9.10 (a) of Concept Problem 9.2,which is asimply supported prismatic beam with a uniformly distributed load w per unit length.
9.10 (a) of Concept Problem 9.2,which is asimply supported prismatic beam with a uniformly distributed load w per unit length.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Page 606.Consider the Figure 9.10 (a) of Concept Problem 9.2,which is asimply supported prismatic beam with a uniformly distributed load w per unit length. Can you think of a different structure that would have the same deflection profile as in c, with the appropriate boundary conditions. Explain in detail but concisely.
![WOX
R₁=TwL
(b)
(a)
D
[x = 0, y=0]
(c)
[x=Ly=0]
(d)
Fig. 9.10 (a) Simply supported
beam with a uniformly distributed
load. (b) Free-body diagram of
segment AD. (c) Boundary conditions.
(d) Point of maximum deflection.
Concept Application 9.2
The simply supported prismatic beam AB carries a uniformly distributed load
w per unit length (Fig. 9.10a). Determine the equation of the elastic curve and
the maximum deflection of the beam.
Draw the free-body diagram of the portion AD of the beam (Fig. 9.10b)
and take moments about D for
M = {wLx - wx²
(1)
Substituting for M into Eq. (9.4) and multiplying both members of this
equation by the constant El gives
Integrating twice in .x,
EI
or
dy
EI ==
dx
dx
El y = -
1
== wx² +
y =
1
=-=-=wXx²* + = 2 ²x² + C₁x + C₂
24
- 7/7wx² + = w²x³² + C₁
1
0=w+w+G₁L
C₁=-wL²
Carrying the values of C, and C₂ back into Eq. (9.4), the elastic curve is
El y=-wx+wLx-wLx
wLx
Observing that y = 0 at both ends of the beam (Fig. 9.10c), let x = 0
and y = 0 in Eq. (4) and obtain C₂ = 0. Then make x = L and y = 0 in the
same equation, so
W
24EI
(-x+ 2Lx¹ - L³x)
W
c =
= 24 E 7 (-16 + 21-²2 - 1²¹²) =
2L
L
8
lylmax =
SwL
384EI
Substituting the value for C, into Eq. (3), we check that the slope of the
beam is zero for x = L/2 and thus that the elastic curve has a minimum at
the midpoint C (Fig. 9.10d). Letting x = L/2 in Eq. (5),
The maximum deflection (the maximum absolute value) is
(2)
5wL
384EI
(3)
(5)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4ef3dc5-68e2-4116-aeaf-14e594acc0f6%2F9bd0fefa-c8b7-43f6-b8bd-b03ffa00992c%2F2mhht4q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:WOX
R₁=TwL
(b)
(a)
D
[x = 0, y=0]
(c)
[x=Ly=0]
(d)
Fig. 9.10 (a) Simply supported
beam with a uniformly distributed
load. (b) Free-body diagram of
segment AD. (c) Boundary conditions.
(d) Point of maximum deflection.
Concept Application 9.2
The simply supported prismatic beam AB carries a uniformly distributed load
w per unit length (Fig. 9.10a). Determine the equation of the elastic curve and
the maximum deflection of the beam.
Draw the free-body diagram of the portion AD of the beam (Fig. 9.10b)
and take moments about D for
M = {wLx - wx²
(1)
Substituting for M into Eq. (9.4) and multiplying both members of this
equation by the constant El gives
Integrating twice in .x,
EI
or
dy
EI ==
dx
dx
El y = -
1
== wx² +
y =
1
=-=-=wXx²* + = 2 ²x² + C₁x + C₂
24
- 7/7wx² + = w²x³² + C₁
1
0=w+w+G₁L
C₁=-wL²
Carrying the values of C, and C₂ back into Eq. (9.4), the elastic curve is
El y=-wx+wLx-wLx
wLx
Observing that y = 0 at both ends of the beam (Fig. 9.10c), let x = 0
and y = 0 in Eq. (4) and obtain C₂ = 0. Then make x = L and y = 0 in the
same equation, so
W
24EI
(-x+ 2Lx¹ - L³x)
W
c =
= 24 E 7 (-16 + 21-²2 - 1²¹²) =
2L
L
8
lylmax =
SwL
384EI
Substituting the value for C, into Eq. (3), we check that the slope of the
beam is zero for x = L/2 and thus that the elastic curve has a minimum at
the midpoint C (Fig. 9.10d). Letting x = L/2 in Eq. (5),
The maximum deflection (the maximum absolute value) is
(2)
5wL
384EI
(3)
(5)
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