Question
Download Solution PDFFind the centroid of laminae shown in figure

Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
- The centroid is the geometric center of a plane figure or lamina. It is the point at which the entire area of the lamina can be assumed to be concentrated for analysis purposes.
- In uniform materials, the centroid corresponds to the center of mass or center of gravity, assuming uniform density and thickness.
Calculations:
Taking reference as the bottom part of the flange.\(A_1=(50\times10)= 500 mm^2\)
\(y_1=20+\frac{50}{2}=45mm\)
\(A_2=(100\times20)= 2000 mm^2\)
\(y_2=\frac{20}{2}=10mm\)
\(\bar y=\frac{A_1y_1+A_2y_2}{A_1+A_2}\)
\(\bar y=\frac{500\times45 + 2000\times45}{2000+500}=\frac{42500}{2500}=17mm\)
The section is symmetric about the vertical axis passing through the center of the web.
Therefore, the centroid lies at the centerline of the web horizontally.
\(\bar x =\frac{100}{2}=50mm\)
The centroid is (50 mm, 17 mm).
Last updated on Jul 1, 2025
-> JKSSB Junior Engineer recruitment exam date 2025 for Civil and Electrical Engineering has been rescheduled on its official website.
-> JKSSB JE exam will be conducted on 31st August (Civil), and on 24th August 2025 (Electrical).
-> JKSSB JE application form correction facility has been started. Candidates can make corrections in the JKSSB recruitment 2025 form from June 23 to 27.
-> JKSSB JE recruitment 2025 notification has been released for Civil Engineering.
-> A total of 508 vacancies has been announced for JKSSB JE Civil Engineering recruitment 2025.
-> JKSSB JE Online Application form will be activated from 18th May 2025 to 16th June 2025
-> Candidates who are preparing for the exam can access the JKSSB JE syllabus PDF from official website of JKSSB.
-> The candidates can check the JKSSB JE Previous Year Papers to understand the difficulty level of the exam.
-> Candidates also attempt the JKSSB JE Mock Test which gives you an experience of the actual exam.