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.

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Chapter2: Loads On Structures
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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)
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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