Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
316 views
in General by (114k points)
closed by
In the case of a beam simply supported at both ends, if the same load instead of being concentrated at centre is distributed uniformly throughout the length, then deflection at centre will get reduced by
1. 1/2 times 
2. 1/4 times
3. 5/8 times
4. 3/8 times

1 Answer

0 votes
by (115k points)
selected by
 
Best answer
Correct Answer - Option 4 : 3/8 times

Concept:

Deflection at centre of a simply supported beam due to point load (W):

\(δ_1=\frac{WL^3}{48EI}\)

Deflection at centre of a simply supported beam due to uniformly distribute load (W = wL):

\(δ_2=\frac{5wL^4}{384EI}=\frac{5WL^3}{384EI}\;\;\;(\because w=\frac{W}{L})\)

Calculation:

Given:

\(\frac{\delta_2}{\delta_1}=\frac{5WL^3}{384EI}\times \frac{48EI}{WL^3}=\frac{5}{8}\)

Hence,

The deflection will be reduced by

\(={\delta _1} - {\delta _2} = \left( {1 - \frac{5}{8}} \right){\delta _1} = \frac{3}{8}{\delta _1}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...