Rc plastic-centroid study material PDF

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Reinforced Concrete Design Module 8 Subject: Reinforced Concrete Design for Architecture Students

1. Title of the Module Plastic Centroid

2. Introduction The plastic centroid of a concrete column is the point through which the resultant axial force on a cross section must act to produce uniform strain at failure. It is obtained by determining the location of the resultant force produced by the concrete and steel, assuming that both are stressed in compression to 0.85f’c and (fy – 0.85f’c) respectively. For symmetrical cross sections, the plastic centroid coincides with the centroid of the section. If the applied axial load does not pass through the plastic centroid, it will create bending moment which will be equal to the product of the axial load multiplied by the distance between the force and the plastic centroid.

3. Learning Outcomes At the end of the course, the students should be able to: a. compute with speed and accuracy the plastic centroid of columns be it approximate or exact method.

4. Learning Content Exact Method: Analyze the compressive force of the steel considering the area displaced by compression bars. Cs = Asfy – 0.85f’c As Cs = As(fy – 0.85f’c ) ⍴𝑔 Problem 1. A short rectangular reinforce concrete column shown is to be part of a long –span rigid frame and will be subjected to high bending moment combined with relatively low axial loads, causing bending about the strong axis. Steel bars are placed unsymmetrically as indicated in the figure. Consider the area displaced by compression bars. f’c = 27MPa

fy = 414.7 MPa

1. Determine the capacity of the column section considering the forces in steel adjusted for concrete stress. 2. Determine the plastic centroid of the section measured from the center of 36 mm ꬾ bars. 3. Determine the eccentricity of the column load assuming the column load is applied at a distance of 300mm from the center of the section towards the right side.

Solution: 1. Capacity of the column section. C1 = As1(fy - 0.85fc’) π C1 = 4 (44)2(3)[414.7 – 0.85(27.6)] C1 = 1784678 N C3 = As3(fy - 0.85fc’) π C3 = 4 (36)2(2)[414.7 – 0.85(27.6)] C3 = 796468 N

C2 = 0.85fc’ (550) (350) C2 = 0.85(27.6) (550) (350) C2 = 4516050 N R = C1 + C2 + C3 R = 1784678 + 4516050 + 796468 R = 7097196 N 2. Plastic centroid:

Rx = C1(475) + C2(275) + C3 (75) 7097196(x) = 1784678(275) + 4516050(275) + 796468(75) X = 302.85 mm b = 302.85 – 75 b = 227.85 mm

3. Eccentricity: c = 302.85 – 275 c =27.85 e = 300 + 27.85 e = 327.85 mm

Problem 2 A tied column 350 mm x 600 mm is reinforced with 2 – 28 mm Ø at the left side and 2 – 36mm f’c = 27.6 MPa, fy = 415MPa.

1. Compute the total axial capacity of the column, considering area of displaced concrete by compression bars. 2. Compute the plastic centroid measured from the center of column 3. Compute the spacing of the 10 mm Ø ties. SOLUTION: 1. Total axial capacity of the column, considering area of displaced concrete by the compression bars. FOR 2 – 28 mm Ø bars C1 = AS1 (fy – 0.85f’c) 𝐶1 =

𝜋 (28)2 (2)(415 − 0.85(27.6)) 4

C1 = 482183.2 N FOR 2 – 28 mm Ø BARS C2 = AS2 (fy – 0.85f’c) 𝐶2 =

𝜋 (36)2 (2)(415 − 0.85(27.6)) 4

C2 = 797078.4 N FOR CONCRETE SECTION C3 = 0.85f’cAc C3 = 0.85(27.6)(350)(600) C3 = 4926600 N CAPACITY OF COLUMN P = C1 + C2 + C3 P = 483183.2 + 797078.4 + 4926600 P = 6206861.6 N = 6206.862 kN 2. Plastic centroid measured from the center of the column.

Px = C1 x1 + C2 x2 + C3 x3 6206861.6x = 482183.2(75) + 797078.4(525) + 4926600(300) x = 311.37 Plastic Centroid, e = 311.37 – 300 = 11.37 mm from center of column 3. Spacing of the 10 mm Ø ties. 1) S = 48 tie diam. S = 48 (10) S = 480 mm o.c. 2) S = 16 bar diam. S = 16(34) S = 544 mm o.c. 3) Least Dimension = 350 mm Use S = 350 mm on centers

5. Teaching and Learning Activities Solve the following problem for mastery of the procedures and specifications:

Problem 1. A tied column 400 mm x 700 mm is reinforced with 2 – 32 mm Ø at the left side and 2 – 32mm f’c = 27.6 MPa, fy = 415MPa.

a. Compute the total axial capacity of the column, considering area of displaced concrete by compression bars. b. Compute the plastic centroid measured from the center of column c. Compute the spacing of the 10 mm Ø ties. Problem 2. A short rectangular reinforce concrete column shown is to be part of a long – span rigid frame and will be subjected to high bending moment combined with relatively low axial loads, causing bending about the strong axis. Steel bars are placed unsymmetrically as indicated in the figure. Do not consider the area displaced by compression bars. f’c = 27MPa

fy = 414.7 MPa

a. Determine the capacity of the column section considering the forces in steel adjusted for concrete stress. b. Determine the plastic centroid of the section measured from the center of 36 mm ꬾ bars. c. Determine the eccentricity of the column load assuming the column load is applied at a distance of 300mm from the center of the section towards the right side.

6. Recommended learning materials and resources for supplementary reading. Reinforced Concrete Design by Gillesania, Plastic Centroid Design of Concrete Structures by Nilson et. al. 7. Flexible Teaching Learning Modality (FTLM) adopted Remote

Asynchronous (modules, exercises, problem sets, etc…) 8. Assessment Task In this part, students are given Self-assessment Questions (SAQs) and asked to consider broader aspects of the different topics taken up. Quizzes have been prepared for this part and can be found in the original module of the author.

9. References Besavilla, V., Reinforced Concrete Design, 2016 Esplana, Dindo Civil Engineering Review Manual, 2015 Gillesania, DIT Reinforced Concrete Design, 3rd Edition, 2015 National Structural Code of the Philippines 2015 Nilson, W. Reinforced Concrete Design, 2010...