Upload
truongcong
View
240
Download
0
Embed Size (px)
Citation preview
EMA 3702
Mechanics & Materials Science
(Mechanics of Materials)
Chapter 9 Deflection of Beams
Homework Answers/Hints
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.1
For a simply supported prismatic beam DE with local moment at D,
please step-by-step determine (a) the equation of the elastic curve,
(b) the maximum deflection, (c) the slope at the two ends.
L
D E M0
y
x
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.1
Draw the FBD for the entire DE
beam,
Consider balance of force
???
Consider balance of moment of D
???
Therefore, RE = ??
For arbitrary section DG from D
Balance of force gives
V = ??
Balance of moment around D gives
?? = ??
Therefore, M (x) = ??
L
D E M0
M M(x)
RE RD
M
D G
RD V
x
y
x
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.1
From previous, M (x) = ??
Therefore, elastic curve satisfies
Integrate for once wrt. x
Integrate again wrt. x
Consider boundary conditions,
x = ?, y = ? C2 = ?; x = ?, y = ?
Therefore, elastic curve
𝐸𝐼𝑑2𝑦
𝑑𝑥2= 𝑀 𝑥 =? ?
𝐸𝐼𝑑𝑦
𝑑𝑥=? ? +𝐶1
𝐸𝐼𝑦 =? ? + 𝐶1𝑥 + 𝐶2
L
D E M0
x = ?, y = ? x = ?, y = ?
? ? =? 𝐶1 =???
𝑦 =𝑀0
6𝐸𝐼𝐿𝑥3 −
𝑀0
2𝐸𝐼𝑥2 +
𝑀0𝐿
3𝐸𝐼𝑥
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.1
Elastic curve:
Maximum deflection occurs when
Therefore, the meaningful solution will be
The corresponding maximum deflection will be
𝑦 =𝑀0
6𝐸𝐼𝐿𝑥3 −
𝑀0
2𝐸𝐼𝑥2 +
𝑀0𝐿
3𝐸𝐼𝑥
𝐸𝐼𝑑𝑦
𝑑𝑥=? ? = 0
𝑥 = 1 −1
3𝐿 = 0.423𝐿
𝑦𝑚𝑎𝑥 =𝑀0
6𝐸𝐼𝐿𝑥3 −
𝑀0
2𝐸𝐼𝑥2 +
𝑀0𝐿
3𝐸𝐼𝑥
𝑦𝑚𝑎𝑥 = 0.0642 ∙𝑀0𝐿
2
𝐸𝐼
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.1
At x = 0
Local slope is
At x = L
Local slope is
𝐸𝐼𝑑𝑦
𝑑𝑥=? ?
𝑑𝑦
𝑑𝑥 𝑥=0
=𝑀0𝐿
3𝐸𝐼
𝐸𝐼𝑑𝑦
𝑑𝑥=? ?
𝑑𝑦
𝑑𝑥 𝑥=𝐿
= −𝑀0𝐿
6𝐸𝐼
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.2
For a simply supported prismatic beam DE with concentrated load at G,
please step-by-step determine (a) the elastic curve y(x) for 0 < x < a and
(b) the deflection at point G.
L=a+b
D E
P a b
G
y
x
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.2
Draw the FBD for the entire beam,
Consider the balance of force
?? = ?
Consider the balance of moment
around G
? = ?
RD = ??
RE = ??
Consider arbitrary section DC to the
left of G, or 0 < x < a, draw FBD
Balance of force gives V = ??
Balance of moment at C gives
L=a+b
D E
P a b
G
RE RD
RD V(x) when 0 < x < a
M(x) x
D E
P
D C M (x) = ? ?
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.2
From previous, when 0 < x < a
M (x) = ?
Integrate wrt. x
Integrate wrt. x again
𝐸𝐼𝑑2𝑦
𝑑𝑥2= 𝑀 𝑥 =?
𝐸𝐼𝑑𝑦
𝑑𝑥=? +𝐶1
𝐸𝐼𝑦 =? ? +C2
L=a+b
D E
P a b
G
RD V(x)
M(x) D C
y
x
when 0 < x < a
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.2
Similarly, when a < x < a+b, for section CE
consider balance of moment around C
Therefore, elastic curve satisfies
Integrate wrt x
Integrate wrt x again,
M (x) = ?
L=a+b
D E
P a b
G
V(x)
M(x) C E
RE
x
𝐸𝐼𝑑2𝑦
𝑑𝑥2= 𝑀 𝑥 =? ?
𝐸𝐼𝑑𝑦
𝑑𝑥=? ? +𝐶3
𝐸𝐼𝑦 =? ? +𝐶3𝑥 + 𝐶4
y
x
L-x
When a < x < a+b
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.2
When 0 < x < a
When a < x < a+b
Consider boundary conditions
x = ?, y = ? C2 = ?
x = a + b, y = 0
L=a+b
D E
P a b
G
x = ?
y = ?
x = ?
y = ?
x = a, yLeft = yRight
x = a, Left = Right
𝐸𝐼𝑦 =𝑏𝑃
6(𝑎 + 𝑏)𝑥3 + 𝐶1𝑥 + 𝐶2
𝐸𝐼𝑦 = −𝑎𝑃
6 𝑎 + 𝑏𝑥3 +
1
2𝑎𝑃𝑥2 + 𝐶3𝑥 + 𝐶4
𝐸𝐼𝑑𝑦
𝑑𝑥=
𝑏𝑃
2(𝑎 + 𝑏)𝑥2 + 𝐶1
𝐸𝐼𝑑𝑦
𝑑𝑥= −
𝑎𝑃
2 𝑎 + 𝑏𝑥2 + 𝑎𝑃𝑥 + 𝐶3
−𝑎𝑃
6 𝑎 + 𝑏𝑎 + 𝑏 3 +
1
2𝑎𝑃 𝑎 + 𝑏 2 + 𝐶3(𝑎 + 𝑏) + 𝐶4 = 0
𝑎𝑃
3𝑎 + 𝑏 2 + 𝐶3(𝑎 + 𝑏) + 𝐶4 = 0 𝑎 + 𝑏 𝐶3 + 𝐶4 = −
𝑎𝑃
3𝑎 + 𝑏 2
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.2
Consider the boundary condition that deflection has to match where the load P is
x = a, yLeft = yR,ight
Therefore
𝐸𝐼𝑦 =𝑏𝑃
6(𝑎 + 𝑏)𝑎3 + 𝐶1a = −
𝑎𝑃
6 𝑎 + 𝑏𝑎3 +
1
2𝑃𝑎3 + 𝐶3𝑎 + 𝐶4
𝑎𝐶1 − 𝑎𝐶3 − 𝐶4 = −𝑎𝑃
6 𝑎 + 𝑏𝑎3 −
𝑏𝑃
6 𝑎 + 𝑏𝑎3 +
1
2𝑃𝑎3
𝑎𝐶1 − 𝑎𝐶3 − 𝐶4 = −𝑎 + 𝑏
6 𝑎 + 𝑏𝑃𝑎3 +
1
2𝑃𝑎3 = −
1
6𝑃𝑎3 +
1
2𝑃𝑎3
𝑎𝐶1 − 𝑎𝐶3 − 𝐶4 =1
3𝑃𝑎3
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.2
Consider the other boundary condition that the slope for the elastic curve has to match
where the load P is
x = a, Left = Right
𝐸𝐼𝑑𝑦
𝑑𝑥=
𝑏𝑃
2(𝑎 + 𝑏)𝑎2 + 𝐶1 = −
𝑎𝑃
2 𝑎 + 𝑏𝑎2 + 𝑎𝑃𝑎 + 𝐶3
𝐶1 − 𝐶3 = −𝑎𝑃
2 𝑎 + 𝑏𝑎2 −
𝑏𝑃
2(𝑎 + 𝑏)𝑎2 + 𝑃𝑎2
𝐶1 − 𝐶3 = −(𝑎 + 𝑏)
2 𝑎 + 𝑏𝑃𝑎2 + 𝑃𝑎2
𝐶1 − 𝐶3 =1
2𝑃𝑎2
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.2
Therefore,
𝐶1 − 𝐶3 =1
2𝑃𝑎2
𝑎𝐶1 − 𝑎𝐶3 − 𝐶4 =1
3𝑃𝑎3
𝑎 + 𝑏 𝐶3 + 𝐶4 = −𝑎𝑃
3𝑎 + 𝑏 2
𝑎𝐶1 − 𝑎𝐶3 =1
2𝑃𝑎3
1
2𝑃𝑎3 − 𝐶4 =
1
3𝑃𝑎3 𝐶4 =
1
6𝑃𝑎3
𝑎 + 𝑏 𝐶3 +1
6𝑃𝑎3 = −
𝑎𝑃
3𝑎 + 𝑏 2
𝑎 + 𝑏 𝐶3 = −1
6𝑃𝑎3 −
𝑎𝑃
3𝑎 + 𝑏 2 𝐶3 = −
1
6 𝑎 + 𝑏𝑃𝑎3 −
𝑎𝑃
3𝑎 + 𝑏
𝐶1 = 𝐶3 +1
2𝑃𝑎2 = −
1
6 𝑎 + 𝑏𝑃𝑎3 −
𝑎𝑃
3𝑎 + 𝑏 +
1
2𝑃𝑎2
𝐶1 = −1
6 𝑎 + 𝑏𝑃𝑎3 −
𝑃𝑎𝑏
3+
1
6𝑃𝑎2
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.2
Therefore, the deflection curve for 0 < x < a will be
𝐸𝐼𝑦 =𝑏𝑃
6 𝑎 + 𝑏 𝐸𝐼𝑥3 + 𝐶1𝑥
𝑦 =1
𝐸𝐼
𝑏𝑃
6 𝑎 + 𝑏𝑥3 + [−
1
6 𝑎 + 𝑏𝑃𝑎3 −
𝑃𝑎𝑏
3+
1
6𝑃𝑎2]𝑥
𝑦 =1
𝐸𝐼
𝑃
6 𝑎 + 𝑏𝑏𝑥3 + [−
𝑃𝑎3
6 𝑎 + 𝑏−
𝑃𝑎𝑏 ∙ 2(𝑎 + 𝑏)
6(𝑎 + 𝑏)+
𝑃𝑎2 ∙ (𝑎 + 𝑏)
6(𝑎 + 𝑏)]𝑥
𝑦 =𝑃
6𝐸𝐼(𝑎 + 𝑏)𝑏𝑥3 + −𝑎3 − 2𝑎2𝑏 − 2𝑎𝑏2 + 𝑎3 + 𝑎2𝑏 𝑥
𝑦 =𝑃
6𝐸𝐼(𝑎 + 𝑏)𝑏𝑥3 + −𝑎2𝑏 − 2𝑎𝑏2 𝑥
𝑦 =𝑏𝑃
6𝐸𝐼(𝑎 + 𝑏)𝑥3 − (𝑎2 + 2𝑎𝑏) 𝑥
𝐶1 = −1
6 𝑎 + 𝑏𝑃𝑎3 −
𝑃𝑎𝑏
3+
1
6𝑃𝑎2
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.2
Therefore, the deflection curve for 0 < x < a will be
When x = a, the y deflection is
𝑦 = −𝑃𝑎2𝑏2
3𝐸𝐼(𝑎 + 𝑏)
𝑦 =𝑏𝑃
6𝐸𝐼(𝑎 + 𝑏)𝑎3 − (𝑎2 + 2𝑎𝑏) 𝑎
𝑦 =𝑏𝑃
6𝐸𝐼(𝑎 + 𝑏)𝑎3 − 𝑎3 − 2𝑎2𝑏
𝑦 = −2𝑎2𝑏2𝑃
6𝐸𝐼(𝑎 + 𝑏)
𝑦 =𝑏𝑃
6𝐸𝐼(𝑎 + 𝑏)𝑥3 − (𝑎2 + 2𝑎𝑏) 𝑥
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.3
For a prismatic cantilever beam DE with D side on roller and subject
to moment M0 and E side fixed, please step-by-step determine (a) the
elastic curve y(x) and (b) reactions at D and E
Draw FBD for the entire beam
Balance of force gives:
? = ?
Balance of moment around E gives
? = ?
Three variables and two equations
a statically indeterminate problem!
Consider arbitrary section DG, draw FBD
Consider balance of moment around G
M (x) = ??
L
D E
M0
M0
RD RE
ME
M0
RD V
M (x)
x
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.3
Internal bending moment M (x) = ??
The elastic curve satisfies
Integrate wrt. x
Integrate wrt. x again
Totally (3+2) = 5 variables; Three boundary conditions, totally (2+3) = 5 equations
x = 0, y = 0
x = L, y = ?
x = L, = ?
𝐸𝐼𝑑2𝑦
𝑑𝑥2= 𝑀 𝑥 =?
𝐸𝐼𝑑𝑦
𝑑𝑥= ? +C1
𝐸𝐼𝑦 =1
6RDx3 –
1
2M0x
2+C1x+C2
L
D E
M0
x = 0, y = ? x = ?, y = ?
x = ?, = ?
C2 = 0
1
6RDL3 –
1
2M0L
2+C1L = ?
1
2RDL2 – M0L+C1
= ?
y
x
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.3
Therefore,
RD = RE
M0 + ME = RDL
1
6RDL
3 –1
2M0L
2+C1L = 0
1
2RDL
2 – M0L+C1 = 0
1
2RDL
3 – M0L2+C1
L = 0
1
6RDL
3 –1
2M0L
2=1
2RDL
3 –M0L2
1
3RDL3=
1
2M0L
2 RD=
3M0
2𝐿
RE=3M0
2𝐿
RD = RE
M0 + ME = RDL 𝑀𝐸 = 𝑅𝐷𝐿 – 𝑀0 =3M0
2𝐿𝐿 − 𝑀0 𝑀𝐸 =
M0
2
1
2RDL2 – M0L+C1
= 0 C1 = M0L −
1
2RDL
2 =M0L −
1
2∙3𝑀0
2𝐿L2 C1
= 𝑀0𝐿
4
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.3
The elastic curve satisfy
𝐸𝐼𝑦 =1
6RDx3 –
1
2M0x
2+C1x C1 =
𝑀0𝐿
4
𝐸𝐼𝑦 =1
6∙3𝑀0
2𝐿x3 –
1
2M0x
2+𝑀0𝐿
4x
RD=3M0
2𝐿
𝐸𝐼𝑦 =𝑀0
4𝐿x3 –
1
2M0x
2+𝑀0𝐿
4x
𝑦 =𝑀0
4𝐸𝐼𝐿x3 –
1
2𝐸𝐼M0x
2+𝑀0𝐿
4𝐸𝐼x
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
Homework 9.4
For cantilever beam with both distributed and concentrated load as
illustrated, please determine the slope and deflection at D end. (Hint: to
treat concentrated load at G alone, section GE will behave like a
cantilever beam under vertical load at the end (case 1 of textbook
Appendix D), while section DG remains straight.)
L
D E
w=P/L
P
G
0.5L
y
x
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
The stress states is superposition of the following two
Homework 9.4
L
D E
w=P/L P
G
0.5L
L
D E
w=P/L
L
D E
P
G
0.5L
PDwDDyyy )()()(
PDwDD)()(
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
For distributed load w, from appendix D case 2
Deflection at D due to distributed load w=P/L
Slope at end D due to due to distributed load
w=P/L
Note that positive slope instead of negative slope is used here due to the origin of the
coordinate system is chosen at D, while x axis is long the direction from D to E
Homework 9.4
L
D E
w=P/L
?)( wD
y
?)( wD
y
x
EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 9 Deflection of Beams
For concentrated load P, from appendix D case 1
Deflection at G due to concentrated load P
Slope at G due to concentrated load P
Note that positive slope instead of negative
slope is used here due to the origin of the
coordinate system is chosen at D, while x
axis is long the direction from D to E
Deflection at G due to concentrated load P
Slope at G due to concentrated load P will be
Homework 9.4
L
D E
P
G
0.5L
LyyPGPGPD
5.0)()()(
?)( PG
y
PGy )(
(G)P
𝑦𝐷 𝑃 =? ? = −5𝑃𝐿3
48𝐸𝐼
= −𝑃𝐿3
24𝐸𝐼
𝜃𝐺 𝑃 =?
𝜃𝐷 𝑃 = 𝜃𝐺 𝑃 =? ?
y
x