negative moments create compression in the web, the sections at their corresponding locations are to be designed as rectangular sections. M u 1.07 t. m ρ ( 300) ( 1.07) 0.9 ( 10)( 18.8) ( 300) ( 10)( 18.8) 1.6 cm A s , use 1φ 10 mm + 1φ 1 mm per rib. M u 0.83 t. m

15 ρ ( 300) ( 0.83) 0.9 ( 10)( 18.8) ( 300) ( 10)( 18.8) 1.4 cm A s , use φ 10 mm per rib. M u 0.7 t. m ρ ( 300) ( ) ( 10)( 18.8) ( 300) ( 10)( 18.8) 1.07 cm A s 0.007, use φ 10 mm per rib. c. Neat sketches showing arrangement of ribs and details of the reinforcement are to be prepared: Figure 14 shows required rib reinforcement and concrete dimensions. (a)

16 (b) Figure 14: (a) Slab reinforcement; (b) sect ion A-A

Design of Reinforced Concrete Structures (II)

Design of Reinforced Concrete Structures (II) Discussion Eng. Mohammed R. Kuheil Review The thickness of one-way ribbed slabs After finding the value of total load (Dead and live loads), the elements are