Hibbeler R.C. Structural Analysis

Подождите немного. Документ загружается.

510 CHAPTER 12 DISPLACEMENT METHOD OF A NALYSIS: MOMENT DISTRIBUTION

Fig. 12–16

12.5 Moment Distribution for Frames:

Sidesway

It has been shown in Sec. 11–5 that frames that are nonsymmetrical or

subjected to nonsymmetrical loadings have a tendency to sidesway. An

example of one such case is shown in Fig. 12–16a. Here the applied

loading P will create unequal moments at joints B and C such that the

frame will deflect an amount to the right. To determine this deflection

and the internal moments at the joints using moment distribution,

we will use the principle of superposition. In this regard, the frame in

Fig. 12–16b is first considered held from sidesway by applying an

artificial joint support at C. Moment distribution is applied and then by

statics the restraining force R is determined. The equal, but opposite,

restraining force is then applied to the frame, Fig. 12–16c, and the

moments in the frame are calculated. One method for doing this last

step requires first assuming a numerical value for one of the internal

moments, say Using moment distribution and statics, the

deflection and external force corresponding to the assumed value

of can then be determined. Since linear elastic deformations occur,

the force develops moments in the frame that are proportional to

those developed by R. For example, if and are known, the

moment at B developed by R will be Addition of

the joint moments for both cases, Fig. 12–16b and c, will yield the actual

moments in the frame, Fig. 12–16a. Application of this technique is

illustrated in Examples 12–6 through 12–8.

Multistory Frames. Quite often, multistory frameworks may have

several independent joint displacements, and consequently the moment

distribution analysis using the above techniques will involve more

computation. Consider, for example, the two-story frame shown in

Fig. 12–17a.This structure can have two independent joint displacements,

since the sidesway of the first story is independent of any displacement¢

= M

1R>R¿2.

R¿M

R¿

R¿¢¿

artificial joint applied

(no sidesway)

(b)

ⴙ



(a)

ⴝ

artificial joint removed

(sidesway)

(c)

12.5 MOMENT DISTRIBUTION FOR FRAMES: SIDESWAY 511

Fig. 12–17

of the second story. Unfortunately, these displacements are not known

initially, so the analysis must proceed on the basis of superposition, in the

same manner as discussed previously. In this case, two restraining forces

and are applied, Fig. 12–17b, and the fixed-end moments are

determined and distributed. Using the equations of equilibrium, the

numerical values of and are then determined. Next, the restraint at

the floor of the first story is removed and the floor is given a

displacement This displacement causes fixed-end moments (FEMs)

in the frame, which can be assigned specific numerical values. By

distributing these moments and using the equations of equilibrium, the

associated numerical values of and can be determined. In a similar

manner, the floor of the second story is then given a displacement

Fig. 12–17d. Assuming numerical values for the fixed-end moments, the

moment distribution and equilibrium analysis will yield specific values of

and Since the last two steps associated with Fig. 12–17c and d

depend on assumed values of the FEMs, correction factors and

must be applied to the distributed moments. With reference to the

restraining forces in Fig. 12–17c and 12–17d, we require equal but

opposite application of and to the frame, such that

Simultaneous solution of these equations yields the values of and

These correction factors are then multiplied by the internal joint

moments found from the moment distribution in Fig. 12–17c and 12–17d.

The resultant moments are then found by adding these corrected

moments to those obtained for the frame in Fig. 12–17b.

Other types of frames having independent joint displacements can be

analyzed using this same procedure; however, it must be admitted that

the foregoing method does require quite a bit of numerical calculation.

Although some techniques have been developed to shorten the

calculations, it is best to solve these types of problems on a computer,

preferably using a matrix analysis. The techniques for doing this will be

discussed in Chapter 16.

C–.C¿

=+C¿R

- C–R

ﬂ

=-C¿R

+ C–R

ﬂ

C–C¿

ﬂ

¢–,

¢¿.



(a)



ⴝ

ⴙ

(b)

(

)

¿

ⴙ

(d)

¿¿¿¿

R¿¿

Determine the moments at each joint of the frame shown in Fig. 12–18a.

EI is constant.

SOLUTION

First we consider the frame held from sidesway as shown in Fig. 12–18b.

We have

The stiffness factor of each span is computed on the basis of 4EI/L or

by using the relative-stiffness factor I/L. The DFs and the moment

distribution are shown in the table, Fig. 12–18d. Using these results,

the equations of equilibrium are applied to the free-body diagrams of

the columns in order to determine and Fig. 12–18e. From the

free-body diagram of the entire frame (not shown) the joint restraint

R in Fig. 12–18b has a magnitude of

An equal but opposite value of must now be applied

to the frame at C and the internal moments computed, Fig. 12–18c.To

solve the problem of computing these moments, we will assume a

force is applied at C, causing the frame to deflect as shown in

Fig. 12–18f. Here the joints at B and C are temporarily restrained from

rotating, and as a result the fixed-end moments at the ends of the

columns are determined from the formula for deflection found on the

inside back cover, that is,

¢¿R¿

R = 0.92 kN

R = 1.73 kN - 0.81 kN = 0.92 kN©F

= 0;

1FEM2

16112

142

152

= 2.56 kN

1FEM2

16142

112

152

=-10.24 kN

EXAMPLE 12.6

512 CHAPTER 12 DISPLACEMENT METHOD OF A NALYSIS: MOMENT DISTRIBUTION

16 kN

1 m 4 m

5 m5 m

(a)

(b)

16 kN

ⴙ

(

)

Joint AB CD

Member AB BA BC CB CD DC

DF 0 0.5 0.5 0.5 0.5 0

FEM

5.12 1.28

0.32

0.02

2.56

0.16

5.12

10.24

0.32

0.64

0.32

0.64

0.02

0.04

1.28

0.08

0.64

0.04

兺M

2.88 5.78 5.78

1.28

2.56

1.28

2.56

0.08

0.16

0.08

0.16

2.72 2.72 1.32

(d)

Dist.

2.72 kNm2.72 kNm

5.78 kNm5.78 kNm

1.32 kNm1.32 kNm2.88 kN m2.88 kN m

 0.81 kND

 0.81 kNA

 1.73 kNA

 1.73 kN

5 m5 m 5 m5 m

(e)(e)

Fig. 12–18

12.5 MOMENT DISTRIBUTION FOR FRAMES: SIDESWAY 513

Since both B and C happen to be displaced the same amount

and AB and DC have the same E, I, and L, the FEM in AB will be the

same as that in DC.As shown in Fig. 12–18f, we will arbitrarily assume

this fixed-end moment to be

A negative sign is necessary since the moment must act counterclockwise

on the column for deflection to the right. The value of associated

with this moment can now be determined. The moment

distribution of the FEMs is shown in Fig. 12–18g. From equilibrium, the

horizontal reactions at A and D are calculated, Fig. 12–18h.Thus, for the

entire frame we require

Hence, creates the moments tabulated in Fig. 12–18g.

Corresponding moments caused by can be determined by

proportion. Therefore, the resultant moment in the frame, Fig. 12–18a,is

equal to the sum of those calculated for the frame in Fig. 12–18b plus the

proportionate amount of those for the frame in Fig. 12–18c. We have

Ans.

Ans.M

=-1.32 +

0.92

56.0

1-802=-2.63 kN

=-2.72 +

0.92

56.0

1-602=-3.71 kN

= 2.72 +

0.92

56.0

1602= 3.71 kN

=-5.78 +

0.92

56.0

1602=-4.79 kN

= 5.78 +

0.92

56.0

1-602= 4.79 kN

= 2.88 +

0.92

56.0

1-802= 1.57 kN

R = 0.92 kN

R¿=56.0 kN

R¿=28 + 28 = 56.0 kN©F

= 0;

-100 kN

R¿¢¿

1FEM2

= 1FEM2

=-100 kN

¢¿,

M =

6EI¢

(f)(f)

¿¿ ¿¿

R¿R¿

100 kNm100 kN m

100 kNm100 kN m 100 kNm100 kN m

100 kNm100 kN m

Joint AB CD

Member AB BA BC CB CD DC

DF 0 0.5 0.5 0.5 0.5 0

FEM

12.5

3.125

0.78

12.5

50 50

12.5

3.125

6.25

0.78

1.56

兺M

80.00 60.00 60.00

(g)

100

6.25

1.56

0.39

100 100

0.195

0.39

12.5

25 25

3.125

6.25

0.78

1.56

60.00

0.195

0.39

60.00 80.00

100

3.125

0.78

0.195

0.39

1.56

6.25

Dist.

60 kNm60 kNm60 kN m60 kNm

80 kNm80 kNm80 kN m80 kNm

 28 kND

 28 kNA

 28 kN

5 m5 m 5 m5 m

(h)(h)

Determine the moments at each joint of the frame shown in Fig. 12–19a.

The moment of inertia of each member is indicated in the figure.

EXAMPLE 12.7

514 CHAPTER 12 DISPLACEMENT METHOD OF A NALYSIS: MOMENT DISTRIBUTION

SOLUTION

The frame is first held from sidesway as shown in Fig. 12–19b.The

internal moments are computed at the joints as indicated in Fig. 12–19d.

Here the stiffness factor of CD was computed using 3EI/L since there is

a pin at D. Calculation of the horizontal reactions at A and D is shown

in Fig. 12–19e. Thus, for the entire frame,

R = 2.89 – 1.00 = 1.89 k©F

= 0;

2 k/ft

10 ft

12 ft

15 ft

(a)

= 1500 in

= 2000 in

= 2500 in

ⴙ

2 k/ft

(b)

(c)

Joint AB CD

Member AB BA BC CB CD DC

DF 0 0.615 0.385 0.5 0.5 1

FEM

14.76 9.24

24

12

3.69

0.713

0.18

7.38

1.84

0.357

2.31

6

0.447

1.16

0.11

0.29

2.31

4.62

0.58

1.16

0.11

0.224

2.31

0.58

0.11

兺M

9.58 19.34 19.34 15.00 15.00 0

(d)

Dist.

10 ft10 ft

19.34 kft19.34 k ft

9.58 kft9.58 kft

 2.89 kA

 2.89 k

15 ft15 ft

15.00 kft15.00 k ft

 1.00 kD

 1.00 k

(e)(e)

Fig. 12–19

12.5 MOMENT DISTRIBUTION FOR FRAMES: SIDESWAY 515

The opposite force is now applied to the frame as shown in Fig. 12–19c.

As in the previous example, we will consider a force acting as shown

in Fig. 12–19f. As a result, joints B and C are displaced by the same

amount The fixed-end moments for BA are computed from

However, from the table on the inside back cover, for CD we have

Assuming the FEM for AB is as shown in Fig. 12–19f,

the corresponding FEM at C, causing the same is found by

comparison, i.e.,

Moment distribution for these FEMs is tabulated in Fig. 12–19g.

Computation of the horizontal reactions at A and D is shown in

Fig. 12–19h. Thus, for the entire frame,

The resultant moments in the frame are therefore

Ans.

Ans.M

=-15.00 +

1.89

12.55

1-23.312=-18.5 k

= 15.00 +

1.89

12.55

123.312= 18.5 k

=-19.34 +

1.89

12.55

140.012=-13.3 k

= 19.34 +

1.89

12.55

1-40.012= 13.3 k

= 9.58 +

1.89

12.55

1-69.912=-0.948 k

R¿=11.0 + 1.55 = 12.55 k©F

= 0;

1FEM2

=-27.78 k

¢¿ = -

1-10021102

6E120002

1FEM2

1152

3E125002

¢¿,

-100 k

1FEM2

3EI¢

3E125002¢¿

1152

1FEM2

= 1FEM2

6EI¢

6E120002¢¿

1102

¢¿.

R¿

Joint AB CD

Member AB BA BC CB CD DC

DF 0 0.615 0.385 0.5 0.5 1

FEM

61.5 38.5 13.89 13.89

4.27

2.96

0.20

100

30.75

2.14

2.67

6.94

1.85

4.81

0.13

0.33

9.625

19.25

0.67

1.34

0.46

0.92

9.625

0.67

0.46

兺M

69.91 40.01 40.01 23.31 23.31 0

(g)

1.48

100

27.78

Dist.

10 ft

40.01 kft

69.91 kft

A¿

 11.0 k

15 ft

23.31 kft

D¿

 1.55 k

(h)

(f)

R¿

100 kft

27.78 kft

100 kft

¿ ¿

(f)

Determine the moments at each joint of the frame shown in

Fig. 12–20a. EI is constant.

EXAMPLE 12.8

516 CHAPTER 12 DISPLACEMENT METHOD OF A NALYSIS: MOMENT DISTRIBUTION

Fig. 12–20

8 k

6 ft 5 ft 5 ft 6 ft

8 ft

10 ft

(a)

20 k

8 k

20 k

(b)

ⴙ

(

)

Joint AB CD

Member AB BA BC CB CD DC

DF 1 0.429 0.571 0.571 0.429 1

FEM

5.71

10

1.63

2.86

0.47

0.82

0.13

0.24

兺M

4.29

1.23

0.35

0.10

5.97 5.97 5.97

4.29

1.23

0.35

0.10

5.97 0

(d)

5.71

1.63

2.86

0.47

0.82

0.13

0.24

Dist.

SOLUTION

First sidesway is prevented by the restraining force R, Fig. 12–20b.The

FEMs for member BC are

Since spans AB and DC are pinned at their ends, the stiffness factor

is computed using 3EI/L. The moment distribution is shown in

Fig. 12–20d.

Using these results, the horizontal reactions at A and D must be

determined. This is done using an equilibrium analysis of each

member, Fig. 12–20e. Summing moments about points B and C on

each leg, we have

Thus, for the entire frame,

R = 3.75 – 3.75 + 20 = 20 k©F

= 0;

5.97 - D

182+ 4162= 0 D

= 3.75 kd+©M

= 0;

-5.97 + A

182- 4162= 0 A

= 3.75 kd +©M

= 0;

1FEM2

81102

=-10 k

ft 1FEM2

81102

= 10 k

4 k

6 ft

8 ft

5.97 kft

4 ft

4 k

6 ft

8 ft

5.97 kft

4 ft

8 k

5 ft 5 ft

20 k R

5.97 kft

4 k 4 k

5.97 kft

(e)

12.5 MOMENT DISTRIBUTION FOR FRAMES: SIDESWAY 517

The opposite force R is now applied to the frame as shown in

Fig. 12–20c. In order to determine the internal moments developed by

R we will first consider the force acting as shown in Fig. 12–20f.

Here the dashed lines do not represent the distortion of the frame

members; instead, they are constructed as straight lines extended to the

final positions and of points B and C, respectively. Due to the

symmetry of the frame, the displacement Furthermore,

these displacements cause BC to rotate. The vertical distance between

and is as shown on the displacement diagram, Fig. 12–20g.

Since each span undergoes end-point displacements that cause the spans

to rotate, fixed-end moments are induced in the spans. These are:

Notice that for BA and CD the moments are negative since clockwise

rotation of the span causes a counterclockwise FEM.

If we arbitrarily assign a value of

then equating in the above formulas yields

These moments are applied to the

frame and distributed, Fig. 12–20h. Using these results, the equilibrium

analysis is shown in Fig. 12–20i. For each leg, we have

Thus, for the entire frame,

The resultant moments in the frame are therefore

Ans.

Ans. M

=-5.97 +

80.74

1-146.802=-42.3 k

= 5.97 +

80.74

1146.802= 42.3 k

=-5.97 +

80.74

1146.802= 30.4 k

= 5.97 +

80.74

1-146.802=-30.4 k

= 0;

-D

182+ 29.36162+ 146.80 = 0

= 40.37 kd+©M

= 0;

-A

182+ 29.36162+ 146.80 = 0

= 40.37 kd +©M

= 0;

1FEM2

= 1FEM2

= 240 k

ft.

¢¿-100 k

ft,

1FEM2

= 1FEM2

6EI11.2¢¿2>1102

1FEM2

= 1FEM2

=-3EI¢¿>1102

,1FEM2

= 1FEM2

1.2¢¿,C¿B¿

BB¿=CC¿=¢¿.

C¿B¿

R¿

(

)

36.9

B¿

¿

C¿

36.9

B, C

¿

C¿

B¿

0.6¿

(g)

6 ft

8 ft

V¿

146.80 kft

29.36 k

A¿

6 ft

8 ft

D¿

10 ft

R¿

146.80 kft

V¿

146.80 kft

(i)

29.36 k

29.36 k 29.36 k

29.36 k

146.80 kft

29.36 k

Joint AB CD

Member AB BA BC CB CD DC

DF 1 0.429 0.571 0.571 0.429 1

FEM

79.94

240

17.15

4.89

1.40

22.82

39.97

6.52

11.41

1.86

3.26

0 146.80 146.80 0

(h)

0.40 0.53

0.93

79.94

240

22.82

39.97

6.52

11.41

1.86

3.26

0.53

0.93

60.06

17.15

4.89

1.40

100

60.06

100

0.40

146.80

146.80

Dist.

兺M

518 CHAPTER 12 DISPLACEMENT METHOD OF A NALYSIS: MOMENT DISTRIBUTION

Prob. 12–13

Prob. 12–14

Prob. 12–15

Prob. 12–16

12–13. Determine the moment at B, then draw the

moment diagram for each member of the frame. Assume

the supports at A and C are pins. EI is constant.

12–15. Determine the reactions at A and D. Assume the

supports at A and D are fixed and B and C are fixed

connected. EI is constant.

PROBLEMS

12–14. Determine the moments at the ends of each

member of the frame. Assume the joint at B is fixed, C is

pinned, and A is fixed. The moment of inertia of each

member is listed in the figure. ksi.E = 29(10

)

*12–16. Determine the moments at D and C, then draw

the moment diagram for each member of the frame.

Assume the supports at A and B are pins and D and C are

fixed joints. EI is constant.

6 m

5 m

8 kN/m

8 k/ft

15 ft

24 ft

2 k/ft

4 k

8 ft

12 ft

 800 in

 550 in

12 ft

9 ft

5 k/ft

12.5 MOMENT DISTRIBUTION FOR FRAMES: SIDESWAY 519

Prob. 12–17

Prob. 12–18

12–17. Determine the moments at the fixed support A

and joint D and then draw the moment diagram for the

frame. Assume B is pinned.

12–19. The frame is made from pipe that is fixed connected.

If it supports the loading shown, determine the moments

developed at each of the joints. EI is constant.

12–18. Determine the moments at each joint of the frame,

then draw the moment diagram for member BCE. Assume

B, C, and E are fixed connected and A and D are pins.

ksi.E = 29(10

)

*12–20. Determine the moments at B and C, then draw

the moment diagram for each member of the frame.

Assume the supports at A, E, and D are fixed. EI is constant.

4 k/ft

12 ft

2 k

3 k

8 ft

0.5 k/ft

 400 in

 600 in

 500 in

24 ft 12 ft

18 kN 18 kN

4 m

4 m 4 m

10 k

2 k/ft

8 ft

12 ft

16 ft

Prob. 12–19

Prob. 12–20