Question
Download Solution PDFस्पैन 20 m और उठाव 4 m के तीन-हिंजित वाले परवलयिक आर्क बाएं समर्थन 'A' से 4 m पर 150 kN का एक सकेंद्रित भार वहन करता है। समर्थन 'A' पर क्रमशः ऊर्ध्वाधर प्रतिक्रिया और क्षैतिज थ्रस्ट की गणना करें।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFगणना:
ΣV = 0, VA + VB = 150 kN
ΣH = 0, HA = HB = H
समर्थन B पर, ΣMB = 0
VA × 20 - 150 × 16 = 0
VA = 120 kN
VB = 150 - 120 = 30 kN
केंद्रीय हिंजित पर, ΣMC = 0
VA × 10 - H × 4 - 150 × 6 = 0
H × 4 = 120 × 10 - 150 × 6
H = 75 kN
तो A पर ऊर्ध्वाधर प्रतिक्रिया और क्षैतिज थ्रस्ट क्रमशः 120 kN और 75kN हैं।
Last updated on May 28, 2025
-> SSC JE notification 2025 for Civil Engineering will be released on June 30.
-> Candidates can fill the SSC JE CE application from June 30 to July 21.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.
-> The Staff Selection Commission conducts the SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.