Vijay Gupta

MEC201Mechanics of Solids

Vijay Gupta

Deflection of Beams

Vijay Gupta

•

s

θθ

Geometry of deflection of beams

•• v

Vijay Gupta

Moment-deflection equation

1𝜌

=𝑑𝜃𝑑𝑠

=𝑀𝑏

𝐸𝐼

θ θ +dθ

dsdθdx

ρdv

ds = ρdθ ⇒ dθ = (1/ρ)ds, θ = dv/dx , dx/ds = cos θ ≈ 1 So:

Vijay Gupta

Deflection of a cantilevered beamPP

PLx

SFD

- P

xBMD

- PLMB = -P(L – x)

𝐸𝐼 𝑑2𝑣

𝑑𝑥2=𝑀𝑏=−𝑃 (𝐿−𝑥 )

𝐸𝐼 𝑑𝑣𝑑𝑥

=−𝑃𝐿𝑥+𝑃 𝑥2

2+𝐶1

𝐸𝐼𝑣=−𝑃𝐿 𝑥2

2+𝑃 𝑥3

6+𝐶1𝑥+𝐶2

Vijay Gupta

PP

PLx

SFD

- P

xBMD

- PL

Deflection of a cantilevered beam

𝐸𝐼𝑣=−𝑃𝐿 𝑥2

2+𝑃 𝑥3

6+𝐶1𝑥+𝐶2

𝐸𝐼 𝑑𝑣𝑑𝑥

=−𝑃𝐿𝑥+𝑃 𝑥2

2+𝐶1

𝑑𝑣𝑑𝑥

=0at 𝑥=0 , so𝐶2=0

𝑣=0at 𝑥=0 , so𝐶1=0

𝐸𝐼𝑣=−𝑃𝐿 𝑥2

2+𝑃 𝑥3

6

𝐸𝐼 𝑑𝑣𝑑𝑥

=−𝑃𝐿𝑥+𝑃 𝑥2

2

Vijay Gupta

Maximum Deflection

-vmax

𝐸𝐼𝑣=−𝑃𝐿 𝑥2

2+𝑃 𝑥3

6

𝐸𝐼 𝑑𝑣𝑑𝑥

=−𝑃𝐿𝑥+𝑃 𝑥2

2

𝑣𝑚𝑎𝑥=−𝑃 𝐿3

3𝐸𝐼

( 𝑑𝑣𝑑𝑥 )𝑚𝑎𝑥

=− 𝑃𝐿2

2𝐸𝐼

Vijay Gupta

Simply- Supported Beam

+PL/4

for x < L/2, M = (P/2) x

for x > L/2, M = (P/2)x - P(x – L/2) = 0

𝐸𝐼 𝑑2𝑣

𝑑 𝑥2=𝑀𝑏

𝐸𝐼 𝑑2𝑣

𝑑 𝑥2= 𝑃𝑥2

𝐸𝐼 𝑑𝑣𝑑𝑥

=− 𝑃 𝑥2

4+𝐶1

𝐸𝐼𝑣=− 𝑃 𝑥3

12+𝐶1 𝑥+𝐶2

Vijay Gupta

Simply- Supported Beam

+PL/4

Similarly, for x > L/2

There are now be four constants and we need four conditions!

𝑑2𝑣𝑑𝑥2

= 1𝐸𝐼 [− 𝑃2 (𝑥− 𝐿)]

𝑑𝑣𝑑𝑥

= 1𝐸𝐼 [− 𝑃2 (𝑥22 −𝐿𝑥)]+𝐶3

𝑣= 1𝐸𝐼 [− 𝑃2 (𝑥36 − 𝐿 𝑥

2

2 )]+𝐶3 𝑥+𝐶4

Vijay Gupta

Simply- Supported Beam

+PL/4

for x < L/2, M = (P/2) x

for x < L/2, M = (P/2)x - P(x – L/2) = 0

And two compatibility conditions

(𝑣 at 𝑥=𝐿/2 ) I= (𝑣 at 𝑥=𝐿 /2 )II

( 𝑑𝑣𝑑𝑥 at 𝑥=𝐿/2)I

=(𝑑𝑣𝑑𝑥 at 𝑥=𝐿 /2)II

𝑣=0at 𝑥=0𝑣=0at 𝑥=𝐿

Two boundary conditions

Vijay Gupta

Deflections

xP

LL/2

Pa b

L

x

δx (x <a) Pbx(L2 – x2 −b2) /6LEIδx (x > a) Pb [(L/b)( x −a)3+(L2 – b2)x−x3] /6LEIδmax Pb(L2 – b2)3/2/9√3LEI At θ(0), θ(L)

δx (x < L/2) Px(3L2 – 4x2) /48EIδmax (L/2) PL3/48EI(0, L)

Vijay Gupta

Deflections

δxδmax (L/2) 5wL4/384EI(0) = (L)L

x

L

wa b

x

δx (x < a)δx (x > a)δ (x = L/2) if a > b if a <bθ (0) θ(L)

Vijay Gupta

Deflection

δxδmax ) MoL2/16EI(0)(L)

x

L

Mo

Vijay Gupta

Pa b

L

x

Deflections

δx (x < a) Px2 (3a −x) /6EIδx (x > a) Pa2 (3x −a) /6EIδmax (x = L) Pa2 (3L −a) /6EIθ(a ≤ x ≤L) Pa2 /2EI

δx (x) Px2 (3L −x) /6EIδmax (L) PL3/3EI(L)x

P

L

Vijay Gupta

Deflections

δxδ (L) wL4/8EI(L)

L

a bx

δx (x< a)δx (x>a)δ (L)(L)

x

L

w

Vijay Gupta

Deflections

δx

δ(L)

(L)

x

L

Mo

Vijay Gupta

Method of Superposition

𝐸𝐼 𝑑2𝑣

𝑑𝑥2=𝑀𝑏

at 𝑥=𝑥𝑜 ,𝑣=𝑣𝑜

at 𝑥=𝑥1 ,𝑑𝑣 /𝑑𝑥=𝜃𝑜

𝐸𝐼𝑑2𝑣2𝑑𝑥2

=𝑀𝑏 ,2

at 𝑥=𝑥𝑜 ,𝑣2=𝑣𝑜 ,2

at 𝑥=𝑥1 ,𝑑𝑣2 /𝑑𝑥=𝜃𝑜 , 2

𝐸𝐼𝑑2𝑣1𝑑𝑥2

=𝑀𝑏 , 1

at 𝑥=𝑥𝑜 ,𝑣1=𝑣𝑜 , 1

at 𝑥=𝑥1 ,𝑑𝑣1/𝑑𝑥=𝜃𝑜 , 1

Vijay Gupta

Method of Superposition

10 N/m10 N

5 Nm=+ 10 N

10 N/m 10 N1 m

Vijay Gupta

Method of Superposition10 N/m

10 N5 Nm

−10 N−5 Nm −10 Nm

−10 N10 N

10 Nm 10 N

Vijay Gupta

Method of Superposition

15 N20 Nm 10 N/m 10 N10 N/m

10 N5 Nm

10 N10 Nm 10 N+ =

−20 N −10 N−10 N −10 N+ =

15 Nm−5 Nm −10 Nm+ =

Vijay Gupta

Method of Superposition

15 N20 Nm 10 N/m 10 N10 N/m

10 N5 Nm

10 N10 Nm 10 N+ =

δxδ (L) wL4/8EI(L)δx (x) Px2 (3L −x) /6EIδmax (L) PL3/3EI(L)δxδ (L) 10/8EI(L)δx (x) 10x2 (3 −x) /6EIδ (L) 10/3EI(L)

𝛿𝑥= (180𝑥2−80 𝑥+10𝑥4 )/24 𝐸𝐼𝛿 (1 )=55/12𝐸𝐼𝜃 (1 )=20 /3𝐸𝐼

Vijay Gupta

Another exampleP P

a aL

=P

P+

Vijay Gupta

Another example

PaLP

a LPa

xb

L

δx (x <a) Pbx(L2 – x2 −b2) /6LEIδx (x > a) Pb [(L/b)( x −a)3+(L2 – b2)x−x3] /6LEIδmax Pb(L2 – b2)3/2/9√3LEI At θ(0), θ(L)

¿𝑃𝑎 (3𝐿2−4𝑎2 )/ 48𝐸𝐼Same for the secondHence total deflection at mid-point =

Vijay Gupta

Another example

A B 𝜃BPb

PA B Ca b

L

P𝜃B b𝜃B

𝛿 ′𝐶𝛿𝐶

=? P

Vijay Gupta

Another example

A B 𝜃BPb Δxδmax ) MoL2/16EI(0)(L)x

L

Mo

𝜃𝐵=𝑃𝑏𝑎 /3𝐸𝐼

a

Vijay Gupta

Another example

A B 𝜃BPb P𝜃B b𝜃B

𝛿 ′𝐶𝛿𝐶

𝜃𝐵=𝑃𝑏𝑎 /3𝐸𝐼

𝛿𝐶❑′=𝑃 𝑏3/3𝐸𝐼

δx (x) Px2 (3L −x) /6EIδmax (L) PL3/3EI(L)x

P

L

b

+

Vijay Gupta

Statically indeterminate beams

P

L

P

L

L

Q

=+

C𝑄=𝑘𝛿𝐶

k

Vijay Gupta

Statically indeterminate beamsP

L

L

Q+

δx (x) Px2 (3L −x) /6EIδmax (L) PL3/3EI(L)x

P

L

C 𝛿 (C) = −5PL3 /12EI𝛿 (L) = −PL3 /3EI

Vijay Gupta

Statically indeterminate beamsP

L

L

Q+ 𝛿 (C) = −5PL3 /12EIC

Pa b

L

xδx (x < a) Px2 (3a −x) /6EIδx (x > a) Pa2 (3x −a) /6EIδmax (x = L) Pa2 (3L −a) /6EIθ(a ≤ x ≤L) Pa2 /2EI

𝛿 (𝐶 )=+𝑄𝐿3/24𝐸𝐼𝜃 (𝐶 )=+𝑄 𝐿2/16𝐸𝐼𝛿 (𝐿 )=+5𝑄𝐿3 /24𝐸𝐼

𝛿 (L) = −PL3 /3EI

Vijay Gupta

Statically indeterminate beamsP

L

L

Q+ 𝛿 (C) = −5PL3 /12EIC

Total deflection of point C is −5PL3 /12EI + −

𝛿 (𝐶 )=+𝑄𝐿3/24𝐸𝐼𝜃 (𝐶 )=+𝑄 𝐿2/16𝐸𝐼𝛿 (𝐿 )=+5𝑄𝐿3 /24𝐸𝐼

Determine Q in terms of P, k, L and EI. Then determine deflections

Vijay Gupta

Statically indeterminate beams

PQ𝛿 3L/4

𝛿B = QL3/3EI

L 3L/4P

L/4A B C

𝛿B = −QL3/3EI + P (3L/4)2 (9L /4) /6EI = −QL3/3EI + 9P L3/32EI

𝛿LQ

Pa b

L

x

δx (x <a) Pbx(L2 – x2 −b2) /6LEIδx (x > a) Pb [(L/b)( x −a)3+(L2 – b2)x−x3] /6LEI

Vijay Gupta

Statically indeterminate beams

PQ𝛿 3L/4

𝛿B = QL3/3EI

L 3L/4P

L/4A B C

𝛿B = −QL3/3EI + P (3L/4)2 (9L /4) /6EI = −QL3/3EI + 9P L3/32EI

𝛿LQ

2QL3/3EI = 9P L3/32EI 𝑄=27𝑃 /64

Vijay Gupta

Statically indeterminate beams

𝜃AMoCB

L LA

MoCB

2LA

CB2L

P

=+

Vijay Gupta

Statically indeterminate beams

A MoC

B2LA

CB2L

P

δxδmax ) MoL2/16EI(0)(L)

x

L

Mo

𝛿𝐵=𝑀𝑜𝐿2/ 4𝐸𝐼

𝜃𝐶=2𝑀𝑜 𝐿/3𝐸𝐼

Vijay Gupta

Statically indeterminate beams

A MoC

B2LA

CB2L

P 𝛿𝐵=−𝑀𝑜 𝐿2/4𝐸𝐼

𝜃𝐶=+2𝑀𝑜 𝐿 /3𝐸𝐼

xP

LL/2

δx (x < L/2) Px(3L2 – 4x2) /48EIδmax (L/2) PL3/48EI(0, L)

PL3 /6EI𝜃𝐶=+𝑃 𝐿2/4 𝐸𝐼

Vijay Gupta

Statically indeterminate beams

A MoC

B2LA

CB2L

P 𝛿𝐵=−𝑀𝑜 𝐿2/4𝐸𝐼

𝜃𝐶=+2𝑀𝑜 𝐿 /3𝐸𝐼

PL3 /6EI𝜃𝐶=+𝑃 𝐿2/4 𝐸𝐼

PL3 /6EI ¿𝑃=−3𝑀𝑜/2𝐿𝐸𝐼

𝜃𝐶=+2𝑀𝑜𝐿3𝐸𝐼

+ 𝑃 𝐿2

4𝐸 𝐼=7𝑀𝑜𝐿24𝐸𝐼