Session I Structural Analysis

Structural Analysis

J.P. Mohsen

Structures

•

Determinate

400 lb. 300 lb. 10 ft 10 ft 10 ft 10 ft B C D E F G H R_L RR 7.5 ft

2j = m + r Truss is determinate

2j m + r indeterminate

2j m + r Unstable

J = number of joints

m= number of members

r = number of reactions

A 5

Determine the force in members BH, BC, and DG of the truss shown. Note that

the truss is composed of triangles 7.5 ft : 10.0 ft : 12.5 ft, so that they are 3:4:5

right angles.

400 lb. 300 lb. 10 ft 10 ft 10 ft 10 ft B C D E F G H R_L R_R 7.5 ft

400 lb. 300 lb. 10 ft 10 ft 10 ft 10 ft A B C D E F G H R_L R_R 7

Analysis of Member BH.

400 lb. 300 lb. 10 ft 10 ft 10 ft 10 ft A B C D E F G H R_L R_R F_BH

0

0

=

=

∑

F

y

F

bh +

400 lb. 300 lb. 10 ft 10 ft 10 ft 10 ft A B C D E F G H R_L R_R 9

Analysis of Member BC.

400 lb. 300 lb. 10 ft 10 ft 10 ft 10 ft A B C D E F G H R_L R_R F_BC C D F_BG B 12.5 ft 7.5 ft 400 lb. = 275 lb.

)

(

733

5

.

7

)

20

(

275

n

compressio

lbs

F

_BC

=

−

=

−

+

∑

M

G

=

0

−

20

R

R

−

7

.

5

F

BC

=

0

400 lb. 300 lb. 10 ft 10 ft 10 ft 10 ft A B C D E F G H R_L R_R 11

Analysis of Member DG.

400 lb. 300 lb. 10 ft 10 ft 10 ft 10 ft A B C D E F G H R_L R_R F_CD C D F_DG 12.5 ft 7.5 ft

400 lb. 300 lb. 10 ft 10 ft 10 ft 10 ft A B C D E F G H R_L R_R F_CD E R_R G C D F F_GF F_DG 10 ft 12.5 ft 7.5 ft

0

=

∑

F

Y

R

_R

−

DG

_Y

=

0

DG

_Y

=

275

lbs

5

3

=

DG

DG

_Y

(

tension

)

lbs

DG

=

458

+ 13

120 kN-m 20 kN/m 60 kN A B C D E 2 m 2 m 2 m 2 m

Draw the shear and moment diagrams for the beam shown. Indicate the

maximum moment.

120 kN-m 20 kN/m 60 kN A F_B C D F_E 2 m 2 m 2 m 2 m 15

120 kN-m 20 kN/m 60 kN A F_B = 100 kN C D F_E = 40 kN 2 m 2 m 2 m 2 m

Solve for the reactions at supports B and E.

∑M

_B

= 0

→ 60(2) + 120 – 6F

E

= 0 →

F

E

= 40 kN

∑F

_Y

= 0 → -60 – 80 + F

E

+ F

B

= 0 → -100 + F

B

= 0 →

F

B

= 100 kN

+

A

100 kN

C D

40 kN

2 m 2 m 2 m 2 m

for segment AB.

0 0 V (kN) -40

( )

(

)

kN

m

kN

m

20

40

2

−

=

−

17

120 kN-m 20 kN/m 60 kN A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 -40 60 V (kN)

Show the change in Shear

at B.

kN

A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 -40 60 20 V (kN)

for segment BC.

( )

(

)

kN

m

kN

m

20

40

2

−

=

−

19

120 kN-m 20 kN/m 60 kN A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 -40 60 20 -40 V (kN)

Show the change in Shear

at C.

kN

60

−

A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 -40 60 20 -40 -40 V (kN)

for segment CE.

( )

(

)

kN

m

kN

m

0

0

4

=

21

120 kN-m 20 kN/m 60 kN A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 -40 60 20 -40 -40 V (kN)

Show the change in Shear

at E.

kN

A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 -40 60 20 -40 -40 V (kN) 23

120 kN-m 20 kN/m 60 kN A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 -40 60 20 -40 -40 V (kN)

Draw the Moment Diagram

for segment AB.

( )(

m

kN

)



=

−

kN

⋅

m











−

40

2

1

40

2

A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 -40 60 20 -40 -40 -40 M (kN-m) 0 ₀ V (kN)

for segment AB.

( )(

m

kN

)



=

−

kN

⋅

m











−

40

2

1

40

2

25

120 kN-m 20 kN/m 60 kN A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 -40 60 20 -40 -40 40 2° V (kN)

Draw the Moment Diagram

for segment BC.

( )(

m

kN

)



+

( )(

m

kN

)

=

kN

⋅

m











80

20

2

2

1

40

2

A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 -40 60 20 -40 -40 -40 -40 M (kN-m) 0 0 40 2° 2° 2° V (kN)

for segment CD.

( )(

2

m

−

40

kN

)

=

−

80

kN

⋅

m

27

120 kN-m 20 kN/m 60 kN A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 -40 60 20 -40 -40 40 80 2° V (kN)

Show the change in

bending moment at D.

m

kn

⋅

A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 -40 60 20 -40 -40 -40 -40 M (kN-m) 0 0 40 80 2° 2° 2° V (kN)

for segment DE.

( )(

2

m

−

40

kN

)

=

−

80

kn

⋅

m

120 kN-m 20 kN/m 60 kN A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 V (kN) -40 60 20 -40 -40 40 80 2°

Completed Moment

Diagram.

A 100 kN C D 40 kN 2 m 2 m 2 m 2 m 0 0 V (kN) -40 60 20 -40 -40 -40 -40 M (kN-m) 0 0 40

80

2° 2° 2°

moment.

m

kn

M

_max

= 80

⋅

31

What are the vertical and horizontal components of deflection at the 30K Load? All members have a cross sectional area of 1 square inch and modulus of elasticity of 29000 ksi.

∑









=

E

A

δ

E= modulus of elasticity of materials A= Cross-sectional area of each member

L= Length of each member

S = member force with proper sign

These are internal member forces due to original loading

Moment Distribution

•

1)

calculate the fixed end moments

•

2) Calculate distribution of moments at the clamped ends of the members

by the rotation of that joint

•

3) Calculate the magnitude of the moments carried over to the other ends

of the members

•

4) The addition or subtraction of these latter moments to the original fixed

ends moments

Fixed End Moments

P

FEM = PL 8 FEM = PL 8 w FEM= wL2 12 FEM= wL2 12 L/2 _L/2 L P a b

.

FEM

-25 + 25 - 50 + 50

K

_AB

=4EI = I K

_BC

=4EI = I

L 10 L 20

Distribution Factor = K_______________

Sum of K for all members at the joint

∑

=

K

K

DF

₁ 1

Distribution Factor = K_______________

Sum of K for all members at the joint

Distribution Factor = K

_BA

_

K

_BA

+ K

_BC

∑

=

K

K

DF

₁ 1

∑

=

K

K

DF

₂ 2 51

K

_AB

=4EI = I K

_BC

=4EI = I

L 10 L 20

1__

10____ = 2/3 D.F. at B for BA

1__ + 1_

10 20

1__

2/3

D. F.

FEM Balancing Joint B -25 + 25 - 50 + 50

1/3

53

K_AB=4EI = I

K_BC=4EI = I

L 10

L 20 Stiffness K

2/3

D. F. FEM Balancing Joint B -25 + 25 - 50 + 50

+ 16.67

+8.33

1/3

Joint B Released

2/3

D. F. FEM Balancing Joint B

C.O.M.

-25 + 25 - 50 + 50 + 16.67 +8.33

+ 8.33

1/3

55

K_AB=4EI = I

K_BC=4EI = I

L 10

L 20 Stiffness K

2/3

D. F. FEM Balancing Joint B C.O.M. -25 + 25 - 50 + 50 +16.67 +8.33

1/3

2/3

D. F. FEM Balancing Joint B C.O.M. Final Moments

-25

+ 25 - 50 + 50 + 16.67 +8.33 +

8.33

+4.17

- 16.67

1/3

57

K_AB=4EI = I

K_BC=4EI = I

L 10

L 20 Stiffness K

2/3

D. F. FEM Balancing Joint B C.O.M. -25 + 25 - 50 + 50 + 16.67 +8.33 + 8.33 +4.17

1/3

2/3

D. F. FEM Balancing Joint B C.O.M. Final Moments -25 + 25 - 50 + 50 + 16.67 +8.33 + 8.33 +4.17 - 16.67 + 41.67

- 41.67

1/3

59

K_AB=4EI = I

K_BC=4EI = I

L 10

L 20 Stiffness K

2/3

D. F. FEM Balancing Joint B C.O.M. -25 + 25 - 50 + 50 + 16.67 +8.33 + 8.33 +4.17

1/3

2/3

D. F. FEM Balancing Joint B C.O.M. Final Moments -25 + 25 - 50 + 50 + 16.67 +8.33 + 8.33 +4.17 --- --- --- --- - 16.67 + 41.67 - 41.67 + 54.17

1/3

61

20 k

-16.67 FT.K 41.67 -41.67 _54.2

1.5 K/FT

63

AE=Zero member

CE=500 N Compression

BD= Zero member

BC=500 N Compression

AC= 707 N Tension

References

•

Hibbeler, C. R., Structural Analysis, 3

rd

Edition,

Prentice Hall, 1995.

•

Chajes, Alexander, Structural Analysis, Prentice

Thank You!

•

Any Questions?

•

Good Luck!