Page 605.Consider the Figure9.9 (a)of Concept Problem 9.1,whichis a prismatic cantilever beam with a single load P applied at point A. First, solve theConcept Problem9.1as stated.b.Now let’s add a downward uniformly distributed load of w per unit length and solve the problem. c.With the results of b,assume now thatthe load P points upward. Under thiscase, what iswas a functionofP in order tohavezero displacement at point A?What is the slope at Points Aand B? d.Can you think of a different structure that would have the samedeflection profileas inc, with theappropriate boundary conditions. Explain in detail but concisely.
Page 605.Consider the Figure9.9 (a)of Concept Problem 9.1,whichis a prismatic cantilever beam with a single load P applied at point A. First, solve theConcept Problem9.1as stated.b.Now let’s add a downward uniformly distributed load of w per unit length and solve the problem. c.With the results of b,assume now thatthe load P points upward. Under thiscase, what iswas a functionofP in order tohavezero displacement at point A?What is the slope at Points Aand B? d.Can you think of a different structure that would have the samedeflection profileas inc, with theappropriate boundary conditions. Explain in detail but concisely.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Page 605.Consider the Figure9.9 (a)of Concept Problem 9.1,whichis a prismatic cantilever beam with a single load P applied at point A.
First, solve theConcept Problem9.1as stated.b.Now let’s add a downward uniformly distributed load of w per unit length and solve the problem.
c.With the results of b,assume now thatthe load P points upward. Under thiscase, what iswas a functionofP in order tohavezero displacement at point A?What is the slope at Points Aand B?
d.Can you think of a different structure that would have the samedeflection profileas inc, with theappropriate boundary conditions. Explain in detail but concisely.
![A
(a)
B
[x = L,8=0]
[x = Ly=0]
B
Fig. 9.9 (a) Cantilever beam with end
load. (b) Free-body diagram of section AC.
(c) Deformed shape and boundary
conditions.
X
Concept Application 9.1
The cantilever beam AB is of uniform cross section and carries a load P
at its free end A (Fig. 9.9a). Determine the equation of the elastic curve
and the deflection and slope at A.
Using the free-body diagram of the portion AC of the beam
(Fig. 9.9b), where C is located at a distance x from end A,
M =-Px
(1)
Substituting for M into Eq. (9.4) and multiplying both members by the
constant El gives
Integrating in x,
EI
which we carry back into Eq. (2):
dy
dx
dy
EI =-Px²+C₁
dx
Now observe the fixed end B where x = L and 0 = dy/dx = 0 (Fig. 9.9c).
Substituting these values into Eq. (2) and solving for C₁ gives
C₁ =PL²
EI
YA =
Integrating both members of Eq. (3).
=-Px
y=
PL²
3EI
El y =-Px²+ PL²x + C₂
But at B, x = L, y = 0. Substituting into Eq. (4),
0=-PL³ ++PL² + C₂
C₂ = -PL²³
-Px²+ PL²
P
6EI
Carrying the value of C₂ back into Eq. (4), the equation of the elastic
curve is
El y=-Px²+PL²x - PL²
(-x+ 3L²x - 21³)
and
The deflection and slope at A are obtained by letting x = 0) in Eqs. (3)
and (5).
0₁ =
(2)
dx
A
(3)
PL²
2EI
(4)
(5)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4ef3dc5-68e2-4116-aeaf-14e594acc0f6%2F0a2bab9a-458e-4101-9723-a2266acff15a%2Fmuihisk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A
(a)
B
[x = L,8=0]
[x = Ly=0]
B
Fig. 9.9 (a) Cantilever beam with end
load. (b) Free-body diagram of section AC.
(c) Deformed shape and boundary
conditions.
X
Concept Application 9.1
The cantilever beam AB is of uniform cross section and carries a load P
at its free end A (Fig. 9.9a). Determine the equation of the elastic curve
and the deflection and slope at A.
Using the free-body diagram of the portion AC of the beam
(Fig. 9.9b), where C is located at a distance x from end A,
M =-Px
(1)
Substituting for M into Eq. (9.4) and multiplying both members by the
constant El gives
Integrating in x,
EI
which we carry back into Eq. (2):
dy
dx
dy
EI =-Px²+C₁
dx
Now observe the fixed end B where x = L and 0 = dy/dx = 0 (Fig. 9.9c).
Substituting these values into Eq. (2) and solving for C₁ gives
C₁ =PL²
EI
YA =
Integrating both members of Eq. (3).
=-Px
y=
PL²
3EI
El y =-Px²+ PL²x + C₂
But at B, x = L, y = 0. Substituting into Eq. (4),
0=-PL³ ++PL² + C₂
C₂ = -PL²³
-Px²+ PL²
P
6EI
Carrying the value of C₂ back into Eq. (4), the equation of the elastic
curve is
El y=-Px²+PL²x - PL²
(-x+ 3L²x - 21³)
and
The deflection and slope at A are obtained by letting x = 0) in Eqs. (3)
and (5).
0₁ =
(2)
dx
A
(3)
PL²
2EI
(4)
(5)
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