Hibbeler R.C. Structural Analysis

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270 CHAPTER 7APPROXIMATE ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES

7.3 Vertical Loads on Building Frames

Building frames often consist of girders that are rigidly connected to

columns so that the entire structure is better able to resist the effects of

lateral forces due to wind and earthquake. An example of such a rigid

framework, often called a building bent, is shown in Fig. 7–4.

In practice, a structural engineer can use several techniques for

performing an approximate analysis of a building bent. Each is based

upon knowing how the structure will deform under load. One technique

would be to consider only the members within a localized region of the

structure. This is possible provided the deflections of the members within

the region cause little disturbance to the members outside the region.

Most often, however, the deflection curve of the entire structure is

considered. From this, the approximate location of points of inflection,

that is, the points where the member changes its curvature, can be specified.

These points can be considered as pins since there is zero moment within

the member at the points of inflection. We will use this idea in this

section to analyze the forces on building frames due to vertical loads, and

in Secs. 7–5 and 7–6 an approximate analysis for frames subjected to

lateral loads will be presented. Since the frame can be subjected to both

of these loadings simultaneously, then, provided the material remains

elastic, the resultant loading is determined by superposition.

Assumptions for Approximate Analysis. Consider a typical

girder located within a building bent and subjected to a uniform vertical

load, as shown in Fig. 7–5a. The column supports at A and B will each

exert three reactions on the girder, and therefore the girder will be

statically indeterminate to the third degree (6 reactions – 3 equations of

equilibrium). To make the girder statically determinate, an approximate

analysis will therefore require three assumptions. If the columns are

extremely stiff, no rotation at A and B will occur, and the deflection

typical building frame

Fig. 7–4

7.3 VERTICAL LOADS ON BUILDING FRAMES 271

curve for the girder will look like that shown in Fig. 7–5b. Using one of

the methods presented in Chapters 9 through 11, an exact analysis

reveals that for this case inflection points, or points of zero moment,

occur at 0.21L from each support. If, however, the column connections at

A and B are very flexible, then like a simply supported beam, zero

moment will occur at the supports, Fig. 7–5c. In reality, however, the

columns will provide some flexibility at the supports, and therefore we

will assume that zero moment occurs at the average point between the

two extremes, i.e., at from each support, Fig. 7–5d.

Furthermore, an exact analysis of frames supporting vertical loads

indicates that the axial forces in the girder are negligible.

In summary then, each girder of length L may be modeled by a simply

supported span of length 0.8L resting on two cantilevered ends, each

having a length of 0.1L,Fig.7–5e. The following three assumptions are

incorporated in this model:

1. There is zero moment in the girder, 0.1L from the left support.

2. There is zero moment in the girder, 0.1L from the right support.

3. The girder does not support an axial force.

By using statics, the internal loadings in the girders can now be

obtained and a preliminary design of their cross sections can be made.

The following example illustrates this numerically.

10.21L + 02>2 L 0.1L

columncolumn

girder

(a)

points of zero

moment

0.21L

fixed supported

(b)

simply supported

(

)

point of

zero

moment

point of

zero

moment

assumed points of zero

moment

0.1L

approximate case

(

)

0.1L

Fig. 7–5

0.1L

model

(e)

0.1L

0.8L

272 CHAPTER 7APPROXIMATE ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES

Fig. 7–6

Determine (approximately) the moment at the joints E and C caused

by members EF and CD of the building bent in Fig. 7–6a.

EXAMPLE 7.3

SOLUTION

For an approximate analysis the frame is modeled as shown in Fig. 7–6b.

Note that the cantilevered spans supporting the center portion of the

girder have a length of Equilibrium requires the

end reactions for the center portion of the girder to be 6400 lb, Fig. 7–6c.

The cantilevered spans are then subjected to a reaction moment of

Ans.

This approximate moment, with opposite direction, acts on the joints

at E and C,Fig.7–6a. Using the results, the approximate moment

diagram for one of the girders is shown in Fig. 7–6d.

M = 1600112+ 6400122= 14 400 lb

ft = 14.4 k

0.1L = 0.11202= 2 ft.

800 lb/ft

20 ft

(a)

800 lb/ft

16 ft

800 lb/ft

2 ft

(b)

16 ft

6400 lb 6400 lb

12 800 lb

1600 lb

6400 lb

14 400 lbft

1600 lb6400 lb

8000 lb 8000 lb

2 ft 2 ft

(c)

M (kft)

x (ft)

14.4

25.6

(

)

7.4 PORTAL FRAMES AND TRUSSES 273

7.4 Portal Frames and Trusses

Frames. Portal frames are frequently used over the entrance of a

bridge

and as a main stiffening element in building design in order to

transfer horizontal forces applied at the top of the frame to the

foundation. On bridges, these frames resist the forces caused by wind,

earthquake, and unbalanced traffic loading on the bridge deck. Portals

can be pin supported, fixed supported, or supported by partial fixity. The

approximate analysis of each case will now be discussed for a simple

three-member portal.

Pin Supported. A typical pin-supported portal frame is shown in

Fig. 7–7a. Since four unknowns exist at the supports but only three

equilibrium equations are available for solution, this structure is statically

indeterminate to the first degree. Consequently, only one assumption

must be made to reduce the frame to one that is statically determinate.

The elastic deflection of the portal is shown in Fig. 7–7b. This diagram

indicates that a point of inflection, that is, where the moment changes

from positive bending to negative bending, is located approximately at

the girder’s midpoint. Since the moment in the girder is zero at this point,

we can assume a hinge exists there and then proceed to determine the

reactions at the supports using statics. If this is done, it is found that the

horizontal reactions (shear) at the base of each column are equal and

the other reactions are those indicated in Fig. 7–7c. Furthermore, the

moment diagrams for this frame are indicated in Fig. 7–7d.

(a)

—

moment

diagram

(d)

 

assumed

hinge

(

)

—

(c)

—

*See Fig. 3–4.

Fig. 7–7

274 CHAPTER 7APPROXIMATE ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES

Fixed Supported. Portals with two fixed supports, Fig. 7–8a, are

statically indeterminate to the third degree since there are a total of six

unknowns at the supports. If the vertical members have equal lengths

and cross-sectional areas, the frame will deflect as shown in Fig. 7–8b.For

this case we will assume points of inflection occur at the midpoints of all

three members, and therefore hinges are placed at these points. The

reactions and moment diagrams for each member can therefore be

determined by dismembering the frame at the hinges and applying

the equations of equilibrium to each of the four parts. The results are

shown in Fig. 7–8c. Note that, as in the case of the pin-connected portal,

the horizontal reactions (shear) at the base of each column are equal.

The moment diagram for this frame is indicated in Fig. 7–8d.

Partial Fixity. Since it is both difficult and costly to construct a

perfectly fixed support or foundation for a portal frame, it is conservative

and somewhat realistic to assume a slight rotation occurs at the supports,

Fig. 7–9a. As a result, the points of inflection on the columns lie

somewhere between the case of having a pin-supported portal, Fig. 7–7a,

where the “inflection points” are at the supports (base of columns), and a

fixed-supported portal, Fig. 7–8a, where the inflection points are at the

center of the columns. Many engineers arbitrarily define the location at

,Fig.7–9b, and therefore place hinges at these points, and also at the

center of the girder.

h>3

(b)

 

assumed

hinges

(a)

—

(c)

—

(d)

moment

diagram

—

Fig. 7–8

7.4 PORTAL FRAMES AND TRUSSES 275

Trusses. When a portal is used to span large distances, a truss may be

used in place of the horizontal girder. Such a structure is used on large

bridges and as transverse bents for large auditoriums and mill buildings.

A typical example is shown in Fig. 7–10a. In all cases, the suspended truss

is assumed to be pin connected at its points of attachment to the columns.

Furthermore, the truss keeps the columns straight within the region of

attachment when the portal is subjected to the sidesway Fig. 7–10b.

Consequently, we can analyze trussed portals using the same assumptions

as those used for simple portal frames. For pin-supported columns,

assume the horizontal reactions (shear) are equal, as in Fig. 7–7c.For

fixed-supported columns, assume the horizontal reactions are equal and

an inflection point (or hinge) occurs on each column, measured midway

between the base of the column and the lowest point of truss member

connection to the column. See Fig. 7–8c and Fig. 7–10b.

The following example illustrates how to determine the forces in the

members of a trussed portal using the approximate method of analysis

described above.

¢,

(

)

(b)

assumed

hinges

—

(a)

(b)

—



—

assumed

hinges

Fig. 7–10

Fig. 7–9

276 CHAPTER 7APPROXIMATE ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES

Determine by approximate methods the forces acting in the members

of the Warren portal shown in Fig. 7–11a.

EXAMPLE 7.4

SOLUTION

The truss portion B, C, F, G acts as a rigid unit. Since the supports

are fixed, a point of inflection is assumed to exist

above A and I, and equal horizontal reactions or shear act at the base

of the columns, i.e., With these

assumptions, we can separate the structure at the hinges J and K,

Fig. 7–11b, and determine the reactions on the columns as follows:

Lower Half of Column

Upper Portion of Column

d+©M

= 0;

-4015.52+ N182= 0

N = 27.5 kN

d+©M

= 0;

M - 3.51202= 0

M = 70 kN

V = 40 kN>2 = 20 kN.©F

= 0;

7 m>2 = 3.5 m

(a)

8 m

4 m

2 m

7 m

4 m

40 kN

3.5 m

40 kN

V  20 kN

3.5 m

V  20 kN

5.5 m

(

)

Fig. 7–11

7.4 PORTAL FRAMES AND TRUSSES 277

Using the method of sections, Fig. 7–11c, we can now proceed to

obtain the forces in members CD, BD, and BH.

Ans.

In a similar manner, show that one obtains the results on the free-

body diagram of column FGI in Fig. 7–11d. Using these results, we can

now find the force in each of the other truss members of the portal

using the method of joints.

Joint

, Fig. 7–11

Ans.

Joint

, Fig. 7–11

Ans.

These results are summarized in Fig. 7–11g.

sin 45° - 38.9 sin 45° = 0

= 0;

75 - 2138.9 cos 45°2- F

= 0 F

= 20 kN 1C2

©F

= 0;

sin 45° - 38.9 sin 45° = 0 F

= 0;

= 27.5 kN 1T2F

= 0;

= 75 kN 1C2-2013.52 - 40122 + F

122= 0d+©M

= 0;

= 38.9 kN 1T2-27.5 + F

sin 45° = 0+

©F

= 0;

40 kN

20 kN

2 m

3.5 m

27.5 kN

2 m

45

(c)

20 kN

2 m

3.5 m

27.5 kN

2 m

45

35 kN

38.9 kN

27.5 kN

(d)

45 45

75 kN

38.9 kN

(e)

45

27.5 kN

38.9 kN

27.5 kN

(f)

40 kN

20 kN

70 kNm

27.5 kN

70 kNm

20 kN

75 kN(C) 20 kN(C) 35 kN(T)

27.5 kN (T)

38.9 kN (T)

27.5 kN (C)

38.9 kN (C)

38.9 kN (T)

(g)

38.9 kN (C)

278 CHAPTER 7APPROXIMATE ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES

Prob. 7–13

6 m

AB CD

EF G H

8 m 6 m6 m

3 kN/m

Prob. 7–14

400 lb/ft

15 ft 20 ft

Prob. 7–15

8 m

5 kN/m

9 kN/m

Prob. 7–16

AB D

HIJ

5 kN/m5 kN/m

3 kN/m

8 m 8 m 8 m

7–13. Determine (approximately) the internal moments

at joints A and B of the frame.

7–15. Determine (approximately) the internal moment at

A caused by the vertical loading.

PROBLEMS

7–14. Determine (approximately) the internal moments

at joints F and D of the frame.

*7–16. Determine (approximately) the internal moments

at A and B caused by the vertical loading.

7.4 PORTAL FRAMES AND TRUSSES 279

Prob. 7–17

20 ft

40 ft 30 ft

0.5 k/ft

1.5 k/ft 1.5 k/ft

Prob. 7–18

15 ft 20 ft

400 lb/ft

1200 lb/ft

7–18. Determine (approximately) the support actions at

A, B, and C of the frame.

*7–20. Determine (approximately) the internal moment

and shear at the ends of each member of the portal frame.

Assume the supports at A and D are partially fixed, such

that an inflection point is located at h/3 from the bottom of

each column.

7–17. Determine (approximately) the internal moments

at joints I and L. Also, what is the internal moment at joint

H caused by member HG?

7–19. Determine (approximately) the support reactions

at A and B of the portal frame. Assume the supports are

(a) pinned, and (b) fixed.

Prob. 7–19

6 m

4 m

12 kN

Prob. 7–20