Beam Deflections Using Double IntegrationStevenVukazich

SanJoseStateUniversity

𝑑"𝑦𝑑𝑥"

=𝑀𝐸𝐼

Recall the Moment-Curvature Relationshipfor Small Deformations

In order to solvethis differential equation for y and 𝜃 we need:• Moment equation (from statics);• Two boundary (or continuity)

conditions on y or 𝜃;• Information on E and I.

PM

w

x

y 𝜃 𝑥 =𝑑𝑦𝑑𝑥

𝑦 𝑥

Example Problem

A

w

x

y

𝑦 𝑥Modulus of Elasticity = EMoment of Inertia = I

B

Find the equation of the elastic curve for the simply supported beam subjected to the uniformly distributed load using the double integration method. Find the maximum deflection. EI is constant.

L

Free Body Diagram of the Beam

Need to find the moment function M(x)

Ax

Ay By

A

w

x

y

B

L

0𝑀1

�

�

= 0+

Free Body Diagram of the Beam

Ax

Ay By

A

w

x

y

B

L

12𝑤𝐿

23𝐿

𝐵; =13𝑤𝐿

Free Body Diagram of the Beam

Ax

Ay By

A

w

x

y

B

L

12𝑤𝐿

23𝐿

𝐴; =16𝑤𝐿

0𝐹;

�

�

= 0+

0𝐹?

�

�

= 0+

Free Body Diagram of the BeamShowing Known Reactions

A

w

x

y

B

L

13𝑤𝐿

16𝑤𝐿

a

a

Cut the beam at a-a to find the moment function M(x)

x

M

V16𝑤𝐿

𝑤 =𝑤𝐿𝑥

Free Body Diagram of Segment to the Left of a–a

x

M

V16𝑤𝐿

𝑅 =12𝑥

𝑤𝐿𝑥 =

𝑤2𝐿𝑥"

23𝑥

13𝑥

a

a

a

a 0𝑀A

�

�

= 0+

𝑀 =𝑤𝐿6𝑥 −

𝑤6𝐿𝑥C

Need Two Boundary Conditions on y or 𝜃

A

w

x

y

𝑦 𝑥Modulus of Elasticity = EMoment of Inertia = I

B

L

𝑦 0 = 0

𝑦 𝐿 = 0

Solve the Differential Equation

A

w

x

y

𝑦 𝑥Modulus of Elasticity = EMoment of Inertia = I

B

L

𝑦 0 = 0

𝑦 𝐿 = 0𝑑"𝑦𝑑𝑥"

=𝑀𝐸𝐼

𝑀 =𝑤𝐿6𝑥 −

𝑤6𝐿𝑥C

For constant EI

𝐸𝐼𝑑"𝑦𝑑𝑥"

= 𝑀

𝐸𝐼𝑑"𝑦𝑑𝑥"

=𝑤𝐿6𝑥 −

𝑤6𝐿𝑥C

𝐸𝐼𝑑"𝑦𝑑𝑥"

=𝑤𝐿6𝑥 −

𝑤6𝐿𝑥C

First Integration

D𝐸𝐼𝑑"𝑦𝑑𝑥"

�

�

𝑑𝑥 = D𝑤𝐿6𝑥 −

𝑤6𝐿𝑥C 𝑑𝑥

�

�

𝐸𝐼𝑑𝑦𝑑𝑥

= −𝑤24𝐿

𝑥F +𝑤𝐿12

𝑥" + 𝐶H

𝐸𝐼𝜃 = −𝑤24𝐿

𝑥F +𝑤𝐿12

𝑥" + 𝐶H

𝑦 0 = 0

𝑦 𝐿 = 0

Second Integration

D𝐸𝐼𝑑𝑦𝑑𝑥

�

�

𝑑𝑥 = D −𝑤24𝐿

𝑥F +𝑤𝐿12

𝑥" + 𝐶H 𝑑𝑥�

�

𝐸𝐼𝑑𝑦𝑑𝑥

= −𝑤24𝐿

𝑥F +𝑤𝐿12

𝑥" + 𝐶H

𝐸𝐼𝑦 = −𝑤

120𝐿𝑥I +

𝑤𝐿36

𝑥C + 𝐶H𝑥 + 𝐶"

Use the two boundary conditions to find C1 and C2

Find Constants C1 and C2

𝑦 0 = 0

𝑦 𝐿 = 0

𝐸𝐼𝑦 = −𝑤

120𝐿𝑥I +

𝑤𝐿36

𝑥C + 𝐶H𝑥 + 𝐶"

𝐸𝐼 0 = −𝑤

120𝐿0 I +

𝑤𝐿36

0 C + 𝐶H 0 + 𝐶"

𝐸𝐼 0 = −𝑤

120𝐿0 I +

𝑤𝐿36

0 C + 𝐶H 0 + 𝐶"

𝐶" = 0

𝐸𝐼 0 = −𝑤

120𝐿𝐿 I +

𝑤𝐿36

𝐿 C + 𝐶H 𝐿 + 𝐶"

𝐶H =𝑤𝐿C

120−𝑤𝐿C

36= −

7𝑤𝐿C

360

Functions for y and 𝜃

𝐸𝐼𝑦 = −𝑤

120𝐿𝑥I +

𝑤𝐿36

𝑥C −7𝑤𝐿C

360𝑥

𝐸𝐼𝜃 = −𝑤24𝐿

𝑥F +𝑤𝐿12

𝑥" −7𝑤𝐿C

360

𝑦 𝑥 =1𝐸𝐼

−𝑤

120𝐿𝑥I +

𝑤𝐿36

𝑥C −7𝑤𝐿C

360𝑥

𝜃 𝑥 =1𝐸𝐼

−𝑤24𝐿

𝑥F +𝑤𝐿12

𝑥" −7𝑤𝐿C

360

Functions for y and 𝜃

A

w

x

y

𝑦 𝑥Modulus of Elasticity = EMoment of Inertia = I

B

L

𝑦 𝑥 =1𝐸𝐼

−𝑤

120𝐿𝑥I +

𝑤𝐿36

𝑥C −7𝑤𝐿C

360𝑥

𝜃 𝑥 =1𝐸𝐼

−𝑤24𝐿

𝑥F +𝑤𝐿12

𝑥" −7𝑤𝐿C

360

𝜃 𝑥 =𝑑𝑦𝑑𝑥

Questions

A

w

x

y

𝑦 𝑥

Modulus of Elasticity = EMoment of Inertia = I

B

L

𝑦 𝑥 =1𝐸𝐼

−𝑤

120𝐿𝑥I +

𝑤𝐿36

𝑥C −7𝑤𝐿C

360𝑥

𝜃 𝑥 =1𝐸𝐼

−𝑤24𝐿

𝑥F +𝑤𝐿12

𝑥" −7𝑤𝐿C

360

𝜃 𝑥 =𝑑𝑦𝑑𝑥

How can we find the maximum displacement?Where does the maximum displacement occur?How can we find the rotation at the supports?

Answers

A

w

x

y

𝑦max

Modulus of Elasticity = EMoment of Inertia = I

B

L𝑦 𝑥 =

1𝐸𝐼

−𝑤

120𝐿𝑥I +

𝑤𝐿36

𝑥C −7𝑤𝐿C

360𝑥

𝜃 𝑥 =1𝐸𝐼

−𝑤24𝐿

𝑥F +𝑤𝐿12

𝑥" −7𝑤𝐿C

360

𝜃 = 0

ymax occurs where 𝜃 = 0

𝜃A is 𝜃(0) 𝜃B is 𝜃(L)

?

𝜃 𝐿

𝜃 0

Point Where Maximum Deflection Occurs

𝑞 =−𝑏 ± 𝑏" − 4𝑎𝑐�

2𝑎

𝜃 𝑥 =1𝐸𝐼

−𝑤24𝐿

𝑥F +𝑤𝐿12

𝑥" −7𝑤𝐿C

360= 𝟎

Set 𝜃 = 0

−124𝑞" +

𝐿"

12𝑞 −

7𝐿F

360= 𝟎

Let 𝑞 = 𝑥" Multiply both sides by TUVW

𝑞 =− 𝐿"12 ±

𝐿F144 − 4 − 1

24 − 7𝐿F360

�

2 − 124

= 𝐿" ± 12𝐿F

144−

16

7𝐿F

360�

= 𝐿" ± 12𝐿"1270

�

𝑞 = 𝐿" ± 4𝐿"130

�= 1 + 4

130� , 𝟏 − 𝟒

𝟏𝟑𝟎� 𝐿"

𝑥 = 1 − 4130�

�𝐿 = 0.51893𝐿

Maximum Deflection

A

w

x

y

𝑦max

Modulus of Elasticity = EMoment of Inertia = I

B

L

𝜃 = 0

𝜃 𝐿

𝜃 0

1 − 4130�

�𝐿 = 0.51893𝐿

𝑦max =1𝐸𝐼

−𝑤

120𝐿0.5193𝐿 I +

𝑤𝐿36

0.5193𝐿 C −7𝑤𝐿C

3600.5193𝐿 = −0.006522

𝑤𝐿F

𝐸𝐼

Slope at Supports

A

w

x

y

Modulus of Elasticity = EMoment of Inertia = I

B

L

𝜃1 = 𝜃 0 =1𝐸𝐼

−7𝑤𝐿C

360= −

7𝑤𝐿C

360𝐸𝐼= −0.01944

𝑤𝐿C

𝐸𝐼

𝜃 𝐿

𝜃 0

𝜃` = 𝜃 𝐿 =1𝐸𝐼

−𝑤24𝐿

𝐿F +𝑤𝐿12

𝐿" −7𝑤𝐿C

360=𝑤𝐿C

45𝐸𝐼= 0.02222

𝑤𝐿C

𝐸𝐼

Compare to Tabulated Solution in AISC Manual

𝑊 =12𝑤𝐿