a portal frame

7/28/2019 A Portal Frame

1/14

5-3 A PortalFrame

Start GOYA-P to find the portal frame1 shown in Fig. 5-3-1. Apply a horizontal force at

the support. You will find diagrams showing distributions of axial force, shear force and

bending moment (Fig. 5-3-2).

Figure 5-3-3a shows possible reactions for the frame as well as the applied force. The

equilibrium of horizontal forces requires Ax DR F= . The equilibrium of vertical forces and

1 A frame is a structure that comprises a number of members. In this section, we consider a

portal frame which comprises three members.

60


2/14

moments requires 0== ByAy RR . Thus, the actual reactions should be as shown in Fig. 5-3-

3b.

As we have done in the previous section, we consider the free-body diagrams shown in Fig.

5-3-4 to obtain axial-force, shear-force, and bending-moment diagrams shown in Fig. 5-3-

2. Note that the free-body diagrams in Figs. 5-3-4b and c yield the same axial forces and

bending moments having the same magnitudes.

Figure 5-3-5 shows the deflected shape of the portal frame. Make a specialn effort to relate

the deflected shape of the frame to its bending moment diagram (Fig. 5-3-2c). Note that the

moment causes tension on the outside and compression on the inside for all three members

of the frame. That condition suggests that all three members should be bent concave

61


3/14

outward. The deflected shape confirms this conclusion. To become a competent structural

engineer, one should always go through this type ofreasoning and relate bending moment

distribution to deflected shape.

Because the bending moment in the beam is uniform and the slope at mid-span is known to

be zero in this symmetric case, the slope of the beam end ( in Fig. 5-3-5b) is determined to

be

2/ 2

0 2 2

LDF LM MLdx

EI EI EI = = = (5.3.1)

The displacement of joint B is obtainedby superposing the contributions of the slope of the

beam (.L) and the flexural deformation of column AB given by Eq. 2.8.14.

3 35

3 6

D DB

F L F Lu L

EI EI= + = (5.3.2)

Note that the deformation is symmetrical about the vertical axis of the frame [???]: i.e.

column CD deforms in the same way as column AB. Therefore, the displacement of support

D is:

62


4/14

352

3

Dx B

F Lu u

EI= = (5.3.3)

The lengthL is assumed as 100 mm in GOYA-P. Change the height of the member section

to 22.9 mm so thatEIwill be 1 x 106 N/mm2, and apply a force ofF= 6 N. You will obtain

mm10)103/(1065 66 ==xu .

Assume that a vertical load in applied on the portal frame as shown in Fig. 5-3-6a. Select

the most plausible deflected-shape among those shown in Figs. 5-3-6b through e.

To obtain the answer, we need to determine the reactions. Figure 5-3-7a shows the possible

reactions. The equilibrium of horizontal forces yields 0=AxR . The equilibrium of vertical

forces and moments requires / 2Ay By yR R F= = . Thus, the reactions should be as shown in

Fig. 5-3-7b. Considering the free-body diagrams shown in Figs. 5-3-7c and d, we obtain the

63


5/14

axial-force, shear-force, and bending-moment diagrams in Fig. 5-3-8. Because the bending

moment is zero in each column, we should reject Figs. 5-3-6 c and d where the columns are

bent. We should also reject Fig. 5-3-6e, in which the beam is concave downward at its ends.

Figure 5-3-6b is the plausible deflected -shape because the beam bends concave upward

and the columns are straight.

Let us determine the magnitude of the deflections and rotations in reference to Fig. 5-3-9.

We note that, for the loads applied, the bending moment in the beam is same as that of a

simply supported beam shown in Fig. 5-3-9. Expressions for the deflection and the end

slope of the beam were given in Section 3-1:

64


6/14

3

48

yF Lv

EI= (3.1.7)

2

16

yF LEI

= (3.1.4)

Because the columns are continuous with the beams at the joints ends, the slopes of the

columns are also as shown in Fig. 5-3-9a. Therefore, the displacement of the roller

support is:

3

28

y

x

F Lu L

EI= = (5.3.4)

You should check the result using GOYA-P.

To have a lateral movement of zero at the support, the inward movement caused by the

lateral force FD (Eq. 5.3.3) should be offset by the outward movement caused by the

vertical force Fy (Eq. 5.3.4). Therefore, we set the right-hand terms of the two equations

equal to one another.

335

3 8

yDF LF L

EI EI= (5.3.5)

from which we obtain the required proportion ofFD andFy.

65


7/14

3

40

D

y

F

F= (5.3.6)

The axial-force, shear-force, and bending-moment diagrams are obtained by superposing

3/40 times the values in Fig. 5-3-2 on those in Fig. 5-3-8 as shown in Figs. 5-3-10b through

d.

The conditions depicted in Fig. 5-3-10a are equivalent to those for in Fig. 5-3-11a where a

portal frame is supported by two pins. The frame deforms as shown in Fig. 5-3-11b to suit

the bending-moment diagram in Fig. 5-3-10d.

66


8/14

Assume that a horizontal load is applied on the portal frame as shown in Fig. 5-3-12a.

Select the most plausible deflected-shape among those in Figs. 5-3-12b through e.

The first step is to determine the reactions. Figure 5-3-13a shows possible reactions. We

note that the height of the frame is equal to its span. The equilibrium of horizontal forces

67


9/14

yields Ax xR F= . The equilibrium of vertical forces and moments requires Ay xR F= and

Ay xR F= . We conclude that the reactions should be as shown in Fig. 5-3-13b.

Considering the free-body diagrams (Figs. 5-3-14a through c), we obtain the axial-force,

shear-force, and bending-moment diagrams shown in Fig. 5-3-15. We look at Fig. 5-3-13c

15c carefully. The moment diagram indicates that the left column and the beam should have

compressive strain on their outside faces. They should bend so that they are concave

outward. The right column, not subjected to moment, should remain straight. We decide

that the plausible deflected shape is the one shown in Fig. 5-3-12e. (Check this using

GOYA-P.)

68


10/14

Let us compute the deformations in reference to Fig. 5-3-16a. Because the two ends of the

beam are connected to the columns, vertical displacements at the two ends of the beam are

negligibly smalle. Therefore, the beam bends as shown in Fig. 5-3-16b. Recall the

following equation:

EI

M

dx

vd=

2

2

(2.8.9)

Substituting the bending moment shown in Fig. 5-3-16b, ( )xM F L x= , and integrating,

we obtain:

2

12

xFdv xxL Cdx EI

= +

(5.3.7) and

69


11/14

2 3

1 22 6

xF x L xv C x C EI

= + +

(5.3.8)

The boundary condition that v = 0 atx = 0 andx =L leads to:

2

13

xF LCEI

= and 02 =C (5.3.9)

Substituting this into Eq. 5.3.6, we obtain the slopes of the beam ends:

2

0 3

xB

x

F Ldv

dx EI

=

= = and2

6

xB

x L

F Ldv

dx EI

=

= = (5.3.10)

The horizontal displacement of the beam (uA in Fig. 5-3-16a) is obtained by adding the

deformation of the left column and the contribution of the rotation:

3 32

3 3

x xB B

F L F Lu L

EI EI= + = (5.3.11)

The displacement of the roller (ux in Fig. 5-3-16a) is obtained by adding the displacement

of the beam and the contribution of the rotation of the right column:

35

6xx B C

F Lu u L

EI= + = (5.3.12)

70


12/14

Figure 5-3-17a shows the deformation of the portal frame subjected to a horizontal force Fx

on the beam. Figure 5-3-17b shows the deformation of the portal subjected to a horizontal

force Fx/2 on the roller support (see Equations 5.3.2 and 5.3.3). Adding the deflections in

the two figures, we obtain Fig. 5-3-17c, which represents the portal frame supported by two

pins and subjected to horizontal force at the top. You should check this using GOYA-P.

Superposing one half of Fig. 5-3-2 and Fig. 5-3-15, we obtain the axial-force, shear-force,

and bending-moment diagrams shown in Figs. 5-3-18a through c. The bending moment

diagram indicates that the portal frame deforms as shown in Fig. 5-3-18d.

71


13/14

Example 5-3-1: Construct the bending-moment diagram for the portal frame shown in

Fig. 5-3-19.

Solution: The reactions and the bending moment caused by the vertical force are shown in

Fig. 5-3-20a. Those caused by the horizontal force are shown in Fig. 5-3-20b. Superposing

these figures, we obtain Fig. 5-3-20c. (DetermineSketch the deflected shape using GOYA-

P.)

72


14/14

Exercise: Take the last digit of your ID#i. Construct the bending moment diagram for

the portal frame shown in Fig. 5-3-19 and sketch the deflected shape.

Hint: Re-examine the process leading to Eq. 5.3.5 very carefully, and project it to the case

where the length of the beam is different from that of the columns.

73

a portal frame

Documents