5.2 RECTANGULAR BEAMS
5.2.3 Beam Design
Reinforcement Ratio
The amount of steel reinforcement is limited. Too much reinforcement, or over-reinforcing will not allow the steel to yield before the concrete crushes and there is a sudden failure. A beam with the proper amount of steel to allow it to yield at failure is said to be under reinforced. [5.2].
Design of Singly Reinforced Beams
Design for Flexure:
Step-1: Compute dead and five loads and multiply them with their respective design factors combinations.
Figure 5.2-8: Load Factor Combinations
Step-2: Compute Bending Moment (Factored) from factored loads using appropriate formula.
Table 5.2-1: Bending Moments
83
Step-3: Note out Maximum positive and negative bending Moments, and Assume suitable dimension b and d for Beam. Assume d as 1 in. for every foot of span length.
Step-4: Compute Reinforcement Ratios.
a). Minimum Reinforcement Ratio:
200 (5.1)
b). Balanced section Reinforcement ratio:
0.85 . 87
87 (5.2)
Here,
∈
∈ ∈
∈ strain in concrete = 0.003
Figure 5.2-9: Stress in Beam
c). Maximum Reinforcement ratio:
∈ ⁄
∈ ∈ (5.3)
∈ = strain in concrete = 0.003
∈ = strain in Steel = 0.005
d). Required Reinforcement Ratio:
0.85 1 1 4
1.7∅ (5.4)
e). Actual provided reinforcement Ratio:
emin < e < emax (5.5)
Step-5: Compute Area of steel, Bar dia, and bar Nos.
As = bd (5.6)
Table 5.2-2: Bar Diameter Chart
85
Bar Nos:
(5.7)
Step-6: Check moment capacity of designed section.
a). Revise provided area of steel.
As = Ab x No.s
b). Compute depth of stress block ‘a’
0.85 (5.8)
then section is safe for bending
Design for Shear:
Step-7: Compute maximum sheer force from factored loads, we get factored sheer force Vu.
Step-8: Compute Vu at d from support. Fig: 5.2.10.
Step-9: Compute shear strength provided by concrete.
∅ V ∅ (5.11)
Here, = 0.75 for compression and shear members.
Figure 5.2-10: Critical Shear
Step-10: From figure 5.2-10, the point at which web reinforcement theoretically is no longer required is [5.3], say x.
2
∅ (5.12)
From support face. However, according to the ACI code, 11.5.5 at least a minimum amount of web reinforcement is required whenever the shear force exceeds is Vc/2. As shown in figure 5.2-10, this applies to a distance [5.3] say x’.
2
∅ /2 (5.13)
From the support face. At least the minimum amount web reinforcement within the distance of “x’ ” from supports, and with “x” the web steel must provide for the shear force corresponding to the shaded area [5.3].
Step-11: Web reinforcement a). Spacing
(5.14)
A= cross-sectional areas of standard stirrup; twice the area of bar.
87
b). Minimum Web Reinforcement
0.75 .
50 .
(5.16)
It is undesirable to space vertical stirrups closer their about 4 in. the size of the stirrups when vertical stirrups are required over a comparatively short distance, it is a good practice to space them uniformly over the entire distance, the spacing being calculated for the point of greater shear (minimum spacing).
Where web reinforcement is required, the code requires it to be spaced so that every 450 line, representing a potential diagonal crack and extending from the mid-depth d/2 of the crossed by atleast one line of reinforcement, the code specifies a maximum spacing of 24 in. when Vs exceeds 4. bd, these spacing’s are shattered.
[5.3]. For usual case of stirrups, with Vs < 4. bd the maximum spacing of stirrups is the smallest of.
0.75 50 (5.16)
2 (5.17)
Smax = 24 in (5.18)
Design of Doubly Reinforced Beams
If a beam cross section is limited because of architectural or other considerations, it may happen that the concrete cannot develop the compression force required to resist the given bending moment. In this case, reinforcement is added in the compression zone, resulting in a so-called doubly reinforced beam, i.e., one with compression as well as tension reinforcement (see Fig. 5.2-11). The use of compression reinforcement has decreased markedly with the use of strength design methods, which account for the
full strength potential of the concrete on the compressive side of the neutral axis.
However, there are situations in which compressive reinforcement is used for reasons other than strength. It has been found that the inclusion of some compression steel will reduce the long term deflections of members In addition, in some cases, bars will be placed in the compression zone for minimum-moment loading or as stirrup-support bars continuous throughout the beam span.
Figure 5.2-11: Doubly reinforced rectangular beam.
If, in a doubly reinforced beam, the tensile reinforcement ratio is less than or equal to b, the strength of the beam may be approximated within acceptable limits by disregarding the compression bars. The strength of such a beam will be controlled by tensile yielding, and the lever arm of the resisting moment will ordinarily be but little affected by the presence of the compression bars. If the tensile reinforcement ratio is larger than b, a somewhat more elaborate analysis is required. In Fig. 5.2-11a, a rectangular beam cross section is shown with compression steel As’ placed a distance d' from the compression face and with tensile steel As, at effective depth d. It is assumed initially that both As’ and As are stressed to fy at failure. The total resisting moment can be thought of as the sum of two parts. The first part, Mn1 is provided by the couple consisting of the force in the compression steel As’ and the force in an equal area of
89
tension steel as shown in Fig. 5.2-11d. The second part, Mn2 is the contribution of the remaining tension steel As – As’ acting with the compression concrete. [5.3].
Compression steel will provide compressive forced in addition to the compressive force in the concreting area [5.1].
Assuming one row of tension bars:
The procedure for designing a rectangular section with compression steel when Mu, fc’, b, d and d are given can be summarized as follows:
Calculate Asmax = maxbd. (maximum steel over singly reinforced section).
Step 2: Calculate Rumax using max, = 0.9.
∅ 1
1.7 (5.21)
Step 3: Calculate the moment strength of the section, Mu1, as singly reinforced, using
max and Rumax bd2.
(5.22) If Mu1 < Mu (the applied moment), the compression steel is needed. Go the next step.
If Mu1 > Mu, the compression steel is not needed. Design the section as singly reinforced section as explained above.
Step 4: The moment to be resisted by compression steel:
Mu2 = Mu – Mu1 (5.23)
Step 5: Calculate As2.
M ∅ A (5.24)
Here,
d = effective depth for tension steel to top fibre.
d’ = effective depth for compression steel to top fibre.
Then, total steel will be:
As = As1 + As2 (5.25)
Step 6: Calculate the stress in the compression steel follows:
a. Calculate
compression steel yields and .
c. Calculate from Mu2 = ∅ A f (d - d’). if f = fy then A = As2 . if f < fy, then A > As2, and A = As2 (fy/f ).
Choose bars for As and As’ to fit within the section width, b. in most cases As bars will be placed in two rows, whereas bars are placed in one row.
91
Step 7: Calculate h = d + 2.5 in. for one row of tension bars and h = d+ 3.5 in. for two rows of tension steel. Round h to the next higher inch. Now check that [-p’( / )] <
max using the new d, or check that Asmax = bd [-p’ ( / )] > As (used).
(5.29)
′ ′
(5.30)
The check may not be needed if max is used in the basic section.
Assuming two rows of Tension Bars:
In the case of two rows of bars, it can be assumed that d = h – 3.5 in. and dt = h – 2.5 in. = d+1.0 in.
Two approaches may be used to design the section:
One approach is to assume a strain at the level of the centroid of the tension steel equal to 0.005 or s = 0.005 ( at d level). In this case, the strain in the lower row of bars is greater than 0.005 t = (d= – c/c) 0.003 > 0.005, which still meets the ACI code limitation. For this case, follow the above steps.
A second approach is to assume a strain t = 0.005 at the level of the lower row of bars, dt in this case, the stain at the level of the centroid of bars is less than 0.005: s
= [(dt –c)/c] 0.003 < 0.005, which is still acceptable. The solution can be summarized as follows. [5.1].
a. Calculate dt = h – 2.5 and then form the strain diagram and calculate c, the depth of the neutral axis.
0.003
0.003 (5.31)
For t = 0.005,
3
8. (5.32)
b. Calculate the compression force in the concrete.
C1 = 0.85 ab = T1 = As1 fy (5.33)
Determine As1. Calculate Mu1 = Asi. fy (d - a/2). 1 = As1/bd, = 0.9.
c. Calculate Mu2 = Mu – Mu1; assume d’ = 2.5 in.
d. Calculate As2: Mu2 = As2, fy (d –d’) = fy, =0.9 Total As = As1 + As2.
e. Check if compression steel yields similar to step 6 above.