http://dx.doi.org/10.11594/nstp.2022.2725
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Conference Paper
Single Reinforced Concrete Design of Simple Beam with Simple Formulation Method
Nia Dwi Puspitasari* , Achmad Dzulfiqar Alfiansyah
Civil Engineering Department, Universitas Pembangunan Nasional "Veteran" Jawa Timur, Surabaya 60294, Indonesia
*Corresponding author: ABSTRACT
A quick and practical method of designing a single-layer reinforced concrete beam will be analyzed. This article provides a new simple formulation to design simple beam- reinforced concrete. The dimension and reinforcement ratio can be generated by using the formulation. The variables that must be known at using the formulation are the concrete strength (f`c), the ultimate uniform load (qu), the yield strength of steel (fy), and the span length of the simple beam (L). There are two main parameters obtained from the formulation including the cross-sectional dimensions (W and H) and reinforcement ratios (ρ). Moreover, the required diameter and amount of steel reinforcement can be converted from the result. The beam dimensions and reinforcement ratio results obtained from this design method are verified by mathematical calculations based on the SNI 03-2847-2019. Designing a single-layer reinforced concrete beam using the simple formulation method provides high precision.
Keywords: Simple beam, quick design, simple formulation E-mail:
Introduction
Beams are structural members the main function is to resist bending moments and shear forces. Reinforced concrete is a composite material in consist of concrete and steel reinforcement.
Concrete has a stronger ability to restrain compression than tension (Galambos, 2016). Otherwise, the steel reinforcement has a stronger ability to restrain tension than compression. Steel material tends to bend when receiving compressive forces. A combination of concrete, that have relatively low tensile strength, with steel reinforcement which consists of higher tensile strength, can generate ductility of reinforced concrete supplemented by incorporating ductile reinforcements (Park & Ruitong, 1988). A combination of steel and concrete can provide a structural member property that can resist both compressive forces and tension or bending moment (Millais, 2005).
The reinforcement material is usually installed in areas where stress and bending moments occur (Beskopylny, 2017). There are two types of reinforcement in reinforced concrete beams, there are single layer and double layer reinforced concrete beams. For single-layer reinforced concrete beams, steel reinforcement is usually used only to reduce the possibility of cracking at the area where the positive moment occurs (Deluce & Vecchio, 2013).
In this task, we will analyze a single-layer reinforced concrete of a simple beam to obtain a quick and practical method for designing the beam (Setiawan, 2010). The procedures and requirements that must be followed when designing a reinforced concrete structure are regulated by SNI 03-2847-2019. Design procedures and assumptions for reinforced concrete structures are arranged in this code. This article provides a new simple formulation to design simple beam- reinforced concrete. The dimension and reinforcement ratio can be generated by using the formulation. The variables that must be known at using the formulation are the concrete strength (f`c), the ultimate uniform load (qu), the yield strength of steel (fy), and the span length of the simple beam (L). The concrete strength of the beam is set up between 25 to 50 MPa. The span
length of the simple beam is set up between 4 to 10 m. And yield strength of steel (fy) is determined at 400 MPa in all analytical calculations in this article. There are two main parameters obtained from the formulation including the cross-sectional dimensions (W and H) and reinforcement ratios (ρ). Moreover, the required diameter and amount of steel reinforcement can be converted from the result.
A simple formulation methods are used to process this data. And validated using the results of numerical calculations. To prove the accuracy of the calculation results, Table 1 shows additional beam data with a simple formulation analysis of several variations of ultimate uniform load (qu), concrete strength (f`c), and span length of the simple beam (L). The beam dimensions and reinforcement ratio results obtained from simple formulation method are validated by numerical calculations based on the SNI 03-2847-2019. From the validation, results can be concluded that simple formulation design method obtained a high precision (Puspitasari &
Aryaseta, 2022).
Material and Methods
There are two main formulations generated from the analysis that can be used as a reference in designing a single layer of the reinforced concrete beam. The first formulation is for determining the dimensions of the beam cross-section and the second formulation is for determining the reinforcement ratio.
Determine beam dimension
The steps to create a formulation as a reference in designing the dimension of a single-layer reinforced concrete beam are as follows in figure 1:
Figure 1. Flowchart to generate the formulation for beam dimensions
The first step in generating formulation is to determine the cross-sectional dimensions of the reinforced concrete beam. In this article, beam width variation (B) is determined between 150 to 800 mm at each deviation of 50 mm. The span length of the simple beam is set up between 4 to 10 m. And yield strength of steel (fy) is determined at 400 MPa in all analytical calculations. The differences in beam length will affect the dimensions of reinforced concrete beams. Minimum beam height requirements have been established to meet deflection requirements based on SNI 03-2847-2019. The minimum beam height requirements for simple beams can be estimated by formulation as follows (Indonesia, 2002):
16
H = L (1)
Furthermore, the expected design criteria for beams are stated to be under reinforced conditions (McCormac & Brown, 2015). In this state, when the concrete reaches its compression yield strength εcu, the tensile reinforcement collapses (fs = fy). This condition can be achieved if:
b (2) StartDetermine vari- ous beam speci- fication (B, f’c,
L)
Determine 0.75ρb
Make a comparison between qu and the
specification (B, f’c, L)
Make a formula- tion to define the
beam dimension Finish
While ρ is the ratio of the installed reinforcement, and ρb is the reinforcement ratio when the balance-reinforced condition is reached (Indonesia, 2002). To avoid brittle fracture, the reinforcement ratio is limited to the following condition (Committee, 2008):
0.75 b
(3)The maximum ratio of the installed reinforcement (ρ) can be obtained by substituting equation (3) into equation (2):
0,85 1 ' 600 0, 75
600
c
y y
f
f f
=
+ (4)
The β1 value ranges between 0.65 to 0.85, depending on the concrete strength. General requirements for the moment capacity of the beam (ϕMn) are regulated by SNI 03-2847-2019 based on the following provisions (Indonesia, 2002):
u n
M
M (5)Next, the moment capacity of the beam (ϕMn) is compared with the ultimate moment value (Mu) for a simple beam loaded with an ultimate uniform load (qu) (Chakrabarty, 1992). For the simple beam, the comparison can be developed as follows:
2
8 2
u
s y
q L a
A f d
=
− (6)The economic ratio between d and B is based on the following equation. With a greater value of d/B, the properties of the beam become stiffer and the load-bearing ability becomes greater (Imran & Hendrik, 2010). The value of d is obtained by reducing the height of the beam with con- crete cover and the diameter of the transverse reinforcement:
1,5 d 2, 0
B (7) The value of a can be obtained by the following formula:
0.85 ' s y
c
a A f
= f B (8)
And the cross-sectional area of the rebar is by the following formula:
As =
Bd (9)Then generate the maximum value of the ultimate uniform load (qu) that can be endured by the beam, based on the reinforcement ratio obtained from equation (6).
2
8. . . . 2 2 0,85. ' .
y y
u
c
B d f B d f
q d
L f B
= −
(10)
Thus the comparison between beam width (B), beam span (L), and ultimate uniform load (qu) from various concrete strengths (f’c) can be gathered in figure 2 to figure 4.
Figure 2. The relationship between beam width (B), beam span (L), and ultimate uniform load (qu) for con- crete strength 25 MPa and 30 MPa
Figure 3. The relationship between beam width (B), beam span (L), and ultimate uniform load (qu) for con- crete strength 35 MPa and 40 MPa
0 100 200 300 400 500 600 700 800
0 1000 2000 3000 4000 5000
Beam Width (B) (mm)
qU(kN/m)
f'c 25 MPa Span Trendline equation L = 10 m : y = 111.83x0.322 L = 9 m : y = 103.1x0.3235 L = 8 m : y = 94.336x0.3248 L = 7 m : y = 85.476x0.3261 L = 6 m : y = 76.047x0.328 L = 5 m : y = 66.777x0.3291 L = 4 m : y = 57.112x0.3301
0 100 200 300 400 500 600 700 800
0 1000 2000 3000 4000 5000
Beam Width (B) (mm)
qU(kN/m)
f'c 30 MPa Span Trendline equation L = 10 m : y = 103.81x0.324 L = 9 m : y = 95.871x0.3252 L = 8 m : y = 87.862x0.3263 L = 7 m : y = 79.729x0.3273 L = 6 m : y = 71.088x0.329 L = 5 m : y = 62.504x0.3298 L = 4 m : y = 53.519x0.3306
0 100 200 300 400 500 600 700 800
0 1000 2000 3000 4000 5000
Beam Width (B) (mm)
qU(kN/m)
f'c 35 MPa Span Trendline equation L = 10 m : y = 99.075x0.3251 L = 9 m : y = 91.591x0.3261 L = 8 m : y = 84.015x0.3271 L = 7 m : y = 76.302x0.328 L = 6 m : y = 68.115x0.3294 L = 5 m : y = 59.933x0.3302 L = 4 m : y = 51.351x0.3309
0 100 200 300 400 500 600 700 800
0 1000 2000 3000 4000 5000
Beam Width (B) (mm)
qU(kN/m)
f'c 40 MPa Span Trendline equation L = 10 m : y = 95.494x0.3258 L = 9 m : y = 88.342x0.3268 L = 8 m : y = 81.087x0.3277 L = 7 m : y = 73.687x0.3285 L = 6 m : y = 65.839x0.3298 L = 5 m : y = 57.96x0.3305 L = 4 m : y = 49.684x0.3311
Figure 4. The relationship between beam width (B), beam span (L), and ultimate uniform load (qu) for concrete strength 45 MPa and 50 Mpa
Determine the reinforcement ratio
The next step is creating a simple formulation as a tool for determining the cross-sectional area of the rebar for simple beam with a single layer of reinforcement.
Figure 5. Flowchart to generate a formulation for the sectional area of reinforcement The following equation is the formula for finding the ultimate moment in a simple beam loaded by ultimate uniform load (qu).
2
8
u u
M =q L (11) s u
y
A M
f jd
= (12) Results and Discussion
Simple formulation method is an alternative way to find the parameters easily and quickly.
The steps for obtaining two types of formulas have been explained based on the previous chapter.
The first formula is used to determine the dimensions of the beam quickly and precisely.
Furthermore, based on each trendline equation on the graph in figure 2 to figure 4, a formulation to predict the beam width can be obtained. The determining variable are ultimate uniform load (qu), concrete strength (f’c) , and span of the beam (L). And the formulation to find the beam width (B) as shown as follows:
(
1,17 0, 0068 c') (21 9,1 )
u0,35(
L 0,035)
B f L q
−
= − + (13)
And the second formula is used to find the required reinforcement area (As). By substituting equation 11 into equation 13, the new equation to find the value of the required section area of the reinforcement (As) will be obtained as follows:
0 100 200 300 400 500 600 700 800
0 1000 2000 3000 4000 5000
Beam Width (B) (mm)
qU(kN/m)
f'c 45 MPa Span Trendline equation L = 10 m : y = 92.746x0.3264 L = 9 m : y = 85.844x0.3273 L = 8 m : y = 78.832x0.3281 L = 7 m : y = 71.67x0.3289 L = 6 m : y = 64.078x0.3301 L = 5 m : y = 56.433x0.3307 L = 4 m : y = 48.391x0.3313
0 100 200 300 400 500 600 700 800
0 1000 2000 3000 4000 5000
Beam Width (B) (mm)
qU(kN/m)
f'c 50 MPa Span Trendline equation L = 10 m : y = 90.64x0.3268 L = 9 m : y = 83.927x0.3276 L = 8 m : y = 77.099x0.3284 L = 7 m : y = 70.118x0.3291 L = 6 m : y = 62.721x0.3303 L = 5 m : y = 55.253x0.3309 L = 4 m : y = 47.392x0.3314
Start
Determine Mu
for Simple Beam
Make a formula-
tion to define As Finish
2
6, 2
u s
y
A q L
= f d (14)
Formulas in equation (13) and equation (14) are essentially related and cannot be used interchangeably. In the use of equations (1) to (14), the units of all variables used are in mm, MPa, kN/m or N/mm.
Example of application
A beam is designed to carry an ultimate uniform load (qu) of 87 kN/m with a beam span of 8 m. Determine the required beam dimensions if the concrete strength used is 45 MPa and the yield strength of the steel is 400 MPa.
Step 1: use the equation (13) to determine the required width of the beam.
(
1,17 0, 0068 45 21 9,1 8)( )
870,35 8(
0,035)
346.65 350B mm mm
−
= − + =
Step 2: Refer to equation (7) to determine the beam height (H). For example, if being used d/b = 1.5, then the value of d is 525 mm. Considering the thickness of the concrete cover and the diameter of the transverse reinforcement, an H value of 590 mm can be used.
Step 3: The last step is to determine the value of the section area of the reinforcement. Based on equation (14).
2
87 8000 2
6, 2 400 525 4277
As = = mm
Validation
Validation is performed by analyzing the moment capacity of the beam and comparing it with the ultimate moment of failure due to ultimate uniform load (qu). The following are commonly performed steps when analyzing a single layer of reinforced concrete beams.
. 4277 400
127, 79 0,85 ' . 0,85 45 350
s y
c
a A f mm
f B
= = =
127, 79
. . 0,9 4277 400 525 709971936 . 710 .
2 2
n s y
M A f d a N mm kN m
=
− = − = =
2 2
87 8 696 .
8 8
u u
M = q L = = kN m
n u
M M
→ 710kN m. 696kN m.The validation result accomplished the general provisions for the moment capacity of the beam regulated by SNI 03-2847-2019,
Applications for various beam specification
The application examples for calculating and validating the described results apply to different variations of ultimate uniform loads (qu), beam spans (L), and concrete strength (f’c). This is performed to determine the level of accuracy that can be achieved with simple formulation design techniques. Computational applications are summarized in Table 1. Computational analysis using the ratio d/b = 1.5.
Table 1. Calculation results of various load fluctuations, beam spans, and concrete quality qu
(kN/m) f’c
(MPa) L (m)
B (mm) equation (13)
As
(mm2) equation (14)
Result Validation
Control ϕMn>Mu
ϕMn
(kN/m)
Mu
(kN/m)
40 30 5 218 1233 126 125 Accomplished
90 35 6 310 2809 408 405 Accomplished
140 40 7 383 4814 867 857 Accomplished
190 45 8 447 7313 1540 1520 Accomplished
240 50 9 505 10348 2462 2430 Accomplished
40 30 7 274 1923 246 245 Accomplished
90 35 8 379 4086 730 720 Accomplished
140 40 9 459 6641 1442 1417 Accomplished
190 45 10 527 9691 2422 2375 Accomplished
Table 1 shows that the calculation results have accomplished the requirements for moment capacity control (ϕMn> Mu). From these results, the simple formulation method allows it to be used in the design of single reinforcement beams.
Conclusion
Single-layer reinforcement of a simple beam has a design constraint. The value of 0.75ρb limit is used as the maximum reinforcement ratio applied to the beam. Therefore, an under-reinforce type of beam with high ductility will be produced. The formula in equation (13) can be used by adjusting the beam length (L), the concrete strength (f’c), and the ultimate uniform load (qu) applied to the beam. The results obtained is the beam width (B) and can be converted to beam height (H) using the d/b ratio in the range of 1.5 to 2.0. Furthermore, the formula in equation (14) can be used by adjusting the beam length (L), the yield strength of the steel (fy), and the ultimate uniform load (qu) applied to the beam. The results obtained are the section area of the reinforcement (As) that can be converted into a certain diameter and the number of reinforcement applied to the beam. Based on the results shown in Table 1, it can be concluded that the simple formulation design method has a high level of accuracy and can be used as an approximation method in the design of single-layer reinforcement of a simple beam.
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