Verification Example - RC Two-Way Slab Design

Tekla Tedds
Zmodyfikowany: 4 lip 2024
2024
Tekla Tedds Tekla Tedds for Word

Verification Example - RC Two-Way Slab Design

Description

This verification example represents the analysis and design of a reinforced concrete two-way slab utilizing Tekla Tedds. This example is based on the Two-way Slab Example 1 of the ACI Reinforced Concrete Design Handbook, A Companion to ACI 318-19, Volume 1: Member Design (Pages 96 through 113). Comparisons and contrasts are tabularized and discussed regarding the results from Tedds and the ACI Design Example.

Problem statement

Design the reinforced concrete two-way slab shown in Figure E1.1 of the ACI Design Example. The interior strip along Grid B will be designed. See Figures E1.10 and E1.11 of the ACI Design Example for designed slab reinforcement.

Tedds calculation

RC two way slab design (ACI318) - Compared using version 1.3.07

Running the example in Tedds

The Tedds verification examples referenced in this document can be run in Tekla Tedds from the Engineering library index, in the Verification Examples\RC two-way slab design (ACI318) folder.

References

ACI Reinforced Concrete Design Handbook, A Companion to ACI 318-19, Volume 1: Member Design

ACI 318-19: Building Code Requirements for Structural Concrete and Commentary

Example information

Concrete:

γconcrete = 150 lb/ft3

f’c = 5.0 ksi

fy = 60 ksi

 

Uniform Loads:

Superimposed dead load, D = 15 psf

Live load, L = 100 psf

 

Column dimensions: 24” x 24”

 

Concrete cover to reinforcement:

Top clear cover: ¾”

Bottom clear cover: ¾”

Notes and assumptions

  • Analysis and design of the two-way slab will only be performed in the X-direction, matching the ACI design example.
  • Bars running in the X-direction are the outer bars in the slab.
  • The slab does not contain beams or drop panels
  • Lateral loads are resisted by shear walls, so the slab is designed for gravity loads only.
  • Diaphragm design is not considered in this example.
  • The design utilizes the Direct Design Method, which details have been removed in ACI 318-19, but the method can still be used per Section 8.2.1
  • Governing load combination for slab design: 1.2D + 1.6L
Comparison of Results between Tedds and ACI Example 1
Component Tedds Result ACI Example 1 % Difference
Minimum slab thickness for deflection control 5.82” 5.8” 0.3%
Flexural results
Total panel moment (MO) 126.8 kip-ft 127 kip-ft 0.2%
Moment distribution of interior panel, negative moment (Mu-) 82.4 kip-ft 83 kip-ft 0.7%
Moment distribution of interior panel, positive moment (Mu+) 44.4 kip-ft 45 kip-ft 1.3%
Column strip design moment, negative moment (Mx-) 61.8 kip-ft

63 kip-ft

1.9%
Column strip design moment, positive moment (Mx+) 26.6 kip-ft 27 kip-ft 1.5%
Middle strip design moment, negative moment (Mx-) 20.6 kip-ft 20 kip-ft 3.0%
Middle strip design moment, positive moment (Mx+) 17.8 kip-ft 18 kip-ft 1.1%
Required steel area based on strength
Column strip negative (Asx,col,neg) 0.342 in2 / ft 0.373 in2 / ft 8.3%a
Column strip positive (Asx,col,pos) 0.144 in2 / ft 0.156 in2 / ft 8.3%a
Middle strip negative (Asx,mid,neg) 0.111 in2 / ft 0.116 in2 / ft 4.5%a
Middle strip positive (Asx,mid,pos) 0.096 in2 / ft 0.104 in2 / ft 7.7%a
Required steel area including consideration of maximum bar spacing, assuming No. 5 bars
Column strip negative (Asx,col,neg) 0.342 in2 / ft 0.373 in2 / ft 8.3%a
Column strip positive (Asx,col,pos) 0.266 in2 / ft 0.266 in2 / ft 0.0%
Middle strip negative (Asx,mid,neg) 0.266 in2 / ft 0.266 in2 / ft 0.0%
Middle strip positive (Asx,mid,pos) 0.266 in2 / ft 0.266 in2 / ft 0.0%
Slab Reinforcement
Column strip negative #5 @ 9” o.c.
Column strip positive #5 @ 14” o.c.
Middle strip negative #5 @ 14” o.c.
Middle strip positive #5 @ 14” o.c.
Two-way shear strength
Shear force (Vu) 69.6 kips 70 kips 0.6%
Design concrete shear strength (φVc) 137.8 kips 137.4 kipsb 0.3%
One-way shear strength
Shear force (Vu) N/A 32 kips 0.6%
Design concrete shear strength (φVc) N/A 88.5 kips 0.3%

Comparison Notes

aSince the bars running in the X-direction are on the outside, the depth, d = 7” - 0.75” - ⅝”/2 = 5.93”. The ACI example uses the average effective depth, d, of 5.6”, which should only be used when reviewing shear. This leads to the discrepancies in the results.

bThe design example provides the shear stress, so once the stress is multiplied by the effective depth, d, and the perimeter of the critical section, bo the shear strength is displayed.

Conclusion

Upon reviewing the results above, the analysis of the reinforced concrete two-way slab within Tedds matches the ACI Design Example 1 (apart from minor differences due to rounding and precision).

Czy to było pomocne?
Wstecz
Dalej