1. Determine the maximum deflection d in a simply supported beam of length L carrying a uniformly distributed load of intensity w, applied over its entire length. 2. For the beam loaded as shown in the Figure, compute the moment of area of the M diagrams between the reactions about both the left and the right reaction. (Hint: Draw the moment diagram by parts from right to left). 500 N 1 m 2 m 1 m 400 N/m R1
1. Determine the maximum deflection d in a simply supported beam of length L carrying a uniformly distributed load of intensity w, applied over its entire length. 2. For the beam loaded as shown in the Figure, compute the moment of area of the M diagrams between the reactions about both the left and the right reaction. (Hint: Draw the moment diagram by parts from right to left). 500 N 1 m 2 m 1 m 400 N/m R1
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Transcribed Image Text:Home Work:
1. Determine the maximum deflection d in a simply supported beam of length L
carrying a uniformly distributed load of intensity w, applied over its entire length.
2. For the beam loaded as shown in the Figure, compute the moment of area of the
M diagrams between the reactions about both the left and the right reaction. (Hint:
Draw the moment diagram by parts from right to left).
500 N
2 m
1 m
1 m
400 N/m
R1
R2
Please solve according to
the exporters of a typical
solution.
1
E*I
3. Moment Diagram by Parts
but
ds = pd0
M
de
M
ds
E*I
1
The construction of moment diagram by parts depends on two basic principles:
1) The resultant bending moment at any section caused by any load system is the
= de =
%3D
E*I
ds
algebraic sum of the bending moment at that section caused by each load acting
separately.
but
ds = dx
(Flat curve)
M
=) M2 =
Σ
= OP **
E * I
MR
1
[M * dx
EM,EMR : Sum of the moment caused by all the forces to the left and right
section respectively.
:. 0 =
E * I
2) The moment effect of any single specified loading is always some variation of
the general equation.
dt = x * de
Į dt
x * de
1
Area = b * h
n+1
1
IBIA =
1
[x(Mdx )
X = -
* b
1BIA =
E * I
n + 2
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