Basics of Reinforced Concrete Structural Design - Engineersdaily
Learning Objective
The purpose of this course is to give the reader an understanding of how the two basic methods of concrete structural analysis compare by example in theory and results. This course also shows the reader how critical the selected loading model is to developing a successful design.

Course Content 

Concrete is a mixture of gravels, water, and cement. The water hydrates the cement and hardened concrete is created. The knowledge of cement is several thousand years old. It is found as mortar to set stones or brick. This method still commonly used today. True reinforced concrete was developed only about 100 years ago. It took the advent of the industrial age where mass production of steel, cement and aggregate mining became common practice to develop reinforced concrete to become the building material of choice. Reinforced concrete is used in bridges, buildings, dams, foundations and even sculptures.

This material has several advantages over steel or timber for many applications. Concrete can easily be molded into any shape. Concrete is hard, durable, and nearly inert and provides excellent corrosion protection for the steel reinforcement. Construction is economical, maintenance is minimal and it is easy to develop moment resisting connections. The only real problems with concrete is that it tends to crack due to curing shrinkage and the tensile strength is only about 10% of its compressive strength and is heavier than steel or timber. In order to fully utilize the best of the concrete properties tensile reinforcement is cast into the concrete. The added reinforcement provides the needed tensile strength to compliment the concrete compressive strength and stiffness. The concrete and steel can share the shear strength.

There are many different types of cement and additives available to give concrete special properties for placing and strength. Superplasticizers are used to optimize water content and still maintain a workable slump. Retarders, water reducing agents and accelerators are also commonly used to accommodate weather conditions and improve workability. The optimum water content for the most common cement, Portland Type II, is about 0.38 pounds of water for each pound of cement. Without additives to increase slump that 0.38 water to cement ratio will not produce enough slump for the concrete to be placed in the forms. Slump is the measure of how fluid the concrete is. Generally concrete needs a slump of about 3 inches to be readily discharged form the transit mixer truck and placed in the forms. Although it is now possible to achieve concrete strengths well in excess of 10,000 psi, it is also expensive requiring a combination of special aggregates, cement, additives and precise quality control. For this reason most structural reinforced concrete is in the 3,000 to 6,000 psi range. Footings and columns are generally in the 3,000 to 4,000 psi range and bridge superstructures are in the 4,000 to 6,000 psi range where weight is a major design factor.

Reinforcing steel, commonly called rebar, has generally 40,000 or 60,000 psi yield strength. In some applications rebar with yield strength of 80,000 psi can be used. Post-tensioning reinforcement is where the concrete is first cured to strength and then tensioned to place the concrete in compression. Prestressing is where the strand or wire is tensioned first and the concrete is then cast. When the concrete cures to the desired strength the prestressing tension is released and the concrete is placed in compression by the shortening strands. Prestressing is common for concrete piles and bridge girders where the products are made in a plant and steam curing is used to accelerate concrete curing. Psot-tensioning is often used in box girder bridges. This post-tensioning and prestressing uses steel with yield strengths of 150,000 to 270,000 psi. The reason most deformed rebar is limited to 60,000 psi is that the concrete would have to excessively deflect and crack to develop the rebar tension.

The prevailing authority for concrete design is the American Concrete Institute (ACI). They are constantly reviewing the latest products and technologies and incorporating them into their Building Code Requirements for Structural Concrete (ACI 318). Because of the increasing design options and products available, the code is becoming more complex and difficult to comprehend, including many cross-references. But for basic structural concrete design the standards have not changed much in the last few years. While the reader is encouraged to obtain the latest ACI 318 code, it is also recommended that the reader get a copy of an older version of the ACI 318 Code, such as the 1963 issue. The reason is that the code is much easier to understand and the fundamentals are basically unchanged. The example below discusses only a few of the many ACI design options and restrictions. This lesson is designed to introduce or refresh the reader to the basic methods and procedures involved in structural concrete design.

Design Methods:

This course will limit discussion to normal reinforced structural concrete using 4,000 psi concrete and 60,000 psi rebar. There are two acceptable methods to design concrete: the working stress method and the ultimate strength method. The ultimate stress method is the one most commonly used. The reasons for this are the ultimate strength method will require substantially less concrete and rebar, and the design calculations are easier. Working stress design model assumes that as the concrete beam bends due to induced moments the strain relationship between the rebar in tension and the concrete in compression remain constant. Ultimate strength design places the rebar in full yield so the strain relationship between reinforcement and concrete is ignored and a rectangular concrete compression block stressed at design strength is formed.
Basics of Reinforced Concrete Structural Design

B equals the concrete beam compression width.
B' equals the concrete beam web width
Fc equals the allowable concrete working stress.
Fc'equals the design ultimate concrete strength.
Ts equals the slab thickness
D equals the depth of the concrete beam from rebar centroid to extreme compression fiber.
X equals the concrete compression depth from the neutral axis to the extreme compression fiber.
A equals the ultimate compression block developed by the yielding concrete.
C equals the total compression load.
T equals the total tension load.
D-X/3 is the moment lever arm for working stress design.
D-A/2 is the moment arm for ultimate stress design.
As is the rebar area

One of the first considerations is cover. Cover is the minimum distance from the nearest face of the concrete to the encased reinforcement. This cover is provides the corrosion protection for the reinforcement and allows the bars to bond to the concrete. Cover also facilitates the flow of concrete around the rebar. ACI 318 7.7 recommends cover from 3/8" to 3" depending of weather exposure and bar size. If the cover is too thin cracks can form parallel and over the bars. This cracking serves to accelerate corrosion and loss of bond. High strength steels used in post and pretensioning need special consideration.

These steels are subject to stress corrosion, which is the fact that under high tension the steel rusts on the order of 10 times more quickly than normal strength steels. These high strength steels are brittle so any sharp surface defect will cause the steel to snap much in the same way glass will easily break along a scratch line.
Seawater and deicing salt are very corrosive and requires covers on the order of 3" or special concrete waterproofing. Bridge decks are a special problem with the combination of high traffic impact, deicing and thin covers all acting to deteriorate the concrete and rebar. Another consideration is allowance for traffic wear and abrasion.

For bending the working stress design requires a concrete safety factor of 2.22 and a rebar safety factor for 60,000 psi rebar of 2.5. The basic ultimate strength design allows a composite safety factor of
U =1.4D+1.7L. Where U is the ultimate design load, D is the dead load (concrete, dirt and other non-dynamic fixed loads); L is the live load (equipment, traffic etc.). 1.4 is the minimum dead load safety factor and 1.7 is the minimum live load safety factor. 

The reasons that the dead load safety factor is much less than the live load safety factor is that dead loads can be calculated with reasonable certainty and they are static (non dynamic). Live load configurations are approximations at best, often dynamic and constantly changing. Live loads will induce vibration and fatigue. Truck can be over loaded or the service of the structure may change in time. The higher live load safety factor is set to account for these indeterminate factors.
Wind (W) and earthquake (E=1.1W) loads are also introduced where:
U = 0.75(1.4D+1.7L+1.7W)
U = 0.75(1.4D+1.7L+1.87E)
U = 0.9D+1.3W
U = 0.9D+1.43E. 

The maximum U generated form all the above equations shall be used as the design criteria. Below are diagrams of the bending models for working stress and ultimate stress design.

Concentrated Loads:
A fairly common industrial design loading is 250 psf. However, industrial loads include trucks, forklifts and cranes, which places concentrated dynamic loads on tire footprints. Once, I was assigned to design a wharf extension at one of our industrial bases. I was given the plans and design criteria of the existing concrete wharf and told to copy the design for the new extension. During the site visit I noticed several signs of substantial distress. The concrete slab had diagonal bending failure cracking and the beams had shear failure cracks. I was surprised since the live load design was 250 psf and the wharf was only about 5 years old. I then studied the actual concentrated loads induced by the equipment and buoys on the deck. I found that while the average load was less than 250 psf, the concentrated loads induced much higher moments and shears than the uniform live load calculated. Needless to say, the new design was much stronger. 

The 24-foot long beam is equal to 2 traffic lanes. A fully loaded 18-wheel truck weighs 72,000 lbs. When a 30% impact factor is added the weight the total becomes 93,600 lbs. The area the truck occupies is at least 40 feet long by 10 feet wide or 400 square feet. The average load is 93,600/400 = 234 psf which is less than the design load of 250 psf. Now lets see how the concentrated axle loads generate moment and shear and compare them to the uniform load generated moment and shear.
Basics of Reinforced Concrete Structural Design

Basics of Reinforced Concrete Structural Design
 Basics of Reinforced Concrete Structural Design

Now we compare between the moment and shear,
The uniform moment is Mw = WL^2/8 = 1,500x24^2/8 = 108,000 ft-lbs
The AASHTO H20-S16-44 weight for a highway truck dual axle is 32,000 lbs plus a 30% impact factor.
P = 1.3x32,000 = 41,600 lbs
The beams are spaced at 6 feet center to center and the dual axles of a truck are spaced about 4 feet center to center. That places one axle directly over the beam the other axle is 2 feet from the adjacent beam.
The distribution of the dual axle load is R = Beam load = P(1/2 + 2/6) = 5P/6 = 5x41,600/6 = 34,667 lbs
The truck induced moment is Mc = RX = 34,667x7 = 242,670 ft-lbs
The truck moment is 2.25 times greater than the uniform moment
When we check for shear we find that the truck weight produces 2.2 times more shear than the uniform load case. The uniform shear is W(L/2-D/12) = 1,500(24/2+15.5/12) = 16,062 lbs
The Truck shear is 34,667 lbs. This huge difference will at least cause major distress to the reinforced concrete structure, if not collapse. 

There are two ways to resolve this difference. The beams can be strengthened to take the truck loading or the beams can be spaced further apart so the truck induced stresses fall within those generated by the uniform loading. The slab must also be designed to support the truck wheel and axle concentrated loads.
For the purposes of comparison of working stress design Vs ultimate strength design we will design only the beam portion of the structural concrete. The slab is commented on only in passing in that punch out shear around the wheels will control the thickness and there will be negative moments in the slab over the beams. It should be noted that shear reinforcement is not considered effective in any member less than 10" thick. This means that most slabs must rely solely on the concrete for shear resistance.

Working Stress Design:
The working stress model assumes a constant increase in compression strain form the neutral axis to the outer compression fiber and no concrete tension. Only the rebar contributes to the moment tension. This is because the concrete is assumed to be fully cracked by the tensile stress. 4,000 psi compressive strength concrete will crack at about 400 psi tension, but to strain with a rebar tension of 24,000 psi; the concrete would have to develop 3,000 psi in tension. The modulus of tension rupture is about 7.5Fc’^0.5. So any small tension contribution made by the concrete is ignored. By conservation of forces, rebar tension equals concrete compression (T = C).

Now as the beam bends the rebar lengthens and the concrete shortens the strain relationship is resolved by the formula: BX^2/2 = NAs(D-X). B is the concrete compression width. X is the concrete compression depth. As is the tensile rebar area. D is the depth of the rebar centroid. N is the ratio of the modulus of elasticity of steel divided by the modulus of elasticity of concrete. The modulus of elasticity of steel is 29,000,000 psi and the modulus of concrete is Ec=33(Gc^1.5)(Fc’^0.5). Gc is the concrete density, usually about 145 pounds per cubic foot. Fc?is the ultimate design strength. For Fc?= 4,000 psi concrete Ec will calculate to about 3,625,000 psi. The ratio N = Es/Ec = 29,000,000/3,625,000 = 8.
For normal weight concrete, i.e.: 145+/- pcf the modulus of elasticity ratio is:
Fc?= 2500 psi, N = 10
Fc?= 3000 psi, N = 9
Fc?= 4000 psi, N = 8
Fc?= 5000 psi, N = 7

To illustrate the differences between the working stress and the ultimate strength design methods an example of each will be developed using the same configuration and loads. Now lets check a reinforced concrete simply supported beam by the working stress method. ACI Table 9.5(a) requires a minimum length to depth ratio of H = L/16, where L is the length between supports in inches. Lets assume L = 24 feet, the beam depth must be H = 12x24/16 = 18 inches. Because of the high moments we will start with a beam H of 24 inches. If we assume 1-1/2 inches of cover (clear distance from the nearest face of concrete to the rebar), a #4 stirrup (?quot; diameter bar, bars are numbered by 1/8" i.e." 5/8" bar is a #5) and #11 (1.375" diameter) main reinforcement. The rebar depth becomes D= H-Cover-Stirrup-1/2 rebar diameter = 24-1.5-0.5-1.375/2 = 21.3 inches. Let’s use the concentrated truck axle load, as it is the worst case. We will also assume the concrete beam is placed at 6-foot centers and is cast integrally with a 6" thick slab. Structural concrete is usually assumed to weight Gr =160 pounds per cubic foot for design purposes. Now we can begin the design process.
Basics of Reinforced Concrete Structural Design

The first step is to establish the basic design criteria. The following selections are made:
Fy = Rebar yield stress = 60,000 psi
Fs = Rebar working design stress = 24,000 psi for 60 grade rebar through #11 bars ( 1-3/8" Diameter)
Fc'= Concrete ultimate concrete strength = 4,000 psi
Fc = Concrete working strength = 0.45Fc?= 0.45x4000 = 1,800 psi
L = Beam length = 24 ?/span>
R = Live load = Truck Axle at 34,667 lbs
Wd = Gr(slab depth x slab width + (beam depth less slab depth) x beam depth = 160(.5x6+(2-0.5)1.5)
Wd = Dead Load = 840 lb/lf
Beam Calculation
The first step in designing this beam is to determine the bending moment at centerline.
Md = Dead Load Moment at centerline = WtL^2/8 =840x24^2/8 = 60,480 ft-lb
Me = Live Load Moment is = RX =34,667x7 = 242,669 ft-lb
Mt = Total Moment = Md + Me = 60,480 + 242,669 = 303,149 ft-lb
Since the slab and beam are integral, the slab can be used to act as a T-beam flange. ACI allows the width contribution of the slab on both sides of the beam to be no more than L/4, B+16T (T = slab thickness), nor one-half of the clear distance to the next beam or web.
Bw = L/4 = 24?4 = 6 feet
Bw = B?+ 16T = 1.5?16x.5 = 9.5 feet or
Bw = Center Spacing = 6?/span>
Therefore the spacing of 6 feet or 72 inches can be used as the beam compression width.
Basics of Reinforced Concrete Structural Design

B equals the concrete beam compression width.
B?equals the concrete beam web width
Fc equals the allowable concrete working stress.
Ts equals the slab thickness
D equals the depth of the concrete beam from rebar centroid to extreme compression fiber.
X equals the concrete compression depth from the neutral axis to the extreme compression fiber.
A equals the ultimate compression block developed by the yielding concrete.
C equals the total compression load.
T equals the total tension load.
D-X/3 equals the moment lever arm for working stress design.
As equals the rebar area
Now we can solve for X, the compression depth.
BX^2/2 = NAs(D-X)
To start the solution process an As (rebar area) must be assumed. A quick way is to take and educated guess at the D-X/3 moment lever arm. D = 21.3" so lets guess at X = 6".
Dj = D-X/3 = 21.3-6/3 = 19.3"
M = FsAsDj or As = 12Mt/FsDj
As = 12x303,149/24,000x19.3 = 7.85 square inches.
Lets try 5 each #11 bars As = 5x1.56 = 7.80 sqin.
BX^2/2+NAsX =NAsD = 72X^2/2+8x7.80X = 8x7.8x21.3, X = 5.27" < Ts =6" okay.
Dj = 21.3-5.27/3 = 19.5 inches.
Ma = AsFsDj/12 = 7.8x24,000x19.5/12 = 304,200 ft-lb > Mt = 303,149 okay

At this point the rebar selection can be refined or the check can be continued. I have these equations set up in and Excel worksheet so the process can be iterated very quickly. Another thing to note is that X is smaller than the slab thickness (Ts) so no concrete compression area is lost. When the compression depth is greater than the T-beam flange thickness, the calculation for the compression area becomes more complex. Another way to eliminate the problem is to add compression rebar. Since our example does not present this problem, we will proceed with the check and design process.

First calculate the rebar stress: Mt = FaAsDj or Fa = Mt/AsDj = 12x303,149/7.8x19.5 = 23,917 psi. This is less than 24,000 psi so the rebar is not overstressed. Rebar tension and concrete compression are equal so:
C = T = AsFa = 7.8x23,917 = 186,553 lb
C = concrete compression = BXFc/2 or Fc = 2C/BX = 2x186,553/72x5.27 = 983 psi, which is less than the allowable compressive stress of 1,800 psi.

Now another check should be made for rebar spacing. The width of the beam web was set at 18 inches with 1.5 inches of cover and #4 stirrups. Therefore the bar must fit into a space of B-2xCover-2xStirrup-1x#11 bar (1.375" diameter). This is 18-2x1.5-2x0.5-1x1.375 = 12.625". 12.6" divided by 4 spaces = 3.16". ACI recommends at least 1 inch clear between bars or 1.33 times the maximum aggregate size. If we assume a 1" aggregate, the spacing can be reduced to 1.375+1.33 = 2.7 inches or 4x2. = 10.8", which means the beam may be narrowed by 12.6-10.8" = 1.8". This narrowing is usually not worth the effort. Also the stirrups may turn out to be larger than the #4 bar assumed. The reason for checking the rebar spacing is important for two reasons. First if the spacing is too close the concrete aggregate will not flow around the bars. This will cause rock pockets (voids) and segregation of the concrete mix. The second reason is proper spacing ensures full bond development between the concrete and rebar.

The next thing to check is shear. The actual failure plane is a diagonal plane at about 45 degrees from the vertical. The failure is actually a tension failure of the concrete. That is why the allowable shear stresses are so low. The allowable shear stress, v = 1.1Fc’^0.5 with no web reinforcement or up to v = 5Fc’^0.5 with stirrups. Stirrups are vertical reinforcement placed around the beam perimeter at the cover depth. The critical shear is calculated at a distance D from the support. In this case the critical shear is:
V = Wd(L/2-D) + R = 840x(24/2-21.3/12) +34,667 = 43,256 lbs.
The shear stress, v = V/B’D = 43,256/18x21.3 = 112.8.9 psi, or 1.8Fc’^0.5. B?is the beam web width.
This is more than the Vc = 1.1x4,000^0.5 = 70 psi, but less than 5Fc’^0.5. It means the shear stress requires stirrups in the beam. The shear carried by the concrete is
Va = VcB’D = 70x18x21.3 = 26,838 lbs. The Stirrup shear is Vs = V-Va = 43,256-26,838 = 16,418 lbs.
The Stirrup area is: Av = VsS/FvD,
S = stirrup spacing, not to exceed D/2 or 21.3/2 = 10.65 inches. We will use 10" for spacing.
Fv = rebar allowable tensile stress = 24,000 psi.
Av = VsS/FvD = 16,418x10/24,000x21.3 = 0.32 sqin.
A #4 bar stirrup area is (2 leg in a stirrup) 2xAb = 2x0.2 = 0.4 sqin. > 0.32 okay
Another solution is to reduce the stirrup spacing to 6 inches and use #3 bars.
Av = VsS/FvD = 16,418x6/24,000x21.3 = 0.19 sqin.
A #3 bar stirrup area is (2 leg in a stirrup) 2xAb = 2x0.11 = 0.22 sqin. > 0.19 okay
When shear reinforcement is required, ACI sets the minimum stirrup area at:
Av = 0.0015B’S = 0.0015x18x6 = 0.162 sqin < 0.22 okay. Even in cases where no stirrups are required by calculation, it is recommended that the minimum stirrup area be applied to full length of the beam.
Therefore #4 bar at 10" c-c or #3 bar at 6" c-c for the entire length of the beam will satisfy the shear requirement. Although the stirrups calculate increased spacing toward the beam center, the concentrated loads produce rectangular blocks of shear well toward the beam centerline. For that reason I prefer to be a little generous with stirrups. They also act to control cracking. 

The next step is to check anchorage or bond. Tests show that the bond between deformed rebar and concrete is almost entirely mechanical. Rust, mill scale, cement splatter and even form oil has little effect on the concrete to bar bonding. In order for rebar to develop its full available strength in concrete the anchorage must be made over a distance or by an end hook. ACI limits the bond stress for bars conforming to ASTM A 305 to 4.8Fc’^0.5/D and no more than 500 psi. These bars must have less than 12" of concrete cast below them. D is the nominal bar diameter in inches. A #11 bar allowable bond stress is
B = 4.8x4,000^0.5/1.375 = 222 psi < 500 psi.
The perimeter of the bar is PiD so the bond per inch = BPiD = 222x3.142x1.375 = 959 lb/in.
The tensile force of the bar is FaAb = 24,000x1.56 = 37,440 lb.
The length of bond is T/BP = 37,440/959 = 39 inches = 3.25 ft > L/4 = 24?4 = 6?okay.

The reason L/4 is a common test is that the rebar moment diagram will always just touch or be larger than the parabolic beam moment diagram. If high moments are expected close to the supports, the linear fall off of the loss of moment due to bond development should be tested. If this distance is critical to the bond length, an end hook can be added to the rebar. Often hooks are standard additives at beam-ends and slab edges to eliminate any concern for bond length and also for crack control. ACI has more complex formulas that take into account rebar coatings, size, location and light weight concrete, which we are not using here. 

Another application is to use these calculations to decrease the number of bars as the moment decreases. The caution is to keep the bar arrangement symmetrical along the length of the beam and to run the two outer most bars the full length. Now the bond length of 3.25 feet must be added to each end of the shortened bars. Unless there are a lot of repetitions, the saving is not being worth the calculation and drafting effort.
The final property to check is deflection. ACI allows the application of normal deflection formulas unless the member is cracked or varies in section along its length. Since this example has a constant section and the stress levels will limit cracking we can use the simple beam formula:
Dc = 5WtL^4/384EcIg is the uniform load deflection formula, where
Dc = Deflection in inches.
For the purpose of simplicity and reasonable accuracy the truck weights are averaged to a uniform load. The truck moment diagram approximates the parabolic uniform load moment diagram shape.
We = Trucks =2R/L = 2x34,667/24 = 2,889 lb/ft
Wd = Concrete dead load = 840 lb/ft
Wt = We + Wd = 2,889 + 840 = 3,729 lb/ft = 311 lb/in
L = 24 ft = 288 in
Ec = Concrete modulus of elasticity = 3,625,000 psi
Ig = the gross concrete moment of inertia (ignore rebar)
Since this is a T-beam we must first find the neutral axis. By setting the control axis at the bottom of the beam the axis is located at a distance Yx = (AtYt + AbYb)/(At + Ab), where
At = the slab area = TsB = 6x72 = 432 sqin
Ab = the beam web area = (H-Ts)B?= (24-6)18 = 324 sqin
Yt = the centroid of the slab area = H-Ts/2 = 24-6/2 = 21 inches
Yb = the centroid of the web = (H-Ts)/2 = (24-6)/2 = 9 inches
Yx = (AtYt+AbYb)/(At+Ab) = (432x21+324x9)/(432+324) = 15.86 inches
Yy = H-Yx = 24-15.86 = 8.14 inches
This places the neutral axis below the slab and web intersection.
Ig = BYy^3/3-(B-B?(Yy-T)^3/3+B’Yx^3/3 =
72x8.14^3/3-(72-18)(8.14-6)^3/3 +18x15.86^3/3 = 36,700 in^4
Now Deflection can be calculated.
Dc = 5x311x288^4/384x3,625,000x36,700 = 0.21 inches

The L/Dc ratio is a measure of flatness, in this case L/Dc = 288/0.175 = 1,646. This is much greater than a common requirement of L/Dc = 360. This calculation does not account for long term creep, which about doubles the initial deflection over time. Shoring or reshoring for floor slabs and beams is generally kept in place for the first 2 weeks to a month since creep or sag in the early stages of curing is much greater than fully cured concrete.

Ultimate Strength Design:

Now we will design the same beam using the ultimate strength design method. Again we will select the same basic design criteria as used for the above working stress design.
Fy = Rebar yield stress = 60,000 psi
Fc?= Concrete ultimate concrete strength = 4,000 psi
L = Beam length = 24 ?/span>
R = Truck dual axle weight = 34,667 lb at 7 feet from each beam end.
Wd = Gr(slab depth x slab width + (beam depth less slab depth) x beam depth = 160(.5x6+(1.5-0.5)1.5)
Wd = Dead Load = 840 lb/lf
U =1.4D+1.7L. We will ignore wind and earthquake for this example calculation.
The first step is to design for the bending moment.
Mu = Ultimate Moment = 1.4WdL^2/8 +1.7RX =1.4x840x24^2/8+1.7x3,4667x7 = 497,328 ft-lb
Basics of Reinforced Concrete Structural Design

B equals the concrete beam compression width.
B?equals the concrete beam web width
Fc?equals the design ultimate concrete strength.
Ts equals the slab thickness
D equals the depth of the concrete beam from rebar centroid to extreme compression fiber.
A equals the ultimate compression block developed by the yielding concrete.
C equals the total compression load.
T equals the total tension load.
D-A/2 equals the moment arm for ultimate stress design.
As equals the rebar area
Mu = KfAsFy(D-A/2)
Kf = reduction coefficient for flexure = 0.9
As = is the rebar area
Fy = rebar steel yield = 60,000 psi
D = depth to rebar = 21.3 inches
A = the compression block depth = AsFy/0.85Fc’B
Again we start the design with an assumption for the moment lever arm.
If we assume A/2 = 0.7", D-A/2 = 21.3-0.7 =20.6 inches.
As = 12Mu/0.9(D-A/2)xFy = 12x497,328/0.9x20.6x60,000 = 5.36 sqin.
2 each #8+ 3 each #10 = 2x0.79 +3x1.27 = 5.39 sqin.
A = AsFy/0.85Fc’B =5.39x60,000/0.85x4,000x72 = 1.32 inches
Mu = KfAsFy(D-A/2)/12 = 0.9x5.39x60,000x(21.3-1.36/2)/12 = 500,623 ft-lb >497,328 ft-lb okay
Now the rebar area to concrete area ratio must not exceed 0.75Pb. This is to insure that the rebar will yield before the concrete fails in compression.
Pb = 0.85KrFc?7,000/Fy(87,000+Fy)
Kr = the reduction fraction of 0.85 up to 4,000 psi and is reduced by 0.05 continuously for each 1,000 psi in excess of 4,000 psi.
0.75Pb = 0.75x0.85x0.85x4,000x87,000/60,000(87,000+60,000) = 0.02138
The actual area ratios is: Pb = As/BD = 5.39/72x21.3 = 0.0035 < 0.02138, okay
The last bending check is to insure the flange (slab) thickness of the T-beam exceeds the depth to the neutral axis: Yn = 1.18QD/Kr
Q = AsFy/BDFc?= 5.39x60,000/72x21.3x4,000 = 0.0527
D = Rebar depth = 21.3 inches
Kr = 0.85
Yn = 1.18QD/Kr =1.18x0.0527x21.3/0.85 = 1.56" < Ts = 6" (slab thickness) okay.

In the event the slab or flange of the beam is not as thick as the depth to the neutral axis, compression rebar must be added. ACI contains formulas for these situations that will not be discussed here. The Kr and Kf reduction factors are used to account for the fact that concrete is brittle so the compression block is not a true rectangle in stress distribution, but tapers off sharply at the neutral axis.
Only 2 each #8 plus 3 each #10 bars are needed for moment tension. The width of the beam need to be only 2 covers + 2 stirrups + 5 bars + 4 clear spaces: 2x1.5"+2x1/2"+2x1+3x1.25+4x1.33 = 15". Before a reduction in beam width in established, checks on shear need to be made. We now select a beam width of 15 inches.
The next step is to check for shear or diagonal tension. The shear is calculated at a distance D from the support. Vu = 1.4Wd(L/2-D/12) +1.7R = 1.4x840(24?2-21.3"/12) +1.7x34,667 = 70,958 lbs
The shear stress carried by the unreinforced web cannot exceed
Vc?= 2KrFc’^0.5 = 2x0.85x4,000^0.5 = 127 psi
Vc = Vu/BD = 70,958/15x21.3 = 222 psi > 127 psi. Therefore shear rebar is required.
The area of the stirrup bar is Av = Vu’S/KrFyD.
Again S is less than D/2 or 21.3/2 = 10.65 inches, use 10 inches
Vu?is the portion carried by the stirrups = (Vc-Vc?BD = (222-127)15x21.3 = 30,382 lbs
Av = 30,382x10/0.85x60,000x21.3 = 0.28 sqin.
#4 bars at 10 inches center to center with an area of 2x0.2 =0.4 sqin. > 0.28 okay
If we set the stirrup spacing at 7.5 inches, Av = 30,382x7.5/0.85x60,000x21.3 = 0.21 sqin.
#3 bars at 7.5 inches center to center with an area of 2x0.11 =0.22 sqin. > 0.21 okay
The minimum stirrup area is Av = 50B’S/Fy = 50x15x10/60,000 = 0.125 sqin, Spaced at D/2
This means that stirrups should be placed the entire length of the beam because this shear calculates to
V =0.85x4000^0.5x15x21.3 = 17,109 lbs, which is less than one half of an axle load and can occur anywhere in the beam. The spacing of the #3 or #4 stirrups for about the middle one-half of the beam
Could be set at 10" c-c.
The way to more efficiently utilize the stirrups is to bundle the main rebar to reduce the beam web width. Also, the web can be flared near the supports so the concrete absorbs more of the shear forces.
The final check is for bond and anchorage. Again the limits ACI places are 9.5Fc’^0.5/Db and 800 psi for ASTM A 305 deformed bars. For a #10 bar (diameter =1-1/4") this is Ba = 9.5x4,000^0.5/1.25 = 483 psi
The bar perimeter is Pb = PiD = 3.142x1.25 = 3.99 inches
The bond per inch length of bar Bb = PbBa = 3.99x483 = 1,927 lb/in
The bar force Tb = FyAb = 60,000x1.27= 76,200 lbs
The anchorage length is La = Tb/Bb = 76,200/1,927 = 39.54 inches << than L/4 = 72" okay
The deflection calculation is the same as was used for the working stress design, except that it would need to be calculated for the narrower beam web width.

Compare Working vs. Ultimate Design Methods:

The comparison between working stress and ultimate strength design methods shows that the ultimate strength method is easier to calculate and reduced the main rebar nearly 50% (7.8 sqin Vs 5.39 sqin) and reduced the beam 20% in width (18" Vs 15"). The only drawback of the ultimate strength method is the actual stresses are not shown. Occasionally it is desirable to know actual stresses.
One such instance is box culvert construction. The culverts are usually cast in segments using a traveler form. Being able to cycle the form as fast as possible is often critical. The forms are expensive to rent and labor intensive to assemble. Therefore, the form is stripped from the roof soffit as soon as the concrete has gained enough strength to be self-supporting. Box culverts are typically designed for a minimum of 10 feet of soil cover or a soil load on the order of 1,200 psf. The roof concrete is usually less than one foot thick, so it weights less than 150 psf. This means the weight of the roof is on the order of 10% of the design load. If the commonly specified concrete obtains 4,000 psi in 28 days the roof soffit form can be stripped when the concrete has obtained 750 psi. This 750 psi strength can generally be easily obtained with an over night cure. Ultimate strength analysis will not tell us what the actually stresses are at the time of early stripping.
Working stress design will allow a detailed analysis to insure no over stress occurs or that required rebar bond lengths be achieved with the low strength and low modulus of elasticity of the concrete. The only other consideration is creep deflection. The creep difference between stripping within 16 hours and leaving the form in place for a week will add a fraction of an inch to the final roof deflection. Usually this is not a concern because it will not be seen nor will it inhibit the design flow. If creep is a concern, reshoring with posts in the center or suspender beams from the top will limit early creep.

The lesson is don’t just pull a loading from a recommended table and hope for the best. Check to find out what the expected storage and equipment loads really are going to be. An under design can be very dangerous and the repair will cost several times the difference to get it right the first time. It can also be very damaging to an otherwise great career.
Once the fundamentals are understood, the author highly recommends the use of a computer to enhance the design process. The design is an iterative process and many repetitive, but standard computations are necessary. Once the computer model is created, the chance for mathematical error is virtually nonexistent. A favorite program medium is Microsoft Excel. This software contains all the mathematical functions needed to perform the calculations. Sketches can be drawn in Excel to illustrate the design model. The row and column format of Excel allows a readable flow of text and formula calculation. The program can be easily modified to suit different situations. Finally Excel has macro and goal seek capabilities where indeterminate equations can be iterated to a solution within a very few seconds.
Unless esthetics controls the design, keep the concrete configuration as straight and uniform as possible. Not only will this simplify the design process, but will also contribute to the construction economics. A straight uniform structure can be built for about one-half the cost of a curved or overly complex configuration.
Although we have only explored a single design example, the fundamentals carry through all concrete analysis. Ultimate strength design is more efficient than working stress design in both design effort and material utilization. It is also important to use realistic load configurations and reasonable mathematical design models.


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