Internal Forces in Planar StructuresStevenVukazich

SanJoseStateUniversity

Consider the Frame

B

A

C

P

w

Roller support

Pin support

From an Equilibrium Analysis, We Can Find the Support Reactions

B

A

C

P

w

Ax

Cx

Ay

!𝑀#

�

�

= 0+

!𝐹)

�

�

= 0+

!𝐹*

�

�

= 0+

Cx

Ax

Ay

Cut the Frame into Two Sections

B

A

C

P

w

Ax

Cx

Ay

q

q

Make a cut at an arbitrary point Q between B and C perpendicular to member BC

Q

Free Body Diagrams of the Sections of the Frame

B

A

C

P

w

Ax

Cx

Ay

q

q

q

q

Q

MQ

VQ

FQ

Q

MQ

VQ

FQ

Three internal forces are developed to ensure equilibrium of each section

F = Axial ForceV = Shear ForceM = Bending Moment

Can use our three equations of equilibrium to find F, V, and M

w

B

A

C

P

w

Ax

Cx

Ay

q

q

q

q

Q

MQ

VQ

FQ

Q

MQ

VQ

FQ

w

Use Equilibrium to Find Internal Forces

!𝑀+

�

�

= 0+

!𝐹)

�

�

= 0+

!𝐹*

�

�

= 0+

MQ

FQ

VQ

x

y

xy

Can use the free-body diagram of either section to find FQ, VQ, and MQ .

Internal Force Example Problem

A beam is supported by a pin support at point A and extends over a roller support at point D. The beam is and subjected to a uniformly distributed load from A to B, a point moment at point C an inclined point load at point E as shown.

Find the internal forces at points q, r, s, and t.

9 ft 2 ft 3 ft 4 ft

5 k

E

23 k-ft2 k/ft

DCBA

34

6 ft1 ft 1 ft 2 ft

q r s t

Find All of the External Forces

FBD of beam

9 ft 2 ft 3 ft 4 ft

5 k

E

23 k-ft2 k/ft

DCBA

3

4

Ax

Ay Dy

2k/ft 9ft = 18k

4.5 ft

5k35

= 3k

5k45

= 4k

5

!𝑀#

�

�

= 0+

Dy = 10 k

!𝐹*

�

�

= 0+

Ay = 13 k

9 ft 2 ft 3 ft 4 ft

5 k

E

23 k-ft2 k/ft

DCBA

3

4

Ax

Ay Dy

2k/ft 9ft = 18k

4.5 ft

5k35

= 3k

5k45

= 4k

5

Find All of the External Forces

!𝐹)

�

�

= 0+

Ax = 4 k

Find All of the External Forces

9 ft 2 ft 3 ft 4 ft

5 k

E

23 k-ft2 k/ft

DCBA

3

4

Ax

Ay Dy

2k/ft 9ft = 18k

4.5 ft

5k35

= 3k

5k45

= 4k

5

External Forces

FBD of beam showing all external forces

9 ft 2 ft 3 ft 4 ft

5 k

E

23 k-ft2 k/ft

DCBA

3

44 k

13 k8 k6 ft

1 ft 1 ft 2 ft

q r s t

Find Internal Forces at Point q

Cut beam at point q

9 ft 2 ft 3 ft 4 ft

5 k

E

23 k-ft2 k/ft

DCBA

3

44 k

13 k8 k6 ft

1 ft 1 ft 2 ft

r s t

q

q

FBDs of Segments Aq and qBCDE

Mq

Vq

2 k/ft

A

4 k

13 k 6 ft

q

q

3 ft 2 ft 3 ft 4 ft

5 k

E

23 k-ft2 k/ft

DCB

3

4

8 k1 ft 1 ft 2 ft

r s t

q

q

Find Internal Forces at Point q

Fq

Fq

Mq

Vq

Can apply equations of equilibrium to either FBD to find internal forces

FBD of Segments Aq

Find Internal Forces at Point q

2 k/ft

2k/ft 6ft = 12k

!𝑀9

�

�

= 0+

Vq = 1 k

!𝐹*

�

�

= 0+Mq = 42 k-ft

Fq = -4 k!𝐹)

�

�

= 0+

Mq

Vq

2 k/ft

A

4 k

13 k 6 ft

q

q

Fq

Mq

VqA

4 k

13 k

6 ft

q

q

Fq3 ft

FBDs of Segments Aq and qBCDE

42 k-ft

Internal Forces at Point q

Confirm that both segments are in equilibrium

42 k-ft4 k

4 k

1 k

2 k/ft

A

4 k

13 k 6 ft

q

q

3 ft 2 ft 3 ft 4 ft

5 k

E

23 k-ft2 k/ft

DCB

3

4

8 k1 ft 1 ft 2 ft

r s t

q

q

1 k

Find Internal Forces at Point r

Cut beam at point r

9 ft 2 ft 3 ft 4 ft

5 k

E

23 k-ft2 k/ft

DCBA

3

44 k

13 k8 k

1 ft

q s t

r

r

FBDs of Segments ABr and rCDE

Mr

Vr

2 k/ft

A

4 k

13 k9 ft

1 ft 3 ft 4 ft

5 k

E

23 k-ft

DC

3

4

8 kt

r

r

Find Internal Forces at Point r

Fr

FrMr

Vr

r

rq

1 ft

s

FBD of Segment ABr

Find Internal Forces at Point r

!𝑀:

�

�

= 0+

Vr = -5 k

!𝐹*

�

�

= 0+

Mr = 31 k-ft

Fr = -4 k

!𝐹)

�

�

= 0+

Mr

Vr

2 k/ft

A

4 k

13 k9 ft

Fr

r

rq

1 ft

Mr

Vr

2 k/ft

A

13 k9 ft

Fr

r

rq

1 ft

2k/ft 9ft = 18k

4.5 ft

4 k

Internal Forces at Point r

2 k/ft

A

4 k

13 k9 ft

r

rq

1 ft

1 ft 3 ft 4 ft

E

23 k-ft

DC

3

4

8 kt

r

rs

Confirm that both segments are in equilibrium

31 k-ft

5 k

5 k

31 k-ft4 k

4 k

Find Internal Forces at Point s

Cut beam at point s

9 ft 2 ft 3 ft 4 ft

5 k

E

23 k-ft2 k/ft

DCBA

3

44 k

13 k8 k

1 ft

qs

t

s

r

FBDs of Segments ABCs and sDE

Ms

Vs

2 k/ft

A

4 k

13 k9 ft

1 ft4 ft

5 k

E

23 k-ft

D

C

3

4

8 k

t

s

Find Internal Forces at Point s

Fs

FsMs

Vs

rq

2 ft

s

2 ft

s

s

FBD of Segment sDE

Find Internal Forces at Point s

!𝑀;

�

�

= 0+

Vs = -5 k

!𝐹*

�

�

= 0+Ms = -7 k-ft

Fs = -4 k!𝐹)

�

�

= 0+

1 ft4 ft

5 k

ED

3

4

8 k

t

s

FsMs

Vs s

3

45k

35

= 3k

5k45

= 4k

5

1 ft4 ft

5 k

ED

8 k

t

s

FsMs

Vs s

2 k/ft

A

4 k

13 k9 ft

1 ft4 ft

5 k

E

23 k-ft

D

C

3

4

8 k

t

s

Internal Forces at Point s

rq

2 ft

s

2 ft

s

s

Confirm that both segments are in equilibrium

7 k-ft

5 k

4 k

5 k

7 k-ft

4 k

Find Internal Forces at Point t

Cut beam at point t

9 ft 2 ft 3 ft 4 ft

5 k

E

23 k-ft2 k/ft

DCBA

3

44 k

13 k 8 k2 ft

qt

t

r s

FBDs of Segments ABCDt and tE

Mt

Vt

2 k/ft

A

4 k

13 k9 ft

2 ft

5 k

E

23 k-ft

DC

3

4

8 k

t

t

Find Internal Forces at Point t

Ft

FtMt

Vt

rq

2 ft 3 ft

st

2 ft

t

FBD of Segment tE

Find Internal Forces at Point t

!𝑀<

�

�

= 0+

Vt = 3 k

!𝐹*

�

�

= 0+Mt = -6 k-ft

Ft = -4 k!𝐹)

�

�

= 0+

3

45k

35

= 3k

5k45

= 4k

55 k

2 ft

5 k

E

3

4

t

t

FtMt

Vt2 ft

Et

t

FtMt

Vt

2 k/ft

A

4 k

13 k9 ft

2 ft

5 k

E

23 k-ft

DC

3

4

8 k

t

t

Internal Forces at Point t

rq

2 ft 3 ft

st

2 ft

t

Confirm that both segments are in equilibrium

6 k-ft

3 k

4 k

3 k

6 k-ft

4 k

2 k/ft

A

4 k

13 k

23 k-ft

DC

8 k

Summary of Results

rq st

t 6 k-ft

3 k4 k

2 k/ft

A

4 k

13 k

23 k-ft

Crq

s

s 7 k-ft

5 k4 k

2 k/ft

A

4 k

13 k

r

rq

31 k-ft

5 k4 k

42 k-ft4 k

2 k/ft

A

4 k

13 k

q

q 1 k