SECTION 9.4
559
Differential Equations of the Deflection Curve
Differential Equations of the Deflection Curve
The beams described in the problems for Section 9.4 have constant
flexural rigidity EI. Also, the origin of coordinates is at the left-hand
end of each beam.
y
M0
Solution 9.4-1 Cantilever beam (couple M0)
SHEAR-FORCE EQUATION (EQ. 9-12 b).
EIv‡ V 0
EIv– C1
1 M M0 EIv– M M0 C1
EIv¿ C1x C2 M0 x C2
B.C.
2 v¿ (0) 0 C2 0
M0 x2
C3
EIv 2
B.C.
B
A
Problem 9.4-1 Derive the equation of the deflection curve for a
cantilever beam AB when a couple M0 acts counterclockwise at the
free end (see figure). Also, determine the deflection B and slope
B at the free end. Use the third-order differential equation of the
deflection curve (the shear-force equation).
x
L
3 v(0) 0 C3 0
M0 x2
v
2 EI
M0 x
v¿ EI
M0 L2
B v(L) (upward)
2 EI
M0 L
uB v¿(L) (counterclockwise)
EI
(These results agree with Case 6, Table G-1.)
B.C.
x
q = q0 sin —
L
Problem 9.4-2 A simple beam AB is subjected to a distributed load
of intensity q q0 sin x/L, where q0 is the maximum intensity of
the load (see figure).
Derive the equation of the deflection curve, and then determine
the deflection max at the midpoint of the beam. Use the fourth-order
differential equation of the deflection curve (the load equation).
y
B
A
L
Solution 9.4-2 Simple beam (sine load)
LOAD EQUATION (EQ. 9-12 c).
EIv–– q q0 sin
EIv‡ q0 ¢
EIv– q0 ¢
B.C.
x
L
L
x
≤ cos
C1
L
L 2
x
≤ sin
C1x C2
L
1 EIv– M
EIv–(0) 0 C2 0
2 EIv–(L) 0 C1 0
L 3
x
EIv¿ q0 ¢ ≤ cos
C3
L
B.C.
EIv q0 ¢
L 4
x
≤ sin
C3x C4
L
B.C.
3 v(0) 0
C4 0
4 v(L) 0 C3 0
q0 L4
x
v 4 sin
L
EI
B.C.
q0L4
L
≤ 4
2
EI
(These results agree with Case 13, Table G-2.)
max v ¢
x
560
CHAPTER 9
Deflections of Beams
Problem 9.4-3 The simple beam AB shown in the figure
has moments 2M0 and M0 acting at the ends.
Derive the equation of the deflection curve, and then
determine the maximum deflection max. Use the third-order
differential equation of the deflection curve (the shear-force
equation).
y
2M0
B
A
M0
x
L
Solution 9.4-3
Simple beam with two couples
3M0
Reaction at support A: RA (downward)
L
Shear force in beam: V RA 3M0
L
3M0
L
1 EIv– M
EIv¿ ∴ C2 M0 L
2
M0 x 2
M0 x
(L 2 Lx x2 ) (L x) 2
2 LEI
2 LEI
M0
v¿ (L x)(L 3x)
2LEI
MAXIMUM DEFLECTION
3M0 x
EIv– C1
L
B.C.
3 v(L) 0
v SHEAR-FORCE EQUATION (EQ. 9-12 b)
EIv‡ V B.C.
EIv–(0) 2M0 C1 2M0
3M0 x2
2M0 x C2
2L
Set v 0 and solve for x:
L
x1 L and x2 3
L
Maximum deflection occurs at x2 .
3
M0 x3
M0 x2 C2 x C3
2L
B.C. 2 v(0) 0 C3 0
max v ¢
EIv 2M0 L2
L
(downward)
≤
3
27 EI
y
Problem 9.4-4 A simple beam with a uniform load is pin supported
at one end and spring supported at the other. The spring has stiffness
k 48EI/L3.
Derive the equation of the deflection curve by starting with
the third-order differential equation (the shear-force equation).
Also, determine the angle of rotation A at support A.
q
B
A
x
48EI
—
k = L3
L
Solution 9.4-4
Beam with a spring support
REACTIONS
y
q
B
A
x
k
qL
RA = —
2
L
DEFLECTIONS AT END B
k
qL
RB = —
2
48EI
L3
B qL4
RB qL
k
2k 96EI
SECTION 9.4
SHEAR-FORCE EQUATION (EQ. 9-12 b)
q
V RA qx (L 2x)
2
q
EIv‡ V (L 2x)
2
q
EIv– (Lx x2 ) C1
2
B.C.
1 EIv– M
2
EIv¿ EIv 561
Differential Equations of the Deflection Curve
C3 0
B.C.
2 v(0) 0
B.C.
3 v(L) B ∴ C2 qL4
96 EI
5qL3
96
qx
(5L3 8Lx2 4x3 )
96EI
q
v¿ (5 L3 24 Lx2 16x3 )
96 EI
5qL3
uA v¿(0) (clockwise)
96EI
v
EIv–(0) 0 C1 0
3
q Lx
x
¢
≤ C2
2 2
3
q Lx3 x4
¢
≤ C2x C3
2 6
12
y
Problem 9.4-5 The distributed load acting on a cantilever beam AB
has an intensity q given by the expression q0 cos x /2L, where q0 is
the maximum intensity of the load (see figure).
Derive the equation of the deflection curve, and then determine
the deflection B at the free end. Use the fourth-order differential
equation of the deflection curve (the load equation).
x
q = q0 cos —
2L
q0
B
A
L
Solution 9.4-5 Cantilever beam (cosine load)
LOAD EQUATION (EQ. 9-12 c)
x
2L
2L
x
EIv‡ q0 ¢ ≤ sin
C1
2L
EIv–– q q0 cos
B.C.
1 EIv‡ V
EIv– q0 ¢
B.C.
EIv–(L) 0
3 v¿(0) 0 C3 0
EIv q0 ¢
B.C.
∴ C1 2q0 L
2L 2
x 2q0 Lx
≤ cos
C2
2L
2 EIv– M
EIv¿ q0 ¢
EIv‡(L) 0
B.C.
∴ C2 2L 4
x q0 Lx3 q0 L2x2
≤ cos
C4
2L
3
4 v(0) 0
v
2L 3
x q0 Lx2 2q0 L2x
≤ sin
C3
2L
16q0L4
4
q0 L
x
¢ 48L3 cos
48L3 33 Lx2 3x3 ≤
4
2L
3 EI
2q0 L4 3
( 24)
34EI
(These results agree with Case 10, Table G-1.)
B v(L) 2q0 L2
∴ C4 x
562
CHAPTER 9
Deflections of Beams
y
Problem 9.4-6 A cantilever beam AB is subjected to a parabolically varying
load of intensity q q0(L2 x 2)/L2, where q0 is the maximum intensity of
the load (see figure).
Derive the equation of the deflection curve, and then determine the
deflection B and angle of rotation B at the free end. Use the fourth-order
differential equation of the deflection curve (the load equation).
L2 x2
q = q0 —
L2
q0
x
B
A
L
Solution 9.4-6 Cantilever beam (parabolic load)
LOAD EQUATION (EQ. 9-12 c)
q0
EIv–– q 2 (L2 x2 )
L
EIv‡ B.C.
1 EIv‡ V
EIv– B.C.
q0 2
x3
¢
L
x
≤ C1
3
L2
4 v(0) 0 C4 0
q0 x2
v
(45L4 40L3x 15L2x2 x4 )
360 L2EI
B.C.
EIv‡(L) 0
∴ C1 2q0L
3
q0 L2x2 x4
2q0L
≤
x C2
2¢
2
12
3
L
2 EIv– M
3 v¿(0) 0 C3 0
q0 Lx3 q0 L2x2
q0 L2x4
x6
≤
C4
EIv 2 ¢
360
9
8
L 24
B.C.
EIv–(L) 0
q0L2
∴ C2 4
q0Lx2 q0L2x
q0 L2x3 x5
≤
C3
EIv¿ 2 ¢
6
60
3
4
L
B v(L) v¿ 19q0 L4
360 EI
q0 x
(15L4 20L3x 10L2x2 x4 )
60L2EI
uB v¿(L) q0L3
15EI
4q0 x
(L x)
q= —
L2
Problem 9.4-7 A beam on simple supports is subjected to a
parabolically distributed load of intensity q 4q0 x(L x)/L2,
where q0 is the maximum intensity of the load (see figure).
Derive the equation of the deflection curve, and then
determine the maximum deflection max. Use the fourthorder differential equation of the deflection curve (the load
equation).
y
B
A
L
Solution 9.4-7 Single beam (parabolic load)
LOAD EQUATION (EQ. 9-12 c)
EIv–– q 4q0 x
4q0
2
2 (L x) 2 (Lx x )
L
L
2q0
(3Lx2 2x3 ) C1
3L2
q0
EIv– 2 (2Lx3 x4 ) C1x C2
3L
EIv‡ B.C.
1 EIv– M
B.C.
2 EIv–(L) 0
EIv¿ EIv–(0) 0 C2 0]
∴ C1 q0 L
3
q0
(5L3x2 5L x4 2x5 ) C3
30L2
B.C.
3 (Symmetry)
EIv v¿ ¢
L
≤0
2
∴ C3 q0L3
30
q0
5L3x3
x6
5
L x5 ≤ C4
2 ¢L x 3
3
30L
4 v(0) 0 C4 0
q0 x
v
(3L5 5L3x2 3Lx4 x5 )
90L2EI
B.C.
max v ¢
61q0L4
L
≤
2
5760 EI
x
SECTION 9.4
563
Differential Equations of the Deflection Curve
Problem 9.4-8 Derive the equation of the deflection curve for a
simple beam AB carrying a triangularly distributed load of maximum
intensity q0 (see figure). Also, determine the maximum deflection max
of the beam. Use the fourth-order differential equation of the deflection
curve (the load equation).
q0
y
A
B
x
L
Solution 9.4-8
Simple beam (triangular load)
LOAD EQUATION (EQ. 9-12 c)
q0 x
q0 x2
EIv–– q EIv‡ C1
L
2L
EIv– B.C.
B.C.
EIv B.C.
MAXIMUM DEFLECTION
q0L
∴ C1 6
Set v 0 and solve for x:
8
x21 L2 ¢ 1 ≤ x1 0.51933L
A 15
q0 x4 q0 L x2
C3
24L
12
max v (x1 ) q0 x5 q0 Lx3
C3x C4
120L
36
3 v(0) 0
q0 L4
EI
(These results agree with Case 11, Table G-2.)
y
Problem 9.4-9 Derive the equations of the deflection curve for an
overhanging beam ABC subjected to a uniform load of intensity q
acting on the overhang (see figure). Also, obtain formulas for the
deflection C and angle of rotation C at the end of the overhang.
Use the fourth-order differential equation of the deflection curve
(the load equation).
EIv–– q 0
EIv‡ C1
EIv– C1 x C2
(0 x L)
(0 x L)
(0 x L)
EIv–(0) 0 C2 0
3L
EIv–– q
¢L x ≤
2
3L
EIv‡ qx C3
¢L x ≤
2
3qL
3L
B.C. 2 EIv‡ V EIv‡ ¢
≤0
∴ C3 2
2
2
qx
3qLx
3L
EIv– C4 ¢ L x ≤
2
2
2
B.C.
1 EIv– M
q0 L4 5 2
8 12
¢ ≤
225EI 3 3 A 15
0.006522
C4 0
Solution 9.4-9 Beam with an overhang
LOAD EQUATION (EQ. 9-12 c)
7q0 L3
360
q0 x
(7L4 10L2x2 3x4 )
360 LEI
q0
v¿ (7L4 30L2x2 15x4 )
360 LEI
EIv–(0) 0 C2 0
2 EIv–(L) 0
∴ C3 4 v(L) 0
v
q0 x3
C1x C2
6L
1 EIv– M
EIv¿ B.C.
q
B
A
C
L
—
2
L
3L
≤0
2
B.C.
3 EIv– M
B.C.
4 EI(v–) Left EI(v–) Right
2
C1L EIv– ¢
2
∴ C4 at x L
2
qL
3qL
9qL
2
2
8
9qL2
8
∴ C1 EIv¿ qLx2
C5 (0 x L)
16
EIv¿ qx3 3qLx2 9qL2x
C6
6
4
8
¢L
qL
8
x
3L
≤
2
x
564
CHAPTER 9
B.C.
Deflections of Beams
5 (v¿) Left (v¿) Right
∴ C6 C5 at x L
B.C.
23qL3
48
(a)
qLx3
EIv C5 x C7 (0 x L)
48
B.C.
6 v(0) 0
EIv qLx 2
(L x2 )
48 EI
v
q(L x)
(7L3 17L2x 10Lx2 2x3 )
48 EI
C v ¢
qL3
2
3
2 2
¢L
x
q0
y
C
A
L
—
2
B
x
L
—
2
Right-hand half (part CB):
L
2
1
xL
2
q0
EIv–– q 0
EIv‡ C1
EIv– C1 x C2
EIv¿ C1 ¢
q0 Lx3 x4
x2
¢
≤ C5 ¢ ≤ C6 x C7
L 6
12
2
4
5
3
q0 Lx
x
x
x2
EIv ¢
≤ C5 ¢ ≤ C6 ¢ ≤ C7x C8
L 24 60
6
2
EIv¿ Left-hand half (part AC): 0 x BOUNDARY CONDITIONS
B.C.
x2
≤ C2 x C3
2
x3
x2
≤ C2 ¢ ≤ C3 x C4
6
2
q0
PART CB q (2x L)
L
q0
EIv–– q (L 2x)
L
q0
EIv‡ (Lx x2 ) C5
L
q0 Lx2 x3
EIv– ¢
≤ C5 x C6
L 2
3
EIv C1 ¢
qL3
3L
≤
2
16 EI
Simple beam (triangular load)
LOAD EQUATION (EQ. 9-12 c)
PART AC
3L
≤
2
3L
≤
2
Problem 9.4-10 Derive the equations of the deflection curve for a
simple beam AB supporting a triangularly distributed load of maximum
intensity q0 acting on the right-hand half of the beam (see figure). Also,
determine the angles of rotation A and B at the ends and the deflection
C at the midpoint. Use the fourth-order differential equation of the
deflection curve (the load equation).
Solution 9.4-10
x
11qL4
3L
≤
2
384 EI
uC v¿ ¢
3
qx
3qLx
9qL x
qL x
C8
24
12
16
2
(0 x L)
¢L
qL3
B.C. 7 v(L) 0 for 0 x L ∴ C5 48
4
7qL4
3L
∴ C8 2
48
v
C7 0
From Eq.(a): C6 8 v(L) 0 for L x 1 EIv‡ V
C1 C5 B.C.
2 EIv– M
C2 0
EI(v‡) AC EI(v‡) BC
q0 L
4
at x L
2
(1)
EIv–(0) 0
(2)
2
q0 L
6
L
for x 2
B.C.
3 EIv¿(L) 0 C5 L C6 B.C.
4 (EIv–) AC (EIv–) CB
C1L C5 L 2C6 q0 L2
6
(3)
(4)
SECTION 9.5
B.C.
5 (v¿) AC (v¿) CB for x L
2
DEFLECTION CURVE FOR PART AC ¢ 0 x C1L2 8C3 C5 L2 4C6 L 8C7 B.C.
6 v(0) 0
B.C.
7 v(L) 0
q0 L3
8
C4 0
(5)
(6)
C5 L3 3C6 L2 6C7 L 6C8 3q0 L
20
L
2
C1L3 24C3 L C5L3 6C6L2 24C7 L 48C8
q0L4
10
(8)
SOLVE EQS. (1) THROUGH (B):
q0 L
C1 24
C4 0
37q0 L3
C2 0 C3 5760
C5 5q0 L
24
2
C6 q0 L
24
L
≤
2
q0Lx
(37L2 40x2 )
5760EI
q0 L
v¿ (37 L2 120x2 )
5760EI
(7)
8 (v)AC (v)CB for x 565
v
uA v¿(0) 4
B.C.
Method of Superposition
C v ¢
37q0 L3
5760 EI
3q0 L4
L
≤
2
1280 EI
DEFLECTION CURVE FOR PART CB ¢
v
L
x L≤
2
q0
[L2x (37 L2 40x2 ) 3(2x L) 5 ]
5760 LEI
v¿ q0
[L2 (37L2 120x2 ) 30(2x L) 4 ]
5760 LEI
uB v¿(L) 53q0 L3
5760EI
67q0L3
q0L4
CB 5760
1920
Substitute constants into equations for v and v¿.
C7 Method of Superposition
The problems for Section 9.5 are to be solved by the method of
superposition. All beams have constant flexural rigidity EI.
P
A
Problem 9.5-1 A cantilever beam AB carries three equally spaced
concentrated loads, as shown in the figure. Obtain formulas for the
angle of rotation B and deflection B at the free end of the beam.
Solution 9.5-1 Cantilever beam with 3 loads
Table G-1, Cases 4 and 5
L 2
2L 2
≤
P¢ ≤
3
3
PL2 7PL2
uB 2EI
2 EI
2 EI
9 EI
P¢
P
P
B
L
—
3
L
—
3
L
—
3
L 2
2L 2
≤
P¢ ≤
L
2L
PL3
3
3
B ¢ 3L ≤ ¢ 3L ≤
6 EI
3
6 EI
3
3 EI
3
5PL
9 EI
P¢
566
CHAPTER 9
Deflections of Beams
Problem 9.5-2 A simple beam AB supports five equally spaced loads
P (see figure).
(a) Determine the deflection 1 at the midpoint of the beam.
(b) If the same total load (5P) is distributed as a uniform load
on the beam, what is the deflection 2 at the midpoint?
(c) Calculate the ratio of 1 to 2.
P
P
P
B
L
—
6
L
—
6
L
—
6
L
—
6
L
—
6
Simple beam with 5 loads
(a) Table G-2, Cases 4 and 6
(b) Table G-2, Case 1 qL 5P
L
≤
6
L 2
1 B 3L2 4 ¢ ≤ R 24 EI
6
L
P¢ ≤
3
L 2
PL3
B 3L2 4 ¢ ≤ R 24 EI
3
48 EI
2 P¢
P
A
L
—
6
Solution 9.5-2
P
(c)
5qL4
25 PL3
384 EI 384 EI
1
11 384
88
1.173
¢
≤
2 144 25
75
11PL3
144 EI
Problem 9.5-3 The cantilever beam AB shown in the figure has an extension
BCD attached to its free end. A force P acts at the end of the extension.
(a) Find the ratio a/L so that the vertical deflection of point B will be zero.
(b) Find the ratio a/L so that the angle of rotation at point B will be zero.
L
A
B
D
a
P
Solution 9.5-3
Cantilever beam with extension
Table G-1, Cases 4 and 6
P
B
A
L
Pa
(a) B PL3 PaL2
0
3EI
2EI
a 2
L 3
(b) uB PL2 PaL
0
2EI
EI
a 1
L 2
C
SECTION 9.5
Problem 9.5-4 Beam ACB hangs from two springs, as shown
in the figure. The springs have stiffnesses k1 and k2 and the beam
has flexural rigidity EI.
What is the downward displacement of point C, which
is at the midpoint of the beam, when the load P is applied?
Data for the structure are as follows: P 8.0 kN,
L 1.8 m, EI 216 kNm2, k1 250 kN/m, and
k2 160 kN/m.
567
Method of Superposition
L = 1.8 m
k1 = 250 kN/m
k2 = 160 kN/m
A
B
C
P = 8.0 kN
Solution 9.5-4
Beam hanging from springs
P 8.0 kN L 1.8 m
EI 216 kN m2 k1 250 kN/m
k2 160 kN/m
Substitute numerical values:
(8.0 kN)(1.8 m) 3
C 48 (216 kN m2 )
ˇ
Stretch of springs:
A P2
k1
B P2
k2
Table G-2, Case 4
PL3
1 P2 P2
C ¢
≤
48 EI 2 k1
k2
ˇ
8.0 kN
1
1
¢
≤
4
250 kNm 160 kNm
4.5 mm 20.5 mm
25 mm
PL3
P 1
1
¢ ≤
48EI 4 k1 k2
Problem 9.5-5 What must be the equation y f (x) of the axis
of the slightly curved beam AB (see figure) before the load is applied
in order that the load P, moving along the bar, always stays at the
same level?
y
P
B
A
L
Solution 9.5-5
Slightly curved beam
Let x distance to load P
downward deflection at load P
Table G-2, Case 5:
P(L x) x 2
Px2 (L x) 2
[L (L x) 2 x2 ] 6LEI
3LEI
Initial upward displacement of the beam must equal .
Px2 (L x) 2
∴ y
3LEI
x
568
CHAPTER 9
Deflections of Beams
Problem 9.5-6 Determine the angle of rotation B and deflection B at
the free end of a cantilever beam AB having a uniform load of intensity q
acting over the middle third of its length (see figure).
q
B
A
L
—
3
Solution 9.5-6
L
—
3
L
—
3
Cantilever beam (partial uniform load)
q intensity of uniform load
Original load on the beam:
SUPERPOSITION: Original load Load No. 1 minus
Load No. 2
Table G-1, Case 2
q
B
A
L
—
3
L
—
3
L
—
3
uB q 2L 3
q L 3
7qL3
¢
≤ ¢ ≤ 6EI 3
6EI 3
162EI
B q 2L 3
q 1 3
2L
L
¢
≤ ¢ 4L ≤
¢ ≤ ¢ 4L ≤
24EI 3
3
24EI 3
3
Load No. 1:
23qL4
648EI
B
2L
—
3
L
—
3
Load No. 2:
B
L
—
3
2L
—
3
Problem 9.5-7 The cantilever beam ACB shown in the figure
has flexural rigidity EI 2.1 106 k-in.2 Calculate the downward
deflections C and B at points C and B, respectively, due to the
simultaneous action of the moment of 35 k-in. applied at point C
and the concentrated load of 2.5 k applied at the free end B.
2.5 k
35 k-in.
A
B
C
48 in.
48 in.
SECTION 9.5
Solution 9.5-7
Cantilever beam (two loads)
P
Mo
A
B
C
L/2
M0 (L2)
L
PL3
¢ 2L ≤ 2EI
2
3EI
3M0L2 PL3
8EI
3EI
( downward deflection)
SUBSTITUTE NUMERICAL VALUES:
C 0.01920 in. 0.10971 in.
0.0905 in.
B 0.05760 in. 0.35109 in.
0.293 in.
Table G-1, Cases 4.6, and 7
B L/2
EI 2.1 106 k-in.2
M0 35 k-in.
P 2.5 k
L 96 in.
C M0 (L2) 2 P(L2) 2
L
¢ 3L ≤
2EI
6EI
2
M0 L2 5PL3
8EI
48EI
( downward deflection)
Problem 9.5-8 A beam ABCD consisting of a simple span BD and an
overhang AB is loaded by a force P acting at the end of the bracket CEF
(see figure).
(a) Determine the deflection A at the end of the overhang.
(b) Under what conditions is this deflection upward? Under what
conditions is it downward?
A
2L
—
3
L
—
3
L
—
2
B
C
F
E
P
a
Solution 9.5-8
Beam with bracket and overhang
(a) DEFLECTION AT THE END OF THE OVERHANG
P
Mo
B
L
—
3
A uB ¢
C
D
2L
—
3
Consider part BD of the beam.
M0 Pa
Table G-2, Cases 5 and 9
P (L3)(2L3)(5L3)
6LEI
uB 569
Method of Superposition
Pa
L2
L2
B 6 ¢ ≤ 3 ¢ ≤ 2L2 R
6LEI
3
9
PL
(10L 9a)
162EI
( clockwise angle)
L
PL2
≤
(10L 9a)
2
324 EI
( upward deflection)
a
10
(b) Deflection is upward when 6
and
L
9
a
10
downward when 7
L
9
D
570
CHAPTER 9
Deflections of Beams
C
Problem 9.5-9 A horizontal load P acts at end C of the bracket ABC
shown in the figure.
(a) Determine the deflection C of point C.
(b) Determine the maximum upward deflection max of member AB.
Note: Assume that the flexural rigidity EI is constant throughout the
frame. Also, disregard the effects of axial deformations and consider only
the effects of bending due to the load P.
Solution 9.5-9
P
H
B
A
L
Bracket ABC
BEAM AB
(a) ARM BC Table G-1, Case 4
M0 PH
C PH 3
PH 3 PH 2L
u8H 3EI
3EI
3EI
PH 2
(L H)
3EI
max
A
B
Mo = PH
(b) MAXIMUM DEFLECTION OF BEAM AB
M0 L2
PHL2
Table G-2, Case 7: max 913EI 913EI
L
Table G-2, Case 7: uB M0 L PHL
3EI
3EI
Problem 9.5-10 A beam ABC having flexural rigidity EI 75 kNm2
is loaded by a force P 800 N at end C and tied down at end A by a
wire having axial rigidity EA 900 kN (see figure).
0.5 m
What is the deflection at point C when the load P is applied?
B
A
C
P = 800 N
0.5 m
0.75 m
D
Solution 9.5-10
Beam tied down by a wire
P
B
A
L1
H
C
L2
M0 PL2
Table G-2, Case 7:
u¿B M0 L1 PL1L2
3EI
3EI
CONSIDER THE STRETCHING OF WIRE AD
D
EI 75 kN m2
P 800 N
EA 900 kN
H 0.5 m L1 0.5 m
L2 0.75 m
CONSIDER BC AS A CANTILEVER BEAM
Table G-1, Case 4:
CONSIDER AB AS A SIMPLE BEAM
PL32
¿C 3EI
¿A (Force in AD) ¢
PL2 H
PL2H
H
≤¢
≤¢
≤
EA
L1 EA
EAL1
DEFLECTION C OF POINT C
C ¿C u¿B (L2 ) ¿A
¢
L2
≤
L1
PL32 PL1L22 PL22H
3EI
3EI
EAL21
SUBSTITUTE NUMERICAL VALUES:
C 1.50 mm 1.00 mm 1.00 mm 3.50 mm