`çäìãåë=fåíÉê~Åíáçå= aá~Öê~ãë
rkfsbopfqv=lc=tfp`lkpfk=pqlrq
`liibdb=lc=p`fbk`bI=qb`eklildvI=bkdfkbbofkdI=^ka=j^qebj^qf`p
ib`qrob=u aêK=g~ëçå=bK=`Ü~ê~ä~ãÄáÇÉë
`çäìãåë=Ó= `çãé~êÉ=j~íÉêá~äë
Let’s try to visualize the effect of strength and the geometry that would correspond to a design for a specific load. Let’s take two very prominent materials:
Steel: E=29000000psi Concrete: E=3600000psi
It is obvious that in order to compensate the strength difference, we will address the geometric form, i.e. the cross sectional area. So a Steel column can be way more slender than a concrete column, just to bear the load.
tÜ~í=fë=qÜÉ=bÑÑÉÅí=lÑ= `çäìãå=päÉåÇÉêåÉëë\
Imagine the effect of purely axial load applied in this element. What do you think will happen? Even if there is no shear or moment applied, do you believe that it will crush from the axial load? The uniformity and homogeneity of the material should be challenged. Even with prefabricated materials that are made under the strictest of regulations, we can expect some slight abnormalities. Those will render the element asymmetrical and stronger on one direction versus another. The formula that determines a column to be slender or not is the following:
k⋅
lu r
M1
≤ 34 − 12
M2
`çäìãåë=Ó= päÉåÇÉêåÉëë
This is a general method that we can roughly apply in order to consider how the connections can effect the strength of the column. More on this will be addressed during the next lecture.
tÜ~í=áë=íÜÉ=bÑÑÉÅí=lÑ= `çäìãå=päÉåÇÉêåÉëë\
Also we need to consider the following:
ACI 10.12.3. defines Mc as the magnified moment and M2 the larger factored end moment of a no sway compression member:
Mc
δns ⋅ M 2
In case our calculations provide minimal result we can apply the minimum eccentricity formula:
e min
The moment magnifier δns is used to estimate the lateral deflection effect. It involves the code modificator Cm which is Cm M1 also given below:δ ≥ 1.0 C 0.6 + 0.4⋅ ≥ 0.4 ns
0.6 + 0.03⋅ h
Pu 1 − .75⋅ P c
m
M2
eçï=_ÉåÇáåÖ=fë=^ééäáÉÇ=lå= `çäìãåë
Once a slight deflection takes place on an axially loaded element, there is more eccentricity generated, which in turn produces a second generation moment, which will result in further deflection, one more round of moment and deflection and so on and so forth, until equilibrium is reached. Looping this process to analyze the deflection and the applied moment over and over may be extraordinarily tedious and the result will not vary tremendously once two or three cycles are reached. Timoshenko resolves this process by multiplying the primary moment by the following formula, which can give us a result that is precise enough 1 for us: M magn := M u ⋅ Pu 1− Pc
bñ~ãéäÉ Calculate the primary moment due to a lateral 20k load and determine th total moment. bw=1ft, h=1.25ft, k=1.0 and lu=18ft fy := 60ksi f'c := 3ksi lu := 18ft Plat := 20kip Pu := 125kip M uini := 0k' k' Vu := 0kip Estimatiing primary moment: Plat lu M u := M u = 90 k' 4
b w := 1ft h := 1.25ft k := 1
Calculating Young's modulus of concrete: f'c Ec := 57000⋅ psi Ec = 3122.02 ksi psi Using Euler's buckling load formula for secondary moment : I :=
b w⋅ h
3
I = 3375 in
12
4
2
Pc :=
π ⋅ Ec ⋅ I
Pc = 2228.956 kip
( k⋅lu )2
M magn := M u ⋅
Pu
Pc
1
1−
M magn = 95.347 k'
tÜ~í=^Äçìí=açìÄäÉ=`ìêî~íìêÉ= ^åÇ=líÜÉê=pÅÉå~êáçë\
What happens in the case of double curvature with equal but reverse moments, or in the case where we have no moment on one end? In the first scenario we have moment and deflection equal to zero and in the second, we have a deflection that is about half of what the amplification factor provides, and a very large moment. Therefore, the code addresses the issue by the use of the modification factor Cm which can vary between 0.4 and 1.0 that is to be used for braced frames without transverse loads. For other cases the value to be taken is 1.0
Cm
0.6 + 0.4⋅
M1 M2
≥ 0.4
`lirjkp=Ó= pqo^fk=afpqof_rqflk=J= bñ~ãéäÉ
Problem Statement:
Determine whether or not a 16*20 in section with 12#10 bars is adequate for Pu=1080k @ minimum eccentricity (Code clause 10.12.2) about the minor axis of the column. The Column height Lu=20.7ft As_10 := 1.27
f'c := 5000 fy := 60000 Lu := 248.4in
b w := 16 h := 20
Using inches and lb for consistency
Processing Data: DL := 680000lbf
DL_factored := DL ⋅1.2
DL_factored = 816 kip
LL := 165000lbf
LL_factored := LL ⋅1.6
LL_factored = 264 kip
Pu := LL ⋅1.6 + DL ⋅1.2
Pu = 1080 kip
Assuming that M1=minM2 to cause compression on the same face such that M1/M2=1and k factor is for elastic connection on a multi-story building: k := 0.8
PD := DL PL := LL
bñ~ãéäÉ=ÅçåíK kLu := k ⋅Lu Determining the M2min:
kLu = 198.72 in
M2min := Pu ⋅[ ( 0.6 + 0.03 ⋅h) ⋅( 1 ⋅in) ]
M2min = 1296 kip ⋅in
OR
M2min = 108 k'
Note: As stated in the handed out code (ACI318 10.12.3.2), the units of 0.6 and (h) (or c1) are taken in inches. Also note that in this case we treat as h (or c1) the short side of the column because we solve for the max moment on the weakest side. The result can be written in any format the user prefers. As k' are defined above, the k' option is provided.
M2 := M2min &
M1 := M2min
Solution: Determining the Cm (factor relating the actual moment diagram of a slender column to an equivalent uniform moment diagram:
M1 M2
Cm := 0.6 + 0.4
Cm = 1
Determining the modulus of elasticity of Concrete: ksi Ec := 57000 f'c Ec = 4030.509 ksi 1000
bñ~ãéäÉ=ÅçåíK Calculating the (βd) ratio of maximum factored axial Dead Load to the total axial load: PD βd := 1.2 ⋅ βd = 0.756 Pu Determining the (Ig) gross moment of Inertia of the element along the minor axis (see problem statement):
bw ⋅h3 4 Ig := ⋅ in 12
( )
4
Ig = 10666.667 in
Solving for the moment of Inertia of the steel rebars using the Ad^2 formula. There shall be 2.5" cover, allowing 11" along the short axis, so the distance of the outer bars shall be 5.5" and the distance of the inner bars shall be 11"/(3*2), or 1.833": d1 := 5.5 d2 := 1.8333
(
n1 := 8
) ( 4)
2
2
Is := As_10 ⋅ n1 ⋅d1 + n2 ⋅d2 ⋅ in
n2 := 4
Es := 29000ksi 4
Is = 324.414 in
Calculating the EI (stiffness) - See ACI code 10.12.2 & 10.12.3: EI :=
( 0.2 ⋅Ec ⋅Ig + Es ⋅Is)
C l l ti
7
( 1 + βd)
th (P )
iti
2
EI = 1.026 × 10 kip ⋅in ll
d
bñ~ãéäÉ=ÅçåíK Calculating the (Pc) critical load: 2
π ⋅EI
Pc :=
( kLu)
Pc = 2563.477 kip
2
Calculating the δns (moment magnification factor - applied to frame columns that are braced against sidesway, reflecting effects of member curvature between ends): Cm
δns := 1−
δns = 2.2817
Pu 0.75 ⋅Pc
Considering the ultimate moment (Mu) to be equivalent to the critical Moment (Mc) now we solve for the Mc: Mc := δns ⋅M2
Mc = 246.4274 k' Pu = 1080 kip
`lirjkp=Ó= pqo^fk=afpqof_rqflk
We see the interaction diagram can determine the capacity of stress that can be applied to a column. Let’s use the Excel sheet provided for in class exercise.
Column design Moment vs Axial load 1400 1200 1000 800 ΦPn (K)
600 400 200 0 -200
0
50
100
150
200
-400 -600 ΦMn (K')
250
300
350
pmfo^i=obfkclo`bjbkq `lk`obqb=`lirjk=abpfdk Problem Statement: Select a cross section for a spirally reinforced column section to support the loads indicated below, using f'c of 5 ksi and grade 60 steel reinforcement. Try to use ρg of 4%.
Processing Data: Factorizing Dead and Live load: Dead load is multiplied by a factor of 1.2 and Live load by a factor of 1.6: PD := 620 kip
PD_factored := PD⋅ 1.2 PD_factored = 744 kip
PL := 328 kip
PL_factored := PL⋅ 1.6
MD := 0 k'
MD_factored := MD⋅ 1.2 MD_factored = 0 k'
ML := 80k'
ML_factored := ML⋅ 1.6 ML_factored = 128 k'
PL_factored = 524.8 kip Pu := PD_factored + PL_factored Pu = 1268.8 kip Mu := MD_factored + ML_factored Mu = 128 k'
f'c := 5ksi fy := 60ksi
Φ := .75
Solution:
ACI 318 - 10.3.6: For spiral ϕ=.75, for tied ϕ=.65 Also, the first factor changes from .85 to .80 for tied.
The length value that is the result of the division of the applied moment by the applied axial load is the eccentricity "ecc" of the column. Almost always, a compression member may assume a moment either because the axial load is not perfectly centered on the column, or because the column will resist portion of the unbalanced moments at the ends of the beams it supports. In order not to confuse the term "e" with the base of the natural logarithm, we can use the term ecc. ecc :=
Mu Pu
ecc = 1.211 in
Eccentricity should not be higher than 10%, so we can set a minimum diameter of one foot
For reinforcement of 3% the letter "ρ" that signifies density is used ρg =
Ast Ag
ρg_max ≤ .08
ρg_min ≥ .01
ρg := .04
The reason we use the ρg is to determine the area of the column. The ACI code gives the following formula for non prestressed members w/spiral reinforcements (ACI 318 10.3.6.1) maxΦPn := Pu Inversing data:
Ag :=
maxΦPn = 0.85⋅ Φ [ 0.85⋅ f'c⋅ ( Ag) + ρg⋅ Ag⋅ ( fy − 0.85⋅ f'c) ]
maxΦPn Φ⋅ 0.85⋅ [ 0.85⋅ f'c + ρg⋅ ( fy − 0.85⋅ f'c) ]
2
Ag = 307.141 in
Since the column shall be circular, we can use the formula of the circle's area to estimate an approximate radius and round it. r :=
Ag π
r = 9.888 in r := 10in
Rounding..
2
Ag := π r
Therefore:
2
Ag = 314.159 in
Using the following formulas now we can determine the steel reinforcement maxΦPn = 0.85⋅ Φ [ 0.85f'c⋅ ( Ag − As) + fy⋅ As] OR
maxΦPn = 0.85⋅ Φ [ 0.85⋅ f'c⋅ ( Ag) + ρg⋅ Ag⋅ ( fy − 0.85⋅ f'c) ] However, none of the above formulas can be inverted for us to use in terms of As which is what we are trying to solve for. Therefore, we can use a system called "solve block" with an initial guess and allow a series of iterations to take place until a solution is found. maxΦPn := Pu As := ρg⋅ Ag
Guess values:
Given
2 As = 12.566 in This seems like a good starting point.
maxΦPn = 0.85Φ⋅ [ 0.85f'c⋅ ( Ag − As) + fy⋅ As]
(ACI 10.3.5.1)
Ast_value := Find ( As) As := Ast_value
2
As = 11.751 in
For rebars smaller than #9 a formula can be used to define the As: Try 12 #9 rebars. The formula will yield slightly imprecise result As should be 12 sq. inches BarSize1 := 11 n1 := 8
BarSize2 := 14 n2 := 0
2 BarSize1 2 BarSize2 2 As := ⋅ n1 + ⋅ n2 π⋅ in 16 16
Verifying.....
maxΦPn := 0.85Φ⋅ [ 0.85f'c⋅ ( Ag − As) + fy⋅ As] ρg :=
As Ag
1.56⋅ 8
= 12.48
2
As = 11.879 in
2
As := 12.48 in
maxΦPn = 1294.722 kip
ρg = 0.0397
This is a very good result. We are only very slightly above our ρg ratio is as we initially aimed for.
o`=qfba=`lirjk=aÉëáÖå Design a tied column cross section to support an axial load of 350 kips, a moment load of 110kip feet, and a shear load of 14kips. All of the above are factorized. The column is in a braced frame with an unsupported length of 10 ft 6 inches. fy := 60ksi f'c := 4ksi l u := 10.5ft Clearcover := 1.5in Pu := 300kip M u := 110k' Vu := 14kip
ϕ := .65
Estimatiing an initial ratio of steel for tied columns ρt: Based on the most efficient ratio that would be between 1% and 2%, we select an initial ratio of 1.5% ρt := .015 Estimating the initial dimensios of the column: The cross sectional area for a tied column is given by the following formula: 5 ⋅ Pu Ag_ini := 2 Ag_ini = 135.8 in 2.21⋅ f'c + 2.6⋅ fy ⋅ ρt − 2.21⋅ f'c⋅ ρt
ACI 318: 10.3.6.2
Given the option that we may design a square base column.....we estimate an initial base value: b ini :=
Ag_ini
b ini = 11.65 in
Given the fact that there are significant moments applied on this column, it would be wise to override the initial calculation that takes only direct loads into account. Let's round it up about 10 -15% on each side:
b ini ⋅ 1.25 in b trial = 14 in b := b trial in
b trial := trunc
2 h := b
Ag := b ⋅ h Ag = 196 in Determining the bar arrangement:
To determine the preferable bar arrangement we compute the ratio of eccentricity to the height "h" of the column: Note: This is "h" in cross section, not the actual column height.
ecc :=
Mu
ecc = 4.4 in Pu According to the figure indicated it will ecc be more appropriate to apply re-bars on = 0.31 h both sides of this column:
Column slenderness can be neglected if: lu M1 k⋅ ≤ 34 − 12⋅ r M2 According to ACI code 10.11.2, the radius of gyration of rectangular columns is 0.3h and .25D for circular columns. Since this is a braced frame k is lesser or equal to1.0 and the ratio of M1 to M2 can vary between +/- 0.5. We can assume that k=1.0 and M1/M2=0.5. Therefore the above relations yield the following results: M1 lu M1 r := .3⋅ h k := 1 = 0.5 k⋅ = 30 34 − 12 = 28 M2 r M2
lu M1 Slendernesscond := if k ⋅ ≤ 34 − 12⋅ , "Neglect column slenderness" , "Design slender column" r M2
Slendernesscond = "Design slender column" I :=
b⋅ h
3
4
I = 3201.33 in
12 f'c
Ec := 57000
psi
⋅ psi
Ec = 3605 ksi
2
Pc :=
π ⋅ Ec ⋅ I
( k⋅ lu )
Pc = 7174.55 kip
2
M magn := M u ⋅
1 Pu
1 − Pc
M magn = 114.8 k'
Computing the "γ" ratio : At this point we need to compute the value gamma (γ) which is the ratio of distance of centroids of outer rows of bars and column dimension perpendicular to the bending axis: We shall assume that the ties are #3 rebars and the longitudinal bars are #7:
d lbar := .875in d tbar := .375in
d lbar h − 2 Clearcover + dtbar + 2 γ := h
γ = 0.6696
We need to point out that the assumption we make about the #7 rebars may prove imprecise, in which case we shall need to reiterate this process. Given the gamma value above we will refer to the ACI interaction diagrams (or use our own system!!!) to define again the ratio of steel ρt.
Pu ϕ⋅ f'c⋅ Ag
M magn ϕ⋅ f'c⋅ Ag ⋅ h
= 0.589
= 0.193
For the above values the interaction diagram for gamma 0.6 gives a rho value of 0.03. The diagram for a gamma value of 0.7 gives a rho value of 0.025. We shall apply linear interpolation to compute the value of rho at gamma found. ACI code defines that rho should lie between 0.01 and 0.08.
ρt is the ratio of total reonforcement area divided by the cross sectional area of a column ( γ − .6) ρt := 0.03 − ( 0.03 − 0.025 ) ⋅ ρt = 0.0265 ( .7 − .6)
( ) Rhohi_condition := if ( ρt > .08 , "High value" , "OK" ) Rholo_condition := if ρt < .01 , "Low value" , "OK"
Selecting reinforcement: As := ρt⋅ Ag
Rholo_condition = "OK" Rhohi_condition = "OK" 2
As = 5.1975 in We can select twelve 6 bars, six on each face but let's verify that spacing will be approved:.
As 2
.44in
= 11.8
n lbar := 6 s :=
2
d lbar := .75in
Albar := 0.44in
(b − 2⋅ Clearcover − 2dtbar − nlbar⋅ dlbar)
s = 1.15 in
n lbar − 1
(
(
)
Spacing := if s < max d lbar , 1in , "Redesign" , "OK"
)
Spacing = "OK"
Design the lap splices:
l d6 :=
1.3d lbar 25
fy psi
f'c
l d6 = 36.99865 in l d6 = 3.08322 ft
psi
So let's reevaluate our rho value: As_fin := 2n lbar⋅ Albar ρ :=
As_fin
2
As_fin = 5.28 in
Bar Designation Number 3 4 5 6 7 8 9 10 11 14 18
Weight per foot (lbf) 0.376 0.668 1.043 1.502 2.044 2.670 3.400 4.318 5.304 7.650 13.600
Diameter db 0.375 0.500 0.625 0.750 0.875 1.000 1.128 1.270 1.410 1.693 2.257
Area As 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.56 2.25 4.00
Perimeter 1.178 1.571 1.963 2.356 2.749 3.142 3.544 3.990 4.430 5.319 7.091
ρ = 0.02694
Ag
Selecting the ties: Based on ACI 7.10.5.2, the least of the following three conditions determines the spacing of the ties: colleast_dim := if ( b < h , b , h ) colleast_dim = 14 in 48⋅ d tbar = 18 in 16⋅ d lbar = 12 in
(
)
TieSpacing := min colleast_dim , 48⋅ d tbar , 16⋅ d lbar
TieSpacing = 12 in
We need to make reference to the subject we addressed on principal stresses to visualize the effect of the following formula. The factor "Nu" represents an axial tension force resulting from the compression. The angle theta "θ" to be used is the critical 45 degrees. θ := 45deg
ϕ := .85
d lbar d := b − 2 ⋅ Clearcover + d tbar + 2
d = 9.5 in
Pu Nu := tan( θ)
Nu = 300 kip
f'c ΦVc := ϕ⋅ 2 ⋅ ⋅ psi⋅ b ⋅ d ⋅ 1 + psi
Nu lbf ΦV = 25.2 kip c Ag 2000 2 in
Check ACI sections 7.10.5, 11.4.5.1 and 11.4.6.3. If Vc
oÉ~ÇáåÖ
Reading:
Required: Furlong, Chapt 7.3 through 7.6 incl. Recommended: McCormac & Nelson, Chapter 11 (pp. 317-333) for this week’s lectures.
