EMA 3702 Mechanics & Materials Science (Mechanics of ... · Therefore, the meaningful solution will...

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


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


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


Homework 9.1

Elastic curve:

Maximum deflection occurs when

Therefore, the meaningful solution will be

The corresponding maximum deflection will be

𝑦 =𝑀0


𝑀0

2𝐸𝐼𝑥2 +

𝑀0𝐿

3𝐸𝐼𝑥

𝐸𝐼𝑑𝑦

𝑑𝑥=? ? = 0

𝑥 = 1 −1

3𝐿 = 0.423𝐿

𝑦𝑚𝑎𝑥 =𝑀0


𝑀0

2𝐸𝐼𝑥2 +

𝑀0𝐿

3𝐸𝐼𝑥

𝑦𝑚𝑎𝑥 = 0.0642 ∙𝑀0𝐿

2

𝐸𝐼


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


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


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) = ? ?


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


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


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


Homework 9.2

Consider the boundary condition that deflection has to match where the load P is

x = a, yLeft = yR,ight

Therefore


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


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


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


Homework 9.2

Therefore, the deflection curve for 0 < x < a will be


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


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


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


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


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


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


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


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)()(


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


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


Homework 9.4

The total deflection at D

The total slope at D

PDwDDyyy )()()(

??)( wD

y

𝑦𝐷 𝑃 =??

𝑦𝐷 = −𝑃𝐿3

8𝐸𝐼−

5𝑃𝐿3

48𝐸𝐼 𝑦𝐷 = −

11𝑃𝐿3

48𝐸𝐼

PDwDD)()(

??)( wD

𝜃𝐷 𝑃 = 𝜃𝐺 𝑃 =? ?

𝜃𝐷 =𝑃𝐿2

6𝐸𝐼+

𝑃𝐿2

8𝐸𝐼 𝜃𝐷 =

7𝑃𝐿2

24𝐸𝐼

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