“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.
Table 1: Rebar Size Table
All empirical equations presented in ACI 318-19 will have unit [lbs, in, psi] unless otherwise noted.
Reinforced concrete is an inherently non-linear material. Nevertheless, we still model and analyze it linearly for convenience with the use of empirically derived equations and secant stiffness modifiers.
6.6.3.1.1 - Secant stiffness modifier are empirically determined. In some sense, it correlates to the degree of cracking and how much the element has “softened”. In lieu of more accurate analysis, the following modifiers can be used:
For service load analysis (deflection, vibration, etc), 6.6.3.2.2 suggests amplifying the above factor by 1.4
No reduction shall occur for area for axial deformation and shear deformation. Should more rigor be required, equations from 6.6.3.1.1 can be used.
19.2.2.1 - Modulus of elasticity of concrete is calculated with empirical equations. Equation (1) is used for light-weight concrete with weight ranging from 90 pcf to 160 pcf. Equation (2) is exclusively for normal-weight concrete (145 pcf). Concrete stress and strain curve is roughly parabolic. For practical purposes, an elastic modulus is defined as point connecting zero stress to 45% compressive strength.
\[E_c = w_c^{1.5} 33\sqrt{f'_c} \tag 1\] \[E_c = 57000 \sqrt{f'_c} \tag 2\]20.2.2.2 - Modulus of elasticity of reinforcing steel is taken as 29,000 psi
19.2.3.1 - Modulus of rupture is the tensile strength of concrete. Despite concrete having a small amount of tensile strength (about 15% of f’c), it must be neglected in axial/flexure calculations per 22.2.2.2.
\[f_r =7.5\lambda \sqrt{f'_c} \tag 3\]19.2.4 - Light-weight concrete has reduced mechanical properties. Therefore, a reduction factor is commonly included in many of the ACI equations. Generally, the following assumption is okay:
Table 2: Light Weight Concrete Factor as a Function of Density
20.2.2.4(a) - Maximum allowed grade of steel reinforcement. Use GR 60 bars as a minimum, GR 80 and GR 100 bars are mostly for seismic applications.
19.2.1.1 - Minimum allowed f’c. Generally advised to stay above 4 ksi. There is no maximum for f’c.
21.2.2.1 - Yield strain of reinforcement is calculated as follows. Note it is not always 0.002. Values differ for higher grade bars.
\[\epsilon_{ty} = \frac{f_y}{E_s} \tag 4\]22.2.2.1 - Maximum compressive strain is empirically determined to be around 0.003 to 0.004 but can go up to 0.008 in some special cases. The canadian code, for example, opts to use a slightly higher value of 0.0035
\[\epsilon_{cu} = 0.003 \tag 5\]22.2.2.3 - Stress distribution in concrete can be represented by a rectangular, parabolic, trapezoid stress block as long as it is in agreement with tests. Actual stress distribution is very complex.
22.2.2.4.1 - The rectangular stress block is characterized by 0.85f’c psi over a depth of “a” as shown in the figure below. “a” is a percentage of neutral axis depth “c”. \(\beta_1\) is empirically derived.
Figure 1: Rectangular Stress Block
\[a=\beta_1 c \tag 6\]\(\beta_1\) is equal to 0.85 for 4 ksi concrete, and decreases linearly by 0.05 per 1 ksi increase up to 0.65 for 8 ksi concrete.
\[\beta_1 = \left\{ \begin{array}\\ 0.85 & \mbox{if } \ f'_c \leq 4 \; ksi \\ 0.85-\frac{0.05(f'_c-4000)}{1000} & \mbox{if } \ 4\;ksi <f'_c < 8 \;ksi \\ 0.65 & \mbox{if } f'_c \geq 8 \;ksi \end{array} \right. \tag 7\] \[\beta_1 = max(0.65, min(0.85, 0.85-\frac{0.05(f'_c-4000)}{1000})) \tag 8\]21.2.1 - Strength reduction factor is different depending on element type and structural action. Refer to table 21.2.1 for full list.
21.2.2 - Strength reduction factor for moment + axial combination depends on strain of tension reinforcement closest to tension face.
Table 3: Resistance Factor for Axial + Moment
Higher resistance factor corresponds to higher ductility. Excessive deflection and cracking is good because it gives warning signs prior to failure. A net tensile strain of 0.005 is considered to provide sufficient ductility for most applications (Moment frames expect up to 0.0075). Net tensile strain under 0.002 is said to be compression-controlled and fails in a brittle manner.
Spiral ties confine the section better and thus we are able to get a higher resistance factor for compression-controlled to transition sections.
ACI 318 is exclusively based on LRFD (Load and Resistance Factored Design).
The basis of all design checks is to ensure a DCR (Demand-Capacity Ratio) of less than 1.0:
\[Capacity > Demand \tag 9\] \[\phi T_n > T_u \tag {10}\] \[DCR = \frac{demand}{capacity} = \frac{T_u}{\phi T_n} < 1.0 \tag {11}\]In a nutshell, without complicating things with probabilistic analysis. The difference between LRFD and ASD (allowable stress design) is that:
22.4.2 - Axial capacity of a concrete section is calculated as 80% to 85% of the sum of compressive strength of concrete + reinforcement. The 20%-25% reduction is meant to account for accidental eccentricity.
\[P_n = 0.8P_o \tag {12}\] \[P_n = 0.8(0.85f'_c (A_g - A_{st})+f_yA_{st}) \tag {13}\]For spiral tied elements, we can use \(0.85P_o\) instead. Note the 0.85f’c is separate from 0.8Po percentage.
Referring to Figure 1 above, moment capacity can be calculated as the internal force (T or C, doesn’t matter because they are equal), multiply by the distance between them; known as the internal lever arm and sometimes denoted as “jd”.
\[M_n = T (d-a/2) \tag {14}\]where:
\[T = f_yA_s\] \[a=\frac{f_yA_s}{0.85f'_cb}\]The full expression:
\[M_n = f_yA_s (d-\frac{f_yA_s}{2\alpha f'_c b}) \tag {15}\]To determine the correct resistance factor (\(\phi\)) to use, net tensile strain can be calculated with similar triangles. Then refer to Table 21.2.2.
\[\epsilon_s = \frac{d-c}{c} \epsilon_{cu} \tag {16}\]Approximate Solutions
Here is a life-changing trick! A very common, and scarily accurate back-of-envelope equation is often used for preliminary design. This is done by assuming an internal lever arm (say 0.9d)
\[\phi M_n = M_u = \phi f_y A_s (jd)\tag {17}\] \[M_u = \frac{(0.9)(60)A_s(0.9)d}{12} \tag {18}\] \[A_s = \frac{M_u}{4.05d} \tag {19}\]For example, say we have a 16”x30” beam experiencing 150 kip.ft of moment. Assume depth of 28 inches. We can literally calculate how many bars are required in our heads. 150/(4.05*28) = 1.32 square inch of steel area => (3) #6 bars.
To verify, if we calculate the moment capacity with (3) #6 bars rigorously, we get 162.7 kip.ft. Magic! We didn’t even need to know the concrete strength…
This approximate equation is predicated on the fact that \(T\) is easy to calculate, but the depth of neutral axis is trickier. Nevertheless, it is pretty much always in the range of 80% to 95% of depth “d”. Here is a list of recommended internal lever arm ratio “jd” :
Couple of notes on the state of practice for concrete design:
Flexure + Axial Interaction
Combined P+M action requires the use of interaction diagrams as shown below. We will discuss the creation of interaction diagrams in another article. P+M points within the onion-shaped surface is OK. Outside the surface is no good. See this SkyCiv article for more information.
Two key observations:
There were some drastic change to the shear capacity equation in ACI 318-19 compared to ACI 318-14. The equation now considers a “size effect” as well as the effect longitudinal reinforcement ratio has on shear capacity. No longer can we use the simple \(2\sqrt{f'_c}\) unless a minimum amount of transverse ties are provided.
22.5.1.1 - Shear capacity of a section comes from concrete and transverse ties.
\[V_n = V_c + V_s \tag {20}\]22.5.5.1 - Shear capacity contribution from concrete
22.5.8.5.3 - Shear capacity contribution from steel is based on a truss analogy. The fraction \(d/s\) can be thought of as how many ties are within the 45 degree crack spaced at distance of “d”.
\[V_s =A_vf_{yt} \frac{d}{s} \tag {26}\]There are several other important design constraints to note:
22.5.2.1 - “d” does not need to be less than 0.8h. This applies when you have multiple layers of tension bar that extends up the depth.
22.5.3.1 - \(f'_c\) is capped at 10 ksi. Should not use higher due to lack of experiment data.
22.5.3.3 - \(f_y\) is capped at 60 ksi regardless of actual steel grade in order to avoid diagonal compression failure
22.5.1.10 - For elements resisting shear in two orthogonal directions, an interaction equation can be used. Alternatively, if shear DCR in any direction is less than 50%, can ignore this interaction.
\[\frac{v_{ux}}{\phi v_{nx}} + \frac{v_{uy}}{\phi v_{ny}} \leq 1.5 \tag {28}\]Punching Shear Demand
R8.4.4.2.3 - Punching shear demand, also known as two-way shear demand is usually converted into stress by dividing factored shear demand (Vu) divide by the effective perimeter area. In addition, unbalanced moment from two-way slabs (such as edge or corner condition), tend to contribute a significant portion of shear stress and should not be omitted. Usually interior and edge columns have only unbalanced moment in one direction, corner columns have unbalanced moment about both directions.
\[v_u = \frac{V_u}{b_o d} + \frac{\gamma_v M_{scx} c}{J_x} + \frac{\gamma_v M_{scy} c}{J_y}\tag {29}\]Figure 2: Two-Way Shear Stress With Unbalanced Moment Contribution
Figure 3: Unbalance Moment Arising From Unequal Span
Figure 4: Equation to Calculate J (Source: CRSI)
For other shapes such as circular column, or trapezoidal section, refer to CRSI ACI 318-19 design guide.
Critical Section
22.6.4.1 - The effective perimeter area (\(b_od\)) is calculated based on an assumed 45 degree breakout cone. However, for simplicity, it can be taken as d/2 away from edge of column. Refer to the figure below for some different conditions.
22.6.4.2 - If the section is reinforced with stud rails or stirrups, then another critical section must be evaluated just outside of the reinforced perimeter. This section will have a reduced capacity.
Figure 5: Critical Section for Punching Shear
Figure 6: Openings near Critical Section
Punching Shear Capacity of Unreinforced Concrete
Originally, the allowable shear stress for two-way shear is taken simply as \(4\sqrt{f'_c}\). However, this was found to be unconservative in certain situations. Clause 22.6.5.2 defines the punching shear capacity as the minimum of three equations.
\[v_{c1} = (4) \lambda_s \lambda \sqrt{f'_c} \tag {32}\] \[v_{c2} =(2 + 4/\beta) \lambda_s \lambda \sqrt{f'_c} \tag {33}\] \[v_{c3} =(2 + \frac{\alpha_sd}{b_o}) \lambda_s \lambda \sqrt{f'_c} \tag {34}\]Generally, we would want to keep punching shear DCR to well within 0.8 due to the brittle nature of failure. There are several options we have for increasing the capacity:
Punching Shear Capacity of Concrete Reinforced With Stirrups
22.6.7.1 - In order for stirrups to be fully anchored and developed, there are some depth requirements. Slab depth to rebar “d” must be at least 6” or 16 times the bar diameter. Therefore, if we were using #4 stirrups with 0.75” cover, the slab has to be at least 8.75 inches for us to utilize stirrups in resisting punching shear.
22.6.6.1 - capacity of stirrups can only be initiated past a certain degree of inclined cracking. Therefore, punching shear capacity of concrete is reduced by half.
\[v_c = 2 \lambda_s \lambda \sqrt{f'_c} \tag {35}\]22.6.6.2 - size factor \(\lambda_s\) may be taken as 1.0 if \(A_v /s \geq 2\sqrt{f'_c} b_o / f_{yt}\)
22.6.7.2 - punching shear capacity provided by stirrup is calculated as:
\[v_s = \frac{A_v f_{yt}}{b_o s} \tag {36}\]Lastly, the critical section outside of the reinforced perimeter must also be checked with equation (35).
Punching Shear Capacity of Concrete Reinforced With Stud Rails
Contrary to stirrups, there are no slab depth requirements for using stud rails. Most likely due to the better anchoring mechanism formed by the headed studs.
22.6.6.1 - similar to stirrups, concrete capacity must be reduced in the presence of shear studs. Capacity is equal to the minimum of the three equations below. Note the only change is reduction of \(4\sqrt{f'_c}\) to \(3\sqrt{f'_c}\)
\[v_{c1} = (3) \lambda_s \lambda \sqrt{f'_c} \tag {37}\] \[v_{c2} =(2 + 4/\beta) \lambda_s \lambda \sqrt{f'_c} \tag {38}\] \[v_{c3} =(2 + \frac{\alpha_sd}{b_o}) \lambda_s \lambda \sqrt{f'_c} \tag {39}\]Important Note: based on research by Professor Parra-Montesinos at the University of Michigan, Ann Arbor. There is evidence of premature failure due to cracks emanating from column corners for orthogonally placed stud rails (article here). It is recommended to place shear studs in a radial manner. Furthermore, a ESR report was published (ESR-2494 and ESR 2708) that recommends reducing the existing code capacity equations by half to \(1.5 \sqrt{f'_c}\). Therefore, until more research is done, it is recommended to reduce the above equations by half.
22.6.8.2 - punching shear capacity provided by stud rails:
\[v_s = \frac{A_v f_{yt}}{b_o s} \tag {40}\]Lastly, the critical section outside of the reinforced perimeter must also be checked with the capacity below:
\[v_c = 2 \lambda_s \lambda \sqrt{f'_c} \tag {41}\]Finally, similar to one-way shear, there are some material property requirements. f’c is limited to 100 ksi per 22.6.3.1, fyt is limited to 60 ksi per 22.6.3.2. Material upper bound strength intended to prevent excessive cracking. Also because of lack of test data.
The theory of shear friction is very empirical and is based on an assumption of a clamping force provided by steel reinforcement perpendicular to shear crack. As a shear crack forms, aggregate interlock prevents a clean separation. The separating action engages the steel reinforcement. Because the structural action is actually tension, the doweling bars must be fully developed on both ends of the crack surface per 22.9.5.1.
22.9.1.1 - Despite adequate one-way shear capacity, a member may still develop unfavorable shear cracks at odd angles. Another possibility is cracks forming at construction/cold joints or interface between dissimilar materials.
22.9.4.2 - shear friction capacity is calculated as shown. Note fy is limited to 60 ksi per 22.9.1.3 regardless of actual steel grade due to lack of test data.
\[V_{vf} = \mu A_{vf} f_y + N_c \tag {42}\]22.9.4.4 - shear friction capacity is capped to an upper-bound value based on experimental studies. The ceiling is calculated as the minimum of three equations for monolithic or roughed slip surface:
\[(0.2f'_c) A_c \tag {43}\] \[(480 + 0.08f'_c) A_c \tag {44}\] \[(1600f'_c) A_c \tag {45}\]Equation (44) will always govern for f’c between 4000 psi and 8000 psi.
For un-roughened surface or concrete against structural steel, upper bounds is the minimum of two equations
\[(0.2f'_c) A_c \tag {46}\] \[(800f'_c) A_c \tag {47}\]Equation (47) will always govern for f’c between 4000 psi and 8000 psi.
22.9.4.6 - For permanent net tension, additional reinforcement area must be added to resist the net tension in addition to the bars for shear friction. This applies to flexural tension as well. In practice, flexural steel is often designed separately from shear friction dowels.
R22.9.4.6 - Moment acting on the slip surface is often ignored because net axial load is zero (shear friction capacity increases in compression zone, and decreases in tension zone). Ideally shear friction dowels should be placed uniformly to minimize crack width.
22.9.4.3 - shear friction dowels can also be placed at an incline angle (\(\alpha\)) but this is extremely rare.
\[V_{vf} = (\mu sin \alpha + cos \alpha) A_{vf} f_y + N_c \tag {48}\]Types of Torsion
Torsion is in general under-discussed in structural design because it is one of the weakest and least stiff load path. Furthermore, aside from tube sections, torsion is usually non-uniform and creates an additional warping stress which is very complex to characterize. Therefore, engineers usually create another load path to external loads causing torsion. Nevertheless, sometimes torsion is unavoidable due to design constraints.
22.7.3 - The first step is to determine what type of torsion is occurring. There are two types:
Figure 7: Two Types of Torsion
Redistribution must be accounted for in analysis. For example, if redistribution occurs in the spandrel beams in figure 7 above, then the slab must be designed for increased positive bending. Technically only 20% of moment redistribution is allowed per 6.6.5.3, but this is due to lack of test data rather than an actual physical observation. In practice, this clause is often disregarded or at least not explicitly calculated. For example, it is common to design slab edge connection to retaining wall as pinned. Concrete structures are highly indeterminate which means multiple load paths exist, allowing such redistribution to occur.
Torsion Demand
22.7.4.1 - Next we need to determine whether or not the torsion demand is high enough to warrant consideration at all. The threshold torsion is calculated as follows for non-prestressed solid cross section.
\[T_{th} = \lambda \sqrt{f'_c} (\frac{A_{cp}^2}{p_{cp}}) \sqrt{1+ \frac{N_u}{4A_g \lambda \sqrt{f'_c}}} \tag {49}\]From here, there are three possibilities:
Torsion Capacity
Torsion design in concrete is based on a space truss analogy where the core is essentially neglected. Torsion demand is resisted by three actions: diagonal compression strut, vertical shear, and horizontal tension. What makes torsion calculation tedious is the fact that it is not decoupled and must be considered in tandem with shear and flexure. Torsion demand seldom exist in isolation.
Figure 8: Torsion Action - 3D Truss Analogy
22.7.7.1 - To prevent excessive cracking and diagonal crushing of concrete. The following limit is set for solid sections and should be check first. Increase section size if not workable.
\[\sqrt{(\frac{V_u}{b_w d})^2 + (\frac{T_u p_h}{1.7A_{oh}^2})^2} \leq \phi (\frac{V_c}{b_w d}+ 8 \sqrt{f'_c}) \tag {51}\]22.7.6.1 - Torsion capacity is calculated as the minimum of the following. The first expression is related to transverse steel, the second expression is related to longitudinal steel.
\[T_n = \frac{2A_o A_t f_{yt}}{s} cot \theta \tag {52}\] \[T_n = \frac{2A_o A_l f_{y}}{p_h} tan \theta \tag {53}\]The above equations may be suitable for analysis of existing members. However, this is almost never used for design because of how intertwined torsion is to flexure and shear. Instead, use the procedure below for design.
1.) Determine transverse steel ratio required for shear:
\[\frac{A_v}{s} = \frac{V_u - \phi V_c}{\phi d f_{yt}} \tag {54}\]2.) Determine transverse steel ratio required for torsion:
\[\frac{A_v}{s} = \frac{T_u}{\phi 2 cot \theta A_o f_{yt}} \tag {54}\]3.) Combine the effect of shear and torsion:
\[(\frac{A_v}{s})_{total} = (\frac{A_v}{s})_{torsion} + (\frac{A_v}{N_{leg}\times s})_{shear} \tag {55}\]4.) Given you have selected a stirrup size, the required spacing is calculated as:
\[s_{req} = \frac{A_{bar}}{(\frac{A_v}{s})_{total}} \tag {56}\]Lastly, all four sides of a member must have longitudinal reinforcement for torsion. The amount of additional longitudinal steel is calculated as:
\[A_l = (\frac{A_v}{s})_{torsion} \times p_h (\frac{f_{yt}}{f_y}) cot^2(\phi) \tag {57}\]A minimum steel area is specified in 9.6.4.3 as the less or two equations:
\[A_{lmin} = \frac{5\sqrt{f'_c}A_{cp}}{f_y} - (\frac{A_t}{s})p_h \frac{f_{yt}}{f_y} \tag {58}\] \[A_{lmin} = \frac{5\sqrt{f'_c}A_{cp}}{f_y} - (\frac{25b_w}{f_{yt}})p_h \frac{f_{yt}}{f_y}\tag {59}\]Divide the steel area above by 4 to spread into four faces of the section. The top and bottom face usually have enough steel, but it is necessary to ensure the side faces have at least \(A_l/4\) as well. To check if flexural steel is enough:
\[A_{s,flexure} = \frac{Mu}{\phi f_y jd} \tag {60}\]if \(A_{s,flexure} + A_l/4 < A_{s,provided}\), then we are okay. Otherwise increase longitudinal steel.
Bearing most commonly occurs when a structural steel member “bears” onto the concrete such as at a steel column base plate. It could also occur between two concrete member such as at a pre-cast beam support, or a shear keys. Typically, concrete bearing failure occurs at around 0.85f’c, which is why you will see the 0.85 coefficient everywhere. If the support area is must larger than the load area, the surrounding concrete provides a confinement effect which results in up to double the bearing capacity.
22.8.3.1 - Design bearing strength is calculated as follows:
\[B_n = \gamma 0.85f'_c \tag {61}\]Figure 9: Bearing Area Definition
25.2.1 - minimum rebar spacing should be the maximum of 1 in, one bar diameter, or (4/3) the diameter of aggregates. This is to ensure concrete can move into spaces between bars. However, practically speaking you should never have bars spaced less than 3 in.
20.5.1.3.1 - minimum cover to face of reinforcing steel is specified as follows:
The first method of calculating development length is a simplified version of method 2 (next section) that preselects some typical \(c_b + K_r / d_b\) values. They are as follows. Regardless of equations used, development length \(l_d\) shall not be less than 12 inches per 25.4.2.1.
Figure 10: Development Length (ld) Factors
25.4.2.4 - Alternatively, you can calculate development length more accurately and a lot less conservatively with the equation below:
\[l_d = \frac{3 f_y \Psi_t \Psi_e \Psi_g \Psi_s}{40 \lambda \sqrt{f'_c} \frac{c_b + K_{tr}}{d_b}} d_b > 12in \tag {65}\]25.4.3.1 - development length of hook bars is calculated as follows, but shall not be less than 6” or \(8d_b\)
\[l_{dh} = \frac{f_y \Psi_e \Psi_r \Psi_o \Psi_c}{55 \lambda \sqrt{f'_c}} d_b^{1.5} > max(6in,8d_b) \tag {67}\]Figure 10: Hook Development Length (ldh) Factors
25.4.4.2 - development length of headed bars (also known as T-heads) is calculated as follows, but shall not be less than 6” or \(8d_b\)
\[l_{dh} = \frac{f_y \Psi_e \Psi_p \Psi_o \Psi_c}{75 \lambda \sqrt{f'_c}} d_b^{1.5} > max(6in,8d_b) \tag {68}\]Figure 11: Headed Bars Development Length (ldt) Factors
25.4.4.1 - furthermore, headed bars cannot be used in light-weight concrete due to lack of data.
For simplicity, lap splice length is typically taken as \(1.3l_d\). A reduction may be permitted for specially detailed stagger splice. See 25.5.2.1.
25.4.9.1 - compression development length (\(l_{dc}\)) is calculated as follows:
\[l_{dh} = \frac{f_y \Psi_r}{50 \lambda \sqrt{f'_c}} d_b > max(8in,0.0003f_y\Psi_r d_b) \tag {70}\]Figure 12: Compression Development Length (ldc) Factors
Here are some useful design charts assuming some of the most common design properties:
Multiply the charted value by the following modification factors if appropriate. However, Ld must not be less than 12” and Ldt/Ldh must not be less than 6”.
Straight Bar Development Length(ld)
Hook Development Length (ldh)
Headed Bar Development Length (ldt)