Assignment. Design of a RC slab, Advanced Structural Design

GLOBAL COLLAPSE

In this part we will we will consider a collapse mechanism on the whole slab. The first step consists in determining the yielding pattern. Since we are dealing with a complex geometry, several assumptions have been made in order to propose a yielding pattern :

• δ- Yield lines are regarded as rigid bodies so the displacement is constant.
• β γ λ- We have three degrees of freedom : , and . In order to simplify the problem, it is transformed in a one degree of freedom problem : λ = 0,5
• The slab is considered as large enough to consider that . In reality• λ > 0,5 because of the clamped section induces γ
• In order to avoid to very complex geometrical computations the value of is set• 0,35 toβ
• Finally is the only degree of freedom remaining. We will then balance the external andβ pinternal works for several values of in order to determine the value of which willu globalp = p (β )be the the minimum value of .u u 35

previous figure shows the assessed yield line pattern.
Variables :

  
a = 7.8 m
  
λ = 0.5
  
γ = 0.35

and
2δ δ 4δ δθ = θ = θ′ = θ =
y x y x• a βa a γa
Internal Work Analysis :
Each yield line induces an internal work given by this formula :
∑ ∑Wi = m θ l + m θ lx x y y
Of course, our steel reinforcement is not isotropic so the resisting moments are not necessarily constant along the yield lines, we have to implement a method in order to compute an average moment in the yield line. Further details on this method are given for the Yield Line (a).
(a) Yield line a 36l ’ la aɑ l ’’aβ﹒a
Limit case : β=0,462 Limit case : β=0,199
β < 0.199
The previous figure shows that are three different states : ;0.199 ≤ β ≤ 0.462 β > 0.462; and finally . The computation of the internal work for the β yield line (a) will be different for each

of these cases. The specific values for are obtainedas follows : 3600675 β = 0.462 = 0.5 ⋅ tanβ = 0.199 = 0.5 ⋅ tan ;lim1 lim21700 2125β < 0.199• β ⋅ a βtan α = ⇒ α = arctan0,5 ⋅ a 0,5675 0.0865a2 2 =l = 0.5 + β l′ = l′′ = l − l′; ;a a a a asin α sin α[ ]l′l′ ′a [ ] a aWi = θ ⋅ ⋅ βa ⋅ × 44.3 + × 87,66⋅ 46.76 + θa x y2 l la a[ ]l′l′ ′2βaa a a⋅ 46.76 + ⋅ × 44.3 + × 87,66= δ 2βa a l la a0.199 ≤ β ≤ 0.462• β ⋅ a βtan α = ⇒ α = arctan0,5 ⋅ a 0,5 371700 0.218a2 2 =l = 0.5 + β l′ = l′′ = l − l′; ;a a a a acos α cos α[ ]l′ l′′a [ ] a aWi = θ ⋅ ⋅ βa ⋅ × 44.3 + × 87,66⋅ 46.76 + θa x y2 l la a[

<l′ l′′2βaa a a⋅ 46.76 + ⋅ × 44.3 + × 87,66= δ 2βa a l la aβ > 0.462• β ⋅ a βtan α = ⇒ α = arctan0,5 ⋅ a 0.51700 0.218a2 2l = 0.5 + β l′ = l′′ = l − l′=; ;a a a a acos α cos α3600 0.462a2 2h = 0.5 + β h′ = = h′′ = h − h′; ;a a a aa sin α sin α>]

[ ][ h′ l′ l′h′ ′ ′a a a a aWi = θ ⋅ × 46,76 + × 94,98 + θ ⋅ βa ⋅ × 44.3 + × 87,66⋅a x y2 h h l la a a a[ ] [ ]h′ h′ l′ l′′ ′2βaa a a a a⋅ × 46,76 + × 94,98 + ⋅ × 44.3 + × 87,66= δ 2βa h h a l la a a aβAll the values for each between 0 and 0.65 are computed into an Excel tab. Less detailswill be provided for the other yield lines given that the approach is exactly

the same.(b) Yield line b β ⋅ a ββ α tan α = ⇒ α = arctan

For each value of , is computed in the same way : 0,5 ⋅ a 0.50 ≤ β ≤ 0.462• [ ]l′′l′a [ ] b bWi = θ ⋅ ⋅ βa ⋅ × 44.3 + × 87,66⋅ 46.76 + θb x y2 l lb b[ ]l′ l′′2βaa b b⋅ 46.76 + ⋅ × 44.3 + × 87,66= δ 2βa a l lb b 38β > 0.462• [ ] [ ]h′ l′h′ ′ ′l′a b b b bWi = θ ⋅ × 46,76 + × 94,98 + θ ⋅ βa ⋅ × 44.3 + × 87,66⋅b x y2 h h l lb b b b[ ] [ ]h′ h′ l′ l′′ ′2βaa b b b b⋅ × 46,76 + × 94,98 + ⋅ × 44.3 + × 87,66= δ 2βa h h a l lb b b b(c) Yield line cγ ⋅ a γ⇒ α′tan α′ = = arctan0,25 ⋅ a 0,25500 0.0641a2 2 =l = 0.25

+ γ l′ = l′′ = l − l′; ;cc c c csin α sin α][ l′l′ ′a [ ]c cWi = θ′ ⋅ × 46,76 + × 94,98 + θ′ ⋅ γa ⋅ 44,3⋅c x y4 l lc c[ ]l′ l′′ 4γaa [ ]c c⋅ × 46,76 + × 94,98 + ⋅ 44,3= δ 4γa l l ac c(d) Yield line dBy symmetry we can asses that :[ ]l′l′ ′a [ ]c cWi = Wi = θ′ ⋅ × 46,76 + × 94,98 + θ′ ⋅ γa ⋅ 44,3⋅d c x y4 l lc c[ ]l′′l′ 4γaa [ ]c c⋅ × 46,76 + × 94,98 + ⋅ 44,3= δ 4γa l l ac c(e) Yield line e [ ] [ ]Wi = θ ⋅ a(1 − γ − β ) ⋅ 42,79 + θ′ ⋅ a(1 − γ − β ) ⋅ 94,98e y y( )= δ 6 ⋅ (1 − γ − β ) ⋅ 87,66 39(neg) Negative yield line ][ 52 × 44,3 + ×

158,72Wi = θ ⋅ 0,65a ⋅neg y 7 7( )[ ]= δ 2 ⋅ 0,65 ⋅ 126,028

We finally obtain: Wi = Wi + Wi + Wi + Wi + Wi + Witot a b c d e neg= Ki(β ) × δ

External Work Analysis:

1 1 )( ) ( 2V = ⋅ β ⋅ δ⋅ a ⋅ βa ⋅ δ = ⋅ a1 3 3 2a δ δaV = ⋅ a(1 − β − γ) = ⋅ (1 − β − γ)2 2 21 1 )( ) ( 2V = ⋅ γ ⋅ δ⋅ a ⋅ γa ⋅ δ = ⋅ a3 3 3

Then we obtain: We = V × ptot tot u )( × p= V + V + V1 2 3 u= Ke(β ) × p × δu

The equality between internal and external works provides us: Ki(β )We = Wi ⇒ p =tot tot u Ke(β ) p = p (β )

The following figure is the plot of u u 4057,00 pu=pu(β)55,0053,0051,0049,00pu=[kN/m2] 47,0045,0043,0041,0039,0037,00 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7β [Ø] p (β )

The collapse load is basically the minimum value taken

yield line (b).This reasoning is of course not perfectly accurate but reasonable enough. On the otherhand, fixing the value of γ to 0,35 is a stronger assumption. With values of γ greater than 0,35 the geometry of the collapse situation would have been completely modified and therefore the external work would have been modified too. It is here hard to assess how much it would have modified the final result that’s why this assumption is considered as “strong”.

LOCAL COLLAPSE

Now that we ensure the slab safety while considering a global collapse mechanism, we will check the slab safety with respect to the design load considering local mechanisms. The following local mechanisms will be considered :

(1)

(2)

(4)

(4)

Each part of the slab will be analyzed with the same methodology that for the global collapse mechanisms. In order to be able to do this, the sections that are cut are considered as clamped sections.

LOCAL MECHANISM 1 42

Variables :

c = 0,25 × a

d =

0,65 × a• 2β =We assume as there is clamped section on a side and a supported one on the• 3opposite side2δ δθ = θ =;y x• c βdInternal Work Analysis:c [ ] [ ]⋅ 46.76 + θWi = θ ⋅ ⋅ βd ⋅ 44.31 x y2 )( 2βdc ⋅ 46.76 + ⋅ 44.3= δ 2βd cWi = Wi (symmetry)2 1 c [ ][ ]Wi = θ ⋅ ⋅ &bet