“Primer” is my personal reference notebook. Building codes and design guides are often dense and voluminous. These auto-didactic notes help me collect, organize, condense technical topics I’m learning as an engineer. I like to think of ‘primer’ as an extension to my brain whose storage capacity is sadly all too limited.
Please understand I made these for myself. Reader discretion is advised. No warranty is expressed or implied by me on the validity of the information presented herein.
Base plates are usually the interface between two different materials (namely steel and concrete). As a result, although technically straight-forward, a comprehensive design considering all potential failure mode can be quite lengthy and tedious.
This goal of this article is to present the theoretical background behind base plate design, from load distribution to the various limit state. By the end, you’ll hopefully have enough know-how to build your own base plate calculation tool, or gain enough technical knowledge such that you can have more confidence using existing tools.
This article is a combination of fundamentals you’ll find in the references below, as well as commentaries from my own experience.
Base plate footprint (width-B, depth-N) should be sized satisfy:
The recommended hole dimensions are shown in the table below. Notice that base plate holes are even larger than OVS holes!
Figure 1: Recommended Anchor Rod Hole Dimension from DG1
Figure 2: Recommended Anchor Rod Hole Dimension Comparison
Base plate holes are huge! This is because constructing them is a pain. Anchor rods are typically cast first into the foundation element, then base plates are then lowered onto the rods. Rods are often slightly off, or bent/damaged during pour. Hence the extra huge holes.
We need to take into account the worst case of a misaligned anchor rod + washer dimension. See illustration below.
Figure 3: Misaligned Rod + Washer Dimension
The table below provides recommended extension lengths based on the worst case of 1.) minimum edge distance and 2.) providing enough clearance for 5/16” weld around washer. Let:
Figure 4: Minimum Edge Distances and Recommended Base Plate Size
In summary, here are my recommendations for preliminary base plate dimensions given a chosen anchor rod diameter.
Recommended Width - B
Knowing the column width (\(b_f\)) and anchor rod diameter of your choosing: (see suggested alpha value in Figure 4)
\[B_{req} = b_f + 2 \alpha\]Recommended Depth - N
Knowing the column depth (\(d\)) and anchor rod diameter of your choosing: (see suggested alpha value in Figure 4)
\[N_{req} = d + 2 \alpha\]Recommended Thickness - t
Required thickness can be approximated using a simple cantilever beam model with uniform load + some fudge factors for conservatism.
For columns with minimal moment demand:
\[\sigma = P_u/A\] \[\phi F_y \frac{t^2}{4} \geq \sigma \frac{(1.1 l)^2}{2}\] \[t_{req} \geq \sqrt{\frac{2.4 \sigma (l)^2}{\phi F_y}}\]For columns with significant moment demand, replace \(\sigma\) with the expression below (bearing capacity equation from ACI 318)
\[\sigma = 0.65(2)(0.85)f'_{c}\]Recommended Anchor Rod
There are three anchor rod grades commonly used.
There are plenty of options for anchor rod diameter but try to stick to diameters listed in Figure 2. Pick diameter in 1/4” increments. Typically I have three buckets:
Do not specify different grade anchors with the same diameter. It gets confusing on-site.
Picking the appropriate anchor rod is somewhat more iterative, but a good starting point would be to do the following:
For gravity columns and columns with very small moment and no net uplift, it is probably okay to go with a smaller anchor rods (say 1” GR 36). These base plates are categorized as “small moment” and the anchor rods don’t experience much, if any, tension demand. They might need to resist some shear if no shear lug is provided. But since the moment is small, shear is likely small too.
For columns classified as “large moment” or having net uplift, we can use the following expression to get somewhere in the ball park. Select anchor grade (see Fu above), number of anchor rods (\(n_{rod}\)), and base plate depth (N). Ignore interior anchors contribution.
Rather than specifying an unique designs for every single base plate in the building, the better approach is to come up with reasonable number of base plate design “groups” from the outset. This will save you a lot of trouble later on when you put your design on the drawings.
No one wants to build a building where every base plate is different. We need to shrink our number of design to a manageable amount anyways. Therefore, create a base plate design schedule first. Something like below from the outset:
Figure 5: Create A Base Plate Schedule From The Outset
Base reactions can be extracted from any analysis software. We are interested in the axial demand, shear demand, and moment demand from the most critical load combination.
\[{P_u, V_u, M_u}\]Column base connections are technically designed to remain elastic. So we need to use the appropriate overstrength load combinations:
\[1.2D + 0.5L \pm 1.0E_{mh}\]The equation above is just for reference purposes. Ideally, your design combination is an envelope of many other load combinations that takes into account:
For Moment Demand, AISC 341-16 Section D.2.6c specifies the required flexure demand as the minimum of expected plastic moment capacity of the column or moment demand from analysis model using overstrength load combination (the second one is usually smaller and thus governs):
\[M_u = min(1.1 R_y M_p, E_{mh})\]If there are shear lugs, an additional moment is generated by the eccentric bearing. Refer to the shear lug section for more information.
\[M_u = M + M_{lug}\]For Shear Demand, AISC 341-16 Section D.2.6b specifies the required shear demand as the minimum of expected shear due to double curvature of the column or shear demand from analysis model using overstrength load combination.
\[V_u = min(2 M_p / H, E_{mh})\]Furthermore, there is a minimum design shear that is often missed. This value may be refined using a story compatibility analysis assuming drift of 0.025H
\[V_{u,min} = 0.7 M_p / H\]Furthermore, we need to consider two “load scenarios”. A high compression case, and a uplift (or low compression) case.
The need to evaluate uplift is self-explanatory. However, even if the column never experiences uplift, it is important to look at low-compression scenario because base plate classification is a function of axial demand. In other words, under lower compression, the base plate may change from small eccentricity where anchor rods don’t experience tension to large eccentricity case where they do.
Since demands are enveloped, it is difficult to say low or high axial demand necessarily corresponds to high or low shear or moment. The easiest and most conservative approach is to just always extract the absolute maximum value for shear and moment.
ALthough it is always easier to design for pinned condition, make sure your jurisdiction allows for such assumption.
The reality is probably somewhere in between (i.e. partial fixity). Refer to the appendix for some discussion on how to calculate a rotational spring stiffness.
Base plates can be classified into two categories:
Figure 6: Small Moment Base Plate
Figure 7: Large Moment Base Plate
The dividing line between the two categories is known as the critical eccentricity.
Although we use the term kern above, it should be noted that AISC Design Guide 1 suggests a less conservative approach:
Figure 8: Kern Comparison
For example, in the chart above, we are looking at a 24” x 48” base plate with moment demand of 500 kip.ft.
The logic behind plastic stress distribution is as follows:
Procedure
Calculate eccentricity (remember to convert moment from kip.ft to kip.in)
\[e = \frac{M_u}{P_u}\]Calculate concrete bearing capacity. See section 3 for more info.
\[1 \leq \alpha = \sqrt{A_2 / A_1} \leq 2\] \[f_{pmax} = \phi \alpha 0.85 f'_c\]Normalize bearing stress from ksi to kip/in along depth of base plate
\[q_{max} = f_{pmax}B\]Calculate critical eccentricity (N is base plate depth)
\[e_{crit} = N/2 - \frac{P_u}{2q_{max}}\]Design of small moment base plate is easy! No anchor rod tension to worry about which means we get to skip all tension-related failure modes.
One caveat is that small moment reduces base plate contact area, thus increasing bearing stress.
\[\epsilon = N/2 - Y/2\]\(\epsilon\) is equal to e for small moment case, rearrange and solve for Y:
\[Y = N - 2e\] \[A_{bearing} = B \times Y\]Again, moment demand is self-resolving. No tension demand on anchor rods.
\[T = 0\]AISC Design Guide 1 provides a closed-form solution that assumes one row of anchor at each end. This is a convenient approach because we have two unknowns (Y and T), and they both can be readily solved using the quadratic equation. However, there are two very important limitations!
Vertical force equilibrium. T and Pu is always positive; bearing resultant always negative.
\[\sum F_{y} = 0\] \[0 = T + P_u - q_{max}Y \tag {eq.A}\]Take moment equilibrium about point B (location of tension anchor). Bearing resultant contribution will always be positive. Pu contribution always negative.
\[\sum M = 0\] \[0 = q_{max}Y (f + N/2 - Y/2) - P_u (f+e) \tag {eq.B}\]Combine eq.A and eq.B, and solve via quadratic equation:
\[Y = (f+N/2) \pm \sqrt{ (f + N/2)^2 - \left( \frac{2P_u(f+e)}{q_{max}} \right) }\]For cases where Pu = 0, the equation above must be modified because e is infinite (denominator is zero). Substitute \(P_u e = M_u\).
Finally, substitute Y value into eq.A to get anchor tension.
\[T = q_{max}Y - P_u\]Be careful of sign convention! Design Guide 1 assumes your axial demand will always be compression (point downward) and took care of the signs for you.
Here is the python implementation of the equations above:
def DG1_closed_form(width, depth, fpc, Mu, Pu, f,
alpha = 0.85, alpha1 = 2.0, phi = 0.65):
"""
Closed-form solution outlined in AISC Design Guide 1.
Assumes one row of anchor rod on each end.
Interior anchor rods ignored.
Arguments:
width = width of base plate (in)
depth = depth of base plate (in)
fpc = concrete strength (ksi)
Mu = moment demand (kip.in). Always positive
Pu = axial demand (kip). Positive is compression
f = distance from center of base plate to exterior anchor
alpha (optional) = 0.85 (rectangular stress block parameter)
alpha1 (optional) = 2.0 (concrete bearing confinement factor)
phi (optional) = 0.65 (concrete resistance factor)
Return:
T = tension force in exterior anchor row
Y = neutral axis depth
"""
# concrete capacity
fpmax = phi * alpha * alpha1 * fpc
qmax = fpmax * width
# check eccentricity
if Pu < 0:
raise RuntimeError("Cannot handle negative Pu case (tension)")
elif Pu == 0:
e = None
isSmall = False
else:
e = Mu/Pu
ecrit = depth/2 - Pu/2/qmax
isSmall = (e < ecrit)
# small eccentricity
if isSmall:
Y = depth - 2*e
T = 0
return T, Y
# large eccentricity
else:
if e == None:
Y = (f + depth/2) - ( (f + depth/2)**2 - (2*Mu)/qmax )**(1/2)
else:
Y = (f + depth/2) - ( (f + depth/2)**2 - 2*Pu*(f+e)/qmax )**(1/2)
T = qmax*Y - Pu
return T, Y
For example, given this 24” x 36” base plate:
Figure 9: Base Plate Example
If we were to use the general methodology (see next section) that takes into account contribution from the middle row:
Ignoring the middle row of anchors is likely overly conservative. Furthermore, the fact that the closed-form solution above only works for compression + moment is a severe flaw.
We can generalize the closed-form solution above using the rigid plate assumption. Also known as “plane section remain plane” if you are dealing with section analysis. Either way, we are essentially assuming a linear strain profile.
Rather than conducting a full moment curvature analysis on a fiber section, the rigid plate methodology effectively turns this into a univariate root-finding problem.
Derivation
Figure 10: Free Body Diagram of Base Plate
Vertical force equilibrium:
\[\sum F_y = 0\] \[0 = \sum T_i + C + P_u \tag{eq.A}\]Individual anchors can be related to each other using similar triangles:
\[\frac{t_i}{e_i} = \frac{t_n}{e_n}\] \[t_i = \frac{e_i}{e_n} t_n\]Summation of anchor forces can be rewritten as:
\[\sum T_i = \sum t_i N_i = \sum \left( \frac{e_i}{e_n}t_n \right) N_i\] \[\sum T_i = (\sum e_i N_i)\frac{t_n}{e_n} \tag{eq.B}\]Suppose we know the depth of neutral axis (\(Y\)), then we can solve for \(t_n\) which is the force in anchor furthest from pivot point. Combining eq.A and eq.B and solve for \(t_n\)
\[t_n = \frac{(-C-P_u)(e_n)}{\sum e_i N_i}\]Once we know \(t_n\), other anchors can be calculated using our linear strain assumption:
\[t_i = \frac{e_i}{e_n} t_n\]The big question is how do we find the depth of neutral axis? The answer is to iteratively search for it using any root-finding algorithm. Or Solver in Excel.
For an assumed neutral axis depth (Y):
Compression bearing resultant: (equals to zero if Y is negative)
\[q_{max} = \phi \alpha 0.85 f'_c\] \[C = max(q_{max}Y, 0)\]Anchor distance from neutral axis: (anchors within neutral axis depth are inactive and cannot take tension or compression, set to 0)
\[e_i = max(x_i - Y, 0)\]Anchor forces:
\[t_n = \frac{(-C-P_u)(e_n)}{\sum e_i N_i}\] \[t_i = \frac{e_i}{e_n} t_n\]Total anchor row along each row:
\[T_i = t_i N_i\]Moment contribution of each row:
\[M_i = T_i x_i\]Force equilibrium is satisfied through our use of similar triangles to calculate \(t_n\)
\[\sum F = 0 = \sum N_i t_i + C + P_u\]Moment equilibrium is only true if we have the correct neutral axis depth.
\[\sum M = 0 = \sum M_i + P_u(N/2) + C (Y/2) + M_u\]Keep trying different Y values until \(\sum M = 0\)
Python Implementation
def rigid_plate_distribution(width, depth, fpc, Mu, Pu, x_i, N_i, beta = 0.80,
alpha = 0.85, alpha1 = 2.0, phi = 0.65):
"""
Determine anchor rod uplift based on rigid base plate assumption.
down = +ve, counter-clockwise = +ve
Arguments:
width = width of base plate (in)
depth = depth of base plate (in)
fpc = concrete strength (ksi)
Mu = moment demand (kip.in). Always positive
Pu = axial demand (kip). Positive is compression
xi = depth of anchors from the edge [x1,x2,...]
Ni = number of anchor along one row [N1,N2,...]
beta (optional) = 0.80 (rectangular stress block parameter)
alpha (optional) = 0.85 (rectangular stress block parameter)
alpha1 (optional) = 2.0 (concrete bearing confinement factor)
phi (optional) = 0.65 (concrete resistance factor)
Return:
T_i = list of total anchor force at each row []
t_i = list of anchor force at each row [] = Ti / Ni
Y_final = depth of neutral axis (in)
sum_F = force equilibrium. Should always return 0
sum_M = moment equilibrium. Should always return 0
"""
# check eccentricity
if Pu <= 0:
e = None
isSmall = False
else:
fpmax = phi * alpha * alpha1 * fpc
qmax = fpmax * width
e = Mu/Pu
ecrit = depth/2 - Pu/2/qmax
isSmall = (e < ecrit)
# small eccentricity
if isSmall:
zeros = [0 for x in x_i]
Y_final = depth - 2*e
return zeros, zeros, Y_final, 0, 0
# large eccentricity
else:
# secant method good for root finding
def secant_method(func,x0=depth/2,x1=depth/2-1,TOL=0.1):
while abs(func(x0)[0])>=TOL:
x2 = x1 - func(x1)[0] * (x1 - x0) / (func(x1)[0] - func(x0)[0])
x0 = x1
x1 = x2
print(f"Trial NA: {x0} \t {x1}")
return x2
# set up equation for root-finding
def equilibrium_equations(Y):
# Y could be negative. In which case no bearing.
comp = min(0, -width*Y*phi*beta*alpha*alpha1*fpc)
e_i = [max(0,x - Y) for x in x_i]
omega_i = [a*b for a,b in zip(N_i,e_i)]
if sum(omega_i) == 0:
t_n=0
else:
t_n = (-Pu-comp) * (e_i[-1])/ sum(omega_i)
t_i = [max(0, a/e_i[-1]*t_n) for a in e_i]
T_i = [a * b for a,b in zip(t_i, N_i)]
M_i = [a * b for a,b in zip(T_i, x_i)]
sum_M = sum(M_i) + Pu*(depth/2) + comp*(Y/2) - Mu
return sum_M, t_i, T_i, comp
# start root-finding
try:
Y_final = secant_method(equilibrium_equations)
except:
raise RuntimeError("Did not converge")
# final load distribution
sum_M, t_i, T_i, comp = equilibrium_equations(Y_final)
sum_F = sum(T_i) + Pu + comp
return T_i, t_i, Y_final, sum_F, sum_M
User notes:
Figure 11: Example Base Plate with 200 Kips Tension
Figure 12: Example Base Plate with 200 Kips Compression
With all that derivation behind us, let’s clear up notations that you’ll see repeatedly:
Figure 13: Concrete Bearing Confinement Factor
For large-moment base plates, Bearing DCR will always be 100% because contact with concrete is minimal and purely for equilibrium of a base plate on the verge of overturning. The contact area is equal to the neutral axis depth and bearing stress is equal to bearing capacity of concrete per our assumptions.
On the other hand, for small-moment base plates:
Determine confinement factor. \(\alpha\) is the confinement factor; it is equal to 2 if concrete confined on all sides. \(c_{amin}\) is the smallest distance to edge of pedestal or footing
\[A_1 = BY\] \[A_2 = (B+2 c_{amin}) \times (Y + 2c_{amin})\] \[\alpha = \sqrt{A_2 / A_1}\] \[1 \leq \alpha \leq 2\]Calculate allowable bearing stress of concrete (resistance factor phi = 0.65):
\[f_{pmax} = \phi \alpha 0.85 f'_c\]Calculate bearing contact area. Recall contact length is reduced by presence of moment:
\[A_{bearing} = BY\]Bearing demand:
\[f_p = P_u / A_{bearing}\]Bearing DCR:
\[DCR = f_p/f_{pmax}\]Some notes:
Figure 14: Base Plate Bending
Check to see if base plate is thick enough to withstand the design bearing stress. If not, our rigid plate assumption is invalid. The procedure is well-documented in AISC design guide but not codified. In essence, we are checking four yield mechanisms:
The 0.8 and 0.95 coefficients you see in figure 14 are empirical. You may wish to adjust it as you see fit (if you’ve added stiffeners for instance). HSS columns would have 0.95bf instead of 0.8bf
Calculate bending lever arm in major and minor direction:
\[m = \frac{N-0.95d}{2}\] \[n = \frac{B-0.8b_f}{2}\]Calculate “lever arm” for bending between flange
\[X = \left( \frac{4db_f}{(d+b_f)^2} \right) (\frac{f_p}{f_{pmax}}) \leq 1\] \[\lambda = \frac{2\sqrt{X}}{1 + \sqrt{1-X} } \leq 1\] \[\lambda n' = \frac{\lambda \sqrt{db_f}}{4}\]Calculate moment demand due to bearing stress (fp) which we’ve determined from the previous section (unit is kip.in/in)
\[M_m = f_p (m^2/2)\] \[M_n = f_p (n^2/2)\] \[M_{n'} = f_p ((\lambda n')^2/2)\]Calculate plate bending DCRs (phi = 0.9. Using plastic section modulus):
\[DCR_m = \frac{ M_{m}B }{\phi F_y B t^2 /4}\] \[DCR_n = \frac{ M_{n}N }{\phi F_y N t^2 /4}\] \[DCR_{n'} = \frac{ M_{n'}N }{\phi F_y N t^2 /4}\] \[DCR_t = \frac{ M_{t}B }{\phi F_y B t^2 /4}\]Alternatively, the moment DCRs above can be expressed as a required thickness (as is the case in AISC design guide 1 which expresses required thickness as a function of bearing stress). A superior method is to express our thickness equation in terms of moment to allow for the tension case to be more easily incorporated.
\[M_{max} = max(M_m, M_n, M_{n'}, M_t)\] \[t_{req} = \sqrt{ \frac{M_{max}}{\phi F_y/4} }\]Some other design notes:
Anchor rod tension capacity can be calculated in two ways. Both will yield similar results. The AISC method is easier to apply as you do not need to calculate reduced diameter at threaded region.
Three anchor rod grades commonly used.
Check tension rupture capacity per AISC (phi = 0.75)
\[A_b = \pi d_b^2 / 4\] \[F_{nt} = 0.75 F_u\] \[\phi r_n = \phi F_{nt} A_b\]Check tension rupture capacity per ACI (phi = 0.75). See below for how to calculate \(A_{se,N}\) exactly
\[A_{se,N} \approx 0.8A_b\] \[f_{uta} = F_u\] \[\phi N_{sa} = \phi A_{se,N} f_{uta}\]Use the minimum of the two:
\[\phi T_n = min(\phi r_n, \phi N_{sa})\]Calculate anchor tension DCR
\[DCR = t / \phi T_n\]\(A_{se,N}\) = net cross section area in threaded region of anchor. It is usually in the range of 0.7 to 0.9Ag. It can be calculated per ASME B1.1 as follows.
\[A_{se,N}= \frac{\pi}{4} (d - \frac{0.9743}{n_t})^2\]nt is the number of thread per inch. Thread geometry for unified coarse (UNC) can be found online. Example
Figure 13: Effective Area of Threaded Rod
Pullout is calculated as the force at the onset of local concrete crushing at the bearing end of the anchor head. This is thought to be the beginning of a pullout failure because of the rapid decrease in stiffness afterwards. In other words, pullout capacity is purely a function of end bearing area, and is not related to embedment depth (i.e. friction neglected).
17.6.3.2.2 - For cast-in headed studs or bolts:
\[N_{pn} = \Psi_{c,p} \times 8A_{brg}f'_c\]17.6.3.2.2 - For hooked ends anchors
\[N_{pn} = \Psi_{c,p} \times 0.9f'_c e_h d_a\]17.10.5.4 - For seismic application, reduce pullout capacity by another 25% (phi = 0.7 for pullout)
\[DCR = \frac{t}{0.75 \phi N_{pn}}\]There are two ways to get bearing area:
Figure 14: Bearing Area of Standard Nuts
This is arguably one of the more complex checks (along with shear breakout), which is why many software offload this check to HILTI PROFIS. We will only look at the most common breakout loading configurations.
Concrete breakout capacity will at most be around 300 kips. If you have much higher tension demand, it would make sense to skip all this complexity and go straight to anchor reinforcement, refer to the end of this section for some recommendations on that.
Figure 15: Concrete Tension Breakout Failure Cone
17.6.2.1b - For an anchor group:
\[N_{cbg} = N_b \times (\frac{A_{Nc}}{A_{Nco}}) (\Psi_{ec,N} \Psi_{ed,N} \Psi_{c,N} \Psi_{cp,N})\]That is a lot of variables. Let’s go through them one by one.
17.6.2.2 - Basic single anchor breakout capacity
This is the basic concrete breakout strength, of a single anchor, in cracked concrete. We will use this value as a starting point and apply several modification factors for other conditions.
\[N_{b} = k_c \lambda_a \sqrt{f'_c} h_{ef}^{1.5}\]where:
17.6.2.3 - Breakout Eccentricity Factor (\(\Psi_{ec,N}\))
Figure 16: Anchor Tension Eccentricity Factors
If the resultant tension on an anchor group is concentric:
\[\Psi_{ec,N} = 1.0\]On the other hand, when a moment is applied to an anchor group, the resultant tension is not concentric meaning some anchors are more stressed in tension than others.
\[\Psi_{ec,N} = \frac{1}{1+\frac{e'_N}{1.5h_{ef}}} <=1.0\] \[e'_N = N/2 - \frac{ \sum T_i x_i }{ \sum T_i }\]17.6.2.4 - Breakout Edge Effect Factor (\(\Psi_{ed,N}\))
If the minimum edge distance is at least \(1.5h_{ef}\), then a full breakout cone can form and no edge reduction factor is necessary.
\[\Psi_{ed,N} = 1.0\]Otherwise, if the anchor or anchor group is close to an edge:
\[\Psi_{ed,N} = 0.7 + 0.3 \frac{ c_{a,min} }{ 1.5h_{ef} }\]\(c_{a,min}\) is the minimum edge distance.
17.6.2.1.2 - If there are 3 or more edges (less than \(1.5h_{ef}\)), such as at the end of a narrow grade beam, the effective embedment depth must also be reduced in all equations that involve embedment depth.
\[h_{ef} = max( c_{a,max}/1.5, s/3)\]where “s” is the maximum spacing between anchors, and \(c_{a,max}\) is the maximum edge distance that is less than \(1.5h_{ef}\)
17.6.2.5 - Breakout Cracking Factor (\(\Psi_{c,N}\))
Modification factor for cracking. Typically assume all concrete are cracked for conservatism unless more rigor is required. Cracked concrete has a modification factor of 1.0.
17.6.2.6 - Breakout Splitting Factor (\(\Psi_{cp,N}\))
For cast-in anchors, or any anchors designed for cracked concrete, the splitting factor can be taken as 1.0.
17.6.2.1.4 - Full Projected Failure Area of Single Anchor (\(A_{Nco}\))
Based on a 35 degree breakout cone, the projected failure area is a square with side length of \(2 \times 1.5h_{ef}\). Note that the failure area is taken as a rectangle rather than an ellipse for simplicity.
\[A_{Nco} = 9h_{ef}^2\]17.6.2.1.1 - Actual Projected Concrete Failure Area (\(A_{Nc}\))
Figure 17: Concrete Breakout Area Example
For anchor group with 1 or 2 edges, the actual projected breakout cone area can be represented in equation form as:
\[X = s_1 + min(c_{a1},1.5h_{ef}) + min(c_{a1},1.5h_{ef})\] \[Y = s_2 + min(c_{a2},1.5h_{ef}) + min(c_{a1},1.5h_{ef})\] \[A_{Nc} = XY \leq nA_{Nco}\]However, the projected area must not exceed \((n A_{nco})\) where n is the number of anchor in the anchor group. \(c_{a1}\) is taken as the minimum edge distance.
For anchor group with 3 or more edges, effective embedment depth has to be reduced per 17.6.2.1.2. In essence, the breakout cone will extend \(1.5h_{ef}\) orthogonally beyond the outermost anchor unless interrupted by an edge.
Putting Everything Together! (finally)
\[N_{cbg} = N_b \times (\frac{A_{Nc}}{A_{Nco}}) (\Psi_{ec,N} \Psi_{ed,N} \Psi_{c,N} \Psi_{cp,N})\]17.10.5.4 - For seismic application, reduce tension breakout capacity by 25%.
Note that the demand this time is the total tension force in the entire base plate assembly, not just a single anchor or a single row! Resistance factor (phi) for breakout is 0.7.
\[DCR = \frac{\sum T}{0.75 \phi N_{cbg} }\]You’ll at most squeeze out about 300 kips capacity here. This is often not enough…
If you have incredibly high tension demand, you have two options. Don’t bother relying on concrete alone, it is not going to work.
Figure 18: Anchor Reinforcement in Blue, Supplementary Reinforcement in Green
17.5.2.1 - Occasionally, it may be impossible to rely on concrete breakout capacity alone. In these cases, we may take breakout capacity as equal to the strength of all anchor reinforcements that can be developed on both ends of the splitting plane. Note we cannot rely on both concrete and steel together (i.e. do not add concrete and anchor steel capacity together)
\[N_{cbg} = 0\] \[\phi N_{n} = \phi f_y A_s\]There are stringent detailing requirements in order to take advantage of anchor reinforcements, here are some of the important ones:
Figure 19: Effective Region for Anchor Reinforcement
DCR can be calculated as:
\[DCR = \frac{\sum T}{\phi N_{n}}\]If that still doesn’t work…
There is a hacky way to go about breakout checks that many practitioners use to circumvent Chapter 17 checks entirely.
This puts us outside the jurisdiction of ACI 318-19 chapter 17 into chapter 22. In fact, ACI 318-19 17.1.2 explicitly states:
“multiple anchors connected to a single steel plate at the embedded end of the anchors […] are not included in the provisions of this chapter [chapter 17]”
There are many advantages to checking punching shear instead of breakout:
You might be asking at this point. What’s the difference? If we extend the anchors deep enough, doesn’t breakout BECOME punching shear? That’s a good observation. Refer to these two excellent paper for more information.
In summary:
Punching shear assumes a 45 degree failure plane; breakout assumes 35 degree. Ultimately this is all just semantics and we are looking at the same phenomenon! (just adjusted to different level of reliability).
This check is commonly missed as it is tucked away in the recesses of ACI 318 chapter 17. This failure mode is only applicable for anchors near an edge (i.e. \(h_{ef} > 2.5 c_{a1}\) )
17.6.4.1 - For a single headed anchor:
\[N_{sb} = 160 c_{a1} \sqrt{A_{brg}} \lambda_a \sqrt{f'_c}\]17.6.4.1.1 - if there is a second edge, \(c_{a2} < 3c_{a1}\), then multiply the above equation by the factor:
\[0.25(1+ \frac{c_{a2}}{c_{a1}}) \tag {24}\]17.6.4.2 - For head anchor group with anchor spacing along the edge \(s< 6c_{a1}\):
\[N_{sbg} = (1+ \frac{s}{6c_{a1}}) N_{sb}\]17.10.5.4 - For seismic application, reduce side face blowout capacity by 25%
\[0.75 \phi N_{sb}\]As for demand, only consider tension in the anchors close to the edge (\(c_{a1}<0.4h_{ef}\)).
\[DCR = \frac{\sum T_{edge}}{0.75 \phi N_{sb}}\]The treatment of shear demand is a very interesting and nuanced topic. There are many detailing complications to consider! In a perfect world, shear check would be a straight forward 0.6FyAs. Unfortunately, the world is not perfect… To quote AISC design guide 1:
“It cannot be emphasized enough that the use of shear in the anchor rods requires attention in the design process to the construction issues associated with column bases.”
Option 1: Let anchor rod resist both tension and shear
Most important things to keep in mind:
Figure 20: Through Thickness Bending For Base Plate Anchors
All things considered, I believe these are the viable options (listed from highest capacity to lowest):
HCAI and DSA may override ACI 318-19 17.7.1.2.1 and force you to check shear WITH leverarm (leaving you only option 1b or 1c)
Option 2: Provide shear lug or other bearing mechanism to decouple tension and shear
Figure 21: Resisting Shear Through Bearing
Shear lugs are becoming more popular especially with the addition of shear lug provisions in ACI 318-19. In the past, bearing mechanisms were provided in many different ways with varying design assumptions. The new provisions in ACI should bring some clarity and consistency in design.
Some of the common bearing methods include:
Two codified methods for shear lug design are currently available. AISC design guide 1 recommends provisions within ACI 349-06 (for concrete design of nuclear structures) but the methodology is quite dated and somewhat unconservative. Instead, it is best to follow the latest provisions in ACI 318-19. (AISC design guide 1 will likely be updated soon to incorporate these ACI equations)
Figure 22: Shear Lugs
Shear lugs are rectangular plates, or steel shape composed of plate-like elements welded to a base plate. They resist shear via a bearing mechanism.
17.11.1.1.2 - Minimum of four anchor rods shall be provided when using a shear lug
17.11.1.1.3 - The use of shear lugs enable separation of shear and tension design, provided that the anchors are not welded to the base plate through washers. In AISC design guide 1, section 2.6 specifically states: “washers should not be welded to base plate, unless they are designed to resist shear”. If anchors are rigidly connected, displacement compatibility enforces a certain amount of shear into the anchor rods which must be accounted for. The applied shear that each anchor carries is calculated as shown:
17.11.1.1.8 - The following dimensional constraints must be satisfied in order to reduce interaction between breakout and anchor in tension.
17.11.1.1.9 - Moment due to shear lug shall be added to the overall design moment in the base plate. Luckily, it is purely a function of shear demand and your shear lug lever arm, so you won’t have to iterate your base plate design:
\[M_{u} = M_{analysis} + M_{lug}\] \[M_{lug} = V_u \times (h_{grout} + h_{sl}/2)\]17.11.2.1 - Bearing capacity of shear lug can be calculated as:
\[V_{brg,sl} = 1.7 f'_c A_{ef,sl} \Psi_{brg,sl}\]17.11.2.2 - \(\Psi_{brg,sl}\) is the bearing factor which accounts for effect of axial load (-ve is tension)
\[\Psi_{brg,sl}= \left\{ \begin{array}\\ 1.0 & \mbox{for no axial load} \\ 1 + \frac{P_u}{n N_{sa}} \leq 1.0 & \mbox{for tension} \\ 1 + \frac{4P_u}{f_c' A_{bp}} \leq 2.0 & \mbox{for compression } \end{array} \right.\]17.11.2.1.1 - \(A_{ef,sl}\) is the effective bearing area an is concisely summarized in the figure below. Note that if stiffeners are added, there are portions of shear lug that are inactive.
Figure 23: Shear Lug Bearing Area
Calculate shear lug bearing DCR (phi = 0.65):
\[DCR = \frac{V_u}{\phi V_{brg,sl}}\]As was the case with tension breakout, shear breakout is long and complicated. A good alternative is to always provide enough shear anchor reinforcements.
It is also worth noting, if you are close enough to an edge, you need to check this limit state even if shear is parallel to the edge!
17.11.3.1 - Shear lug breakout capacity is calculated in the exact same way as regular shear breakout. The figure below summarizes the breakout area.
Figure 24: Shear Lug Shear Breakout Area
17.7.2.1b - For an anchor group:
\[V_{cbg} = V_b \times (\frac{A_{Vc}}{A_{Vco}}) (\Psi_{ec,N} \Psi_{ed,N} \Psi_{c,N} \Psi_{h,N})\]17.7.2.1c - Note that shear breakout may also occur for shear acting parallel to an edge. The breakout capacity in these situations is determined by assuming load acting perpendicular to the edge, then multiply by 2. Edge factor should be taken as 1.
\[V_{cbg,parallel} = 2 \times V_{cbg}\]The equation above is essentially the same as tension breakout with a few subscript changes. Let’s go through them one by one.
17.7.2.2 - Basic single anchor shear breakout capacity
17.7.2.2.1 - This is the basic concrete breakout strength, of a single anchor, in cracked concrete. We will use this value as a starting point and apply several modification factors for other conditions. Use the ceiling value equation for shear lug.
\[V_{b} = (9) \times \lambda_a \sqrt{f'_c} (c_{a1})^{1.5}\]17.7.2.3 - Shear Breakout Eccentricity Factor (\(\Psi_{ec,V}\))
Assume shear is always concentric for shear lug design.
\[\Psi_{ec,V} = 1.0\]17.7.2.4 - Breakout Edge Effect Factor (\(\Psi_{ed,V}\))
Shear breakout is always near an edge. What we are adjusting here is a second edge \(c_{a2}\) (if any)
If the \(c_{a2} > 1.5c_{a1}\) then no reduction is necessary.
\[\Psi_{ed,V} = 1.0\]Otherwise:
\[\Psi_{ed,V} = 0.7 + 0.3 \frac{ c_{a2} }{ 1.5c_{a1} }\]17.7.2.5 - Breakout Cracking Factor (\(\Psi_{c,V}\))
Modification factor for cracking in shear breakout depends on rebar condition near the edge:
Figure 25: Shear Breakout Supplemental Reinforcement (in green)
17.7.2.6 - Member Thickness Factor (\(\Psi_{h,V}\))
Breakout capacity in shear is directly proportional to member thickness (\(h_a\)). This factor accounts for this effect:
\[\Psi_{h,V} = \sqrt{\frac{1.5c_{a1}}{h_a}}\]17.7.2.1.3 - Full Projected Failure Area of Single Anchor (\(A_{Vco}\))
Based on a 35 degree breakout cone, the projected failure area is a rectangle with side length of \((2)1.5c_{a1}\) and depth of \(1.5c_{a1}\)
\[A_{Vco} = 4.5c_{a1}^2\]17.7.2.1.1 - Actual Projected Concrete Failure Area (\(A_{Vc}\))
The actual breakout area must be adjusted depending on the perpendicular edge or anchor groups. However, the projected area must not exceed \((n A_{Vco})\) where n is the number of anchor in the anchor group. See figure below for some example calculations.
Figure 26: Concrete Breakout Area Example
17.7.2.1.2 - If the anchor is located in a narrow section where both its member thickness, and \(c_{a2}\) are less than \(1.5c_{a1}\), then \(c_{a1}\) must be adjusted as the maximum of:
Finally, the DCR can be calculated (phi = 0.7):
\[V_{cbg} = V_b \times (\frac{A_{Vc}}{A_{Vco}}) (\Psi_{ec,N} \Psi_{ed,N} \Psi_{c,N} \Psi_{h,N})\] \[DCR = \frac{V_u}{\phi V_{cbg}}\]Again, if capacity from concrete alone is not enough, add anchor reinforcements:
\[\phi V_{n} = \phi f_y A_s\] \[DCR = \frac{V_u}{\phi V_n}\]Figure 27: Shear Lug Bending
Lastly, shear lug must be thick enough to resist the bearing-induced bending. Remember to include thickness of grout layer in the cantilever!
\[L_b = h_{grout} + h_{sl}/2\] \[M_{u,lug} = V_u (L_b)\]Capacity calculated using plastic section modulus.
\[Z = b t_{sl}^2/4\] \[\phi M_n = \phi f_y Z\]Shear lug bending DCR is calculated as shown. Note the shear lug can be CJP welded or fillet welded to the bottom. Weld design is outside the scope of this article.
\[DCR=\frac{M_{u,lug}}{\phi M_n}\]Again, you may add stiffeners if you wish. The section properties can be calculated using any section property software or through equations online. Another option is to use a steel shape (such as an HSS section), but additional AISC checks would apply in those situations.
ACI 318 chapter 17 is silent on shear with lever arm (commonly seen in cladding attachments). For more discussion on this subject, refer to the textbook “Anchorage in Concrete Construction” by Eligehausen, Malle, and Silva (2006). The research findings therein are codified in the European Organization for Technical Approval (ETAG 001 Annex C). This is the methodology used by HILTI PROFIS.
Rather than imposing additional normal stress due to flexure, the anchor bending capacity is converted to an equivalent shear capacity
\[V_S^M = \alpha_M M_s / L_b\]Where:
Figure 28: Lever Arm Distance
Figure 29: Single and Double Curvature
ETAG 001 Annex C treatment of Grout Layer
ACI does briefly mention the treatment of shear with grout pads. Per 17.7.1.2.1 the calculated steel shear strength must be reduced by 0.8 (regardless of thickness of grout). However, as we will see shortly, this is less conservative than the European equivalent standards by at least a factor of 3!
The conservatism from the European code primarily comes from the concern that thicker grout pads may spall, leading the front anchors transferring shear primarily via bending. Section 14.4.5.2.2 of the Eligehausen textbook provides the recommendation that lever arm may be neglected only if all the following conditions are satisfied:
For comparison purposes, a 1.5” diameter, GR 55 anchor rod, with 2” grout layer and 2” plate has the following nominal capacity in each code:
Refer back to Section 11 for some nuances and options at your disposal. The DCR can be calculated as follows (phi = 0.65):
\[DCR = \frac{V_u / N}{\phi V_S^M}\]17.7.3 - Concrete pry out strength is simply calculated as a multiple of tension breakout strength (typically 2x).
For an anchor group:
\[V_{cpg} = k_{cp} N_{cpg}\]Where:
DCR is calculated as follows (phi = 0.7):
\[DCR = \frac{V_u}{\phi V_{cpg}}\]Shear and tension cannot be decoupled without bearing mechanisms as discussed in Section 11. We need to check the combined tension and shear interaction equations both in AISC and ACI.
There are two documents offering guidance on base connection spring stiffness:
The ASCE 41 method offers a quick, low-fidelity estimate that may be good enough for some applications. The ATC method is much more nuanced and require very lengthy calculations of component deformations.
ASCE 41-17 Method
Assume a yield rotation of 0.003 to 0.005 rad (refer to code section for more information).
\[\theta_y = 0.003 \mbox{ to } 0.005\]Calculate expected yield moment capacity assuming plate bending governs. You may need to use solver, or iteratively increase moment demand until base plate bending DCR = 1.0
\[M_y = f(\mbox{base plate bending limit state})\]The spring rotational stiffness is then calculated as:
\[K_y = M_y / \theta_y\]NIST Nonlinear Modeling Guideline Chapter 4.5.8
Figure A.1: Base Plate Component Deformations
Spring stiffness can be calculated as yield moment divide by yield rotation.
\[k_{rot} = \frac{M_y}{\theta_y}\]Yield moment is equal to the minimum design moment based on limit state of a.) plate bending (bearing side) b.) plate bending (tension side) c.) tension rupture of anchor rods. Similar to the ASCE 41 method above, you might need to iteratively increase moment demand until one of the DCRs reach 1.0.
\[M_y = min(M_y^{pl,c}, M_y^{pl,t}, M_y^{rod})\]Now calculate the individual component deformations which include a.) axial lengthening of anchor rods, b.) bending deformation of base plate, and c.) compression shortening of concrete
\[\Delta_{rod} = \frac{T_{rod}L_{rod}}{A_{rod}E_{rod}}\] \[\Delta_{plate} = (PL^3 / 3EI ) + (PL / AG)\] \[\Delta_{concrete} = \frac{f_{max}}{E_{concrete}} d_{footing}\]The deformations can be converted to an effective rotation:
\[\theta_y = (\Delta_{rod} + \Delta_{t,plate} + \Delta_{c,plate} + \Delta_{concrete} ) / (s + N/2)\]A couple of notes: