Approximate Lateral Load Analysis by Portal Method

Portal Frame

Portal frames, used in several Civil Engineering structures like buildings, factories, bridges have the primary

purpose of transferring horizontal loads applied at their tops to their foundations. Structural requirements

usually necessitate the use of statically indeterminate layout for portal frames, and approximate solutions are

often used in their analyses.

(i) (ii) (iii) (iv)

Portal Frame Structures

Assumptions for the Approximate Solution

In order to analyze a structure using the equations of statics only, the number of independent force

components must be equal to the number of independent equations of statics.

If there are n more independent force components in the structure than there are independent equations of

statics, the structure is statically indeterminate to the nth degree. Therefore to obtain an approximate solution

of the structure based on statics only, it will be necessary to make n additional independent assumptions. A

solution based on statics will not be possible by making fewer than n assumptions, while more than n

assumptions will not in general be consistent.

Thus, the first step in the approximate analysis of structures is to find its degree of statical indeterminacy

(dosi) and then to make appropriate number of assumptions.

For example, the dosi of portal frames shown in (i), (ii), (iii) and (iv) are 1, 3, 2 and 1 respectively. Based on

the type of frame, the following assumptions can be made for portal structures with a vertical axis of

symmetry that are loaded horizontally at the top

1. The horizontal support reactions are equal

2. There is a point of inflection at the center of the unsupported height of each fixed based column

Assumption 1 is used if dosi is an odd number (i.e., = 1 or 3) and Assumption 2 is used if dosi 1.

Some additional assumptions can be made in order to solve the structure approximately for different loading

and support conditions.

3. Horizontal body forces not applied at the top of a column can be divided into two forces (i.e.,

applied at the top and bottom of the column) based on simple supports

4. For hinged and fixed supports, the horizontal reactions for fixed supports can be assumed to be four

times the horizontal reactions for hinged supports

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Portal Frame

Portal frames, used in several Civil Engineering structures like buildings, factories, bridges have the primary

purpose of transferring horizontal loads applied at their tops to their foundations. Structural requirements

usually necessitate the use of statically indeterminate layout for portal frames, and approximate solutions are

often used in their analyses.

(i) (ii) (iii) (iv)

Portal Frame Structures

Assumptions for the Approximate Solution

In order to analyze a structure using the equations of statics only, the number of independent force

components must be equal to the number of independent equations of statics.

If there are n more independent force components in the structure than there are independent equations of

statics, the structure is statically indeterminate to the n

degree. Therefore to obtain an approximate solution

of the structure based on statics only, it will be necessary to make n additional independent assumptions. A

solution based on statics will not be possible by making fewer than n assumptions, while more than n

assumptions will not in general be consistent.

Thus, the first step in the approximate analysis of structures is to find its degree of statical indeterminacy

(dosi) and then to make appropriate number of assumptions.

For example, the dosi of portal frames shown in (i), (ii), (iii) and (iv) are 1, 3, 2 and 1 respectively. Based on

the type of frame, the following assumptions can be made for portal structures with a vertical axis of

symmetry that are loaded horizontally at the top

1. The horizontal support reactions are equal

2. There is a point of inflection at the center of the unsupported height of each fixed based column

Assumption 1 is used if dosi is an odd number (i.e., = 1 or 3) and Assumption 2 is used if dosi 1.

Some additional assumptions can be made in order to solve the structure approximately for different loading

and support conditions.

3. Horizontal body forces not applied at the top of a column can be divided into two forces (i.e.,

applied at the top and bottom of the column) based on simple supports

4. For hinged and fixed supports, the horizontal reactions for fixed supports can be assumed to be four

times the horizontal reactions for hinged supports

Example

Draw the axial force, shear force and bending moment diagrams of the frames loaded as shown below.

(i) (ii)

Solution

(i) For this frame, dosi = 3 3 + 4 3 4 = 1; i.e., Assumption 1 H

= H

= 10/2 = 5 k

M

= 0 10 10 V

15 = 0 V

= 6.67 k

F

= 0 V

+ V

= 0 V

= 6.67 k

Reactions AFD (k) SFD (k) BMD (k-ft)

(ii) dosi = 3 3 + 6 3 4 = 3

Assumption 1 H

= H

= 10/2 = 5 k, Assumption 2 BM

= BM

BM

= 0 H

5 + M

= 0 M

= 25 k-ft; Similarly BM

= 0 M

= 25 k-ft

M

= 0 25 25 + 10 10 V

15 = 0 V

= 3.33 k

F

= 0 V

+ V

= 0 V

= 3.33 k

Reactions AFD (k) SFD (k) BMD (k-ft)

25 k-ft

6.67 k

5 k 5 k

E

F

D

C

B

A

10 k

3.33 k 3.33 k

5 k 5 k

10 k

25 k-ft

D

B C

A

10 k

Analysis of Multi-storied Structures by Portal Method

Approximate methods of analyzing multi-storied structures are important because such structures are

statically highly indeterminate. The number of assumptions that must be made to permit an analysis by

statics alone is equal to the degree of statical indeterminacy of the structure.

Assumptions

The assumptions used in the approximate analysis of portal frames can be extended for the lateral load

analysis of multi-storied structures. The Portal Method thus formulated is based on three assumptions

1. The shear force in an interior column is twice the shear force in an exterior column.

2. There is a point of inflection at the center of each column.

3. There is a point of inflection at the center of each beam.

Assumption 1 is based on assuming the interior columns to be formed by columns of two adjacent bays or

portals. Assumption 2 and 3 are based on observing the deflected shape of the structure.

Example

Use the Portal Method to draw the axial force, shear force and bending moment diagrams of the three-storied

frame structure loaded as shown below.

E

A

Beam BMD (k-ft) Beam SFD (k) Column AFD (k)

Column shear forces are at the ratio of 1:2:2:1.

Shear force in (V) columns IM, JN, KO, LP are

[18 1/(1 + 2 + 2 + 1) =] 3

, [18 2/(1 + 2 + 2 + 1) =] 6

respectively. Similarly,

V

, V

; and

V

, V

Bending moments are

M

, M

M

, M

M

, M

M

I

F

B

N

J

G

C

O

K

H

P

L

D

The rest of the calculations follow from the free-body diagrams

Column SFD (k)

Column BMD (k-ft) Beam AFD (k)

Problems on Lateral Load Analysis by Portal Method

1. The figure below shows the shear forces (kips) in the interior columns of a two-storied frame. Use the

Portal Method to calculate the corresponding

(i) applied loads P

and P

, (ii) column bending moments, (iii) beam axial forces.

2. The figure below shows the applied loads (F

, F

) and shear force (V

) in column EF of a two-storied

frame. If F

= 10 k, and V

= 5 k, use the Portal Method to calculate the

(i) applied load F

, (ii) maximum column bending moments.

3. For the structure shown in Question 2, use the Portal Method to calculate the lateral loads F

, F

if the

axial forces in beams AD and BE are 10 kips and 15 kips respectively.

4. For the structure shown below, use the Portal Method to

(i) draw the bending moment diagrams of the top floor beams AB and BC

(i) calculate the applied load F

if the maximum bending moment in column EH is 30 k-ft.

5. The figure below shows the exterior column shear forces (kips) in a four-storied fame.

Calculate (i) the applied loads, (ii) beam shear forces.

P

F

D

E

V

F

A

C

B

I

G

H

F

D

C

B

F

= 10 k

E

A

Results from ‘Exact’ Structural Analysis

E

A

Beam BMD (k-ft)

M

I

F

B

N

J

G

C

O

K

H

P

L

D

All members have equal cross-sections

Beam AFD (k)

Beam SFD (k)

Column AFD (k)

Column BMD (k-ft)

Column SFD (k)

Beam BMD (k-ft)

Beam AFD (k)

Beam SFD (k)

Column AFD (k)

Column BMD (k-ft)

Column SFD (k)

Interior columns have twice the area of exterior columns

Problems on Lateral Load Analysis by Cantilever Method

1. The figure below shows the axial forces (kips) in the exterior columns of a two-storied frame.

If the cross-sectional area of column ABC is twice the area of the other columns, use the Cantilever

Method to calculate the corresponding applied loads P

and P

2. For the structure shown below, use the Cantilever Method to calculate the lateral loads F

, F

if the shear

forces in beams AB and DE are 10 kips and 15 kips respectively. Assume all the columns have the same

cross-sectional area.

3. Use the Cantilever Method to draw the axial force, shear force and bending moment diagrams of the

three-storied structure loaded as shown below.

4. Figure (a) below shows the exterior column axial forces (kips) in a three-storied fame.

Use the Cantilever Method to calculate (i) the applied loads, (ii) beam bending moments, (iii) column

bending moments. Assume all the columns to have equal cross-sectional areas.

Fig. (a) Fig. (b)

5. Figure (b) above shows the column shear forces (kips) in a three-storied fame.

Calculate the column BM, beam BM, beam SF and column AF.

Also check if they satisfy the conditions for Cantilever Method (for equal column areas).

E

A

M

I

F

B

N

J

G

C

O

K

H

P

L

D

Columns with area 2A

C

B

A

P

J

H I

K

F

D

C

E

G

F

A

B

Example

Analyze the three-storied frame structure loaded as shown below using the approximate location of hinges to

draw the axial force, shear force and bending moment diagrams of the beams and columns.

1 k/

E

A

Beam BMD (k-ft)

The maximum positive and negative beam moments and

shear forces are as follows.

For the 15 beam, M

(+)

= 18 k-ft

M

( )

= 10.13 k-ft

V

( )

= 1 15/2 = 7.5 k

For the 10 beam, M

(+)

= 8 k-ft

M

( )

= 4.5 k-ft

V

( )

= 1 10/2 = 5 k

Axial Force P in all the beams = 0

M

I

F

B

N

J

G

C

O

K

H

P

L

D

The rest of the calculations follow from the free-body diagrams

Beam SFD (k) Column AFD (k)

Column BMD (k-ft)

Column SFD (k)

Beam AFD (k)

Beam BMD (k-ft)

Using the ACI coefficients (for pattern loading)

Beam SFD (k)

Column AFD (k)

Approximate Analysis of Bridge Portal and Mill Bent

Bridge Portals and Mill Bents

Portals for bridges or bents for mill buildings are often arranged in a manner to include a truss between two

flexural members. In such structures, the flexural members are continuous from the foundation to the top and

are designed to carry bending moment, shear force as well as axial force. The other members that constitute

the truss at the top of the structure are considered pin-connected and to carry axial force only.

Such a structure can be statically indeterminate to the first or third degree, depending on whether the

supports are assumed hinged of fixed. Therefore, the same three assumptions made earlier for portal frames

can be made for the approximate analysis of these structures also; i.e., for a load applied at the top

1. The horizontal support reactions are equal

2. There is a point of inflection at the center of the unsupported height of each fixed based column

Example

In the bridge portal loaded as shown below, draw the bending moment diagrams of columns AB and CD.

F

E

D

C

A

B

Bridge Portal

Mill Bent

1 k/ft

Assuming the total load to be applied equally (i.e., 25/2 =

12.5 k and 12.5 k) at A and B, the horizontal reactions are

H

= 12.5 + 12.5/2 = 18.75 k, H

= 12.5/2 = 6.25 k

Also, BM = 0 at the midpoint of the free height; i.e., at 15/

= 7.5 from the bottom.

M

+ 18.75 7.5 7.5 7.5/2 = 0 M

= 112.5 k-ft

M

+ 6.25 7.5 = 0 M

= 46.88 k-ft

M

= 0 112.5 46.88 + 25 12.5 V

V

= 7.66 k

F

= 0 V

+ V

= 0 V

= 7.66 k

H

V

M

V

M

112.5 k-ft

7.66 k

18.75 k

46.88 k-ft

7.66 k

6.25 k

R

BMD (k-ft) of AB

BMD (k-ft) of CD

Problems on Approximate Analysis of Bridge Portal, Mill Bent and Truss

1. In the mill bent shown below, use the portal method to calculate the axial forces in members BG and EH

and draw the shear force and bending moment diagrams of ABC and DEF.

2. In the mill bent shown below,

(i) Use the Portal Method to draw the bending moment diagram of the member KLM.

(ii) Calculate the forces in EG and FH, assuming them to take equal share of the sectional shear.

3. In the bridge portal shown below, compression in member DG is 10 kips. Use the Portal Method to

(i) calculate the load w per unit length, assuming the diagonal members to share the sectional shear

force equally.

(ii) draw the BMD and SFD of the member FGH for the value of w calculated in (i).

B

L

I

J

F

H

G

C

E N

M

K

D

A

10 k

w kip per ft

B

H

G

F

C

A

D

E

G H

1 k/ft

D

E

F

C

B

A

10 k

4. In the structure shown below,

(i) Use the Portal Method to calculate the reactions at support A, G and draw the BMD of ABC.

(ii) Calculate the forces in members CD, BE, CF, assuming diagonal members to take tension only.

5. In the bridge portal shown below,

(i) Use the Portal Method to calculate the reactions at support A and force in member BE.

(ii) Calculate the forces in members GI and FH, assuming diagonal members to take tension only.

B

D

I

H

G

C

A

10 k

5 k

E

5 k

F

K

I

H

F

G

10 k

B

1 k/ft

E

C

D

M

L

J

A

Deflection of Truss due to Temperature Change and Misfit

In addition to external loads, a truss joint may deflect due to change in member lengths (i.e., become longer

or shorter than its original length) caused by change in temperature or geometrical misfit of any truss

member (being longer or shorter than its specified length).

In Eq. (5); i.e., = u. dL …...…………………(5)

the tem dL (elongation of a truss member) can also be due to temperature change or fabrication defect of any

truss member.

The change in length due to increase in the temperature T is = T L …...…………………(8)

where = Coefficient of thermal expansion; i.e., change of length of a member of unit length due to unit

change of temperature, T = Change of temperature of a member of length L.

Adding to it a geometric misfit (due to fabrication defect) of L, the total elongation of a truss member

dL = N

L/EA + T L + L …...…………………(9)

from which the equation for truss deflection [Eq. (5)] becomes

= N

dL = N

(N

L/EA + T L + L) …...…………………(10)

Example

Calculate the vertical deflection of joint B of the truss ABCDEF shown below due to

(i) temperature rise of 30 F in the bottom cord members AB and BC,

(ii) fabrication defects resulting in vertical members BF and CE to be made 0.25 shorter than specified

[Given: Coefficient of thermal expansion = 5.5 10

/ F, for all the truss members].

(i) For members AB and BC, = 5.5 10

/ F, T = 30 F, L = 20 ft = 240 in

dL = T L = (5.5 10

) (30) (240) = 0.0396 in

Ignoring zero force members,

B,v

= (0.0396) ( -0.5 ) + (0.0396) ( -0.5 ) = 0.0396 in

(ii) For members BF and CE, dL = 0.25 in

Ignoring zero force members,

B,v

= ( 0.25) ( -1 ) + ( 0.25) ( 0 ) = 0.25 in

Support Settlement

Settlement of supports due to consolidation or instability of the subsoil/foundation is a major reason of

deflection of structures. There is a fundamental difference between the effect of support settlement on

statically determinate and indeterminate structures. While it causes deflection due to geometrical changes

only in statically determinate structures, it induces internal stresses in statically indeterminate structures

(which may even be more significant than the forces due to external loads).

The effect of support settlement on statically indeterminate structures is dealt separately but the following

figure shows the deflected shape of the truss ABCDEF shown above due to settlement of support C.

A

B

C

F E

D

dL (in) N 1

Problems on Truss Deflection by the Method of Virtual Work

Assume EA/L = 500 k/ft, = 5.5 10

/ F for the following trusses.

1. Calculate

E,h

due to

(i) The external load

(ii) T = 50 F for CD and CE. 12

2. Calculate

B-C(rel)

due to 15 k 15 k

(i) The external loads

(ii) L = 0.5 for CD and CE. C D E

A B

3. Calculate

A-C(rel)

due to 10 k C

(i) The external loads

(ii) T = 50 F for AB and AD.

B D

10 k

4. Calculate

C,v

and

C,h

due to the external loads.

5. Calculate

B,v

and

D(along AD)

due to the external loads.

20 k

10 k

A

B

C

D

A

B

C

D

E

10 k

A

G

F

E

D

C

H

B k

Deflection due to Combined Flexural, Shear and Axial Deformations

EA = 400 10

k, GA

k, EI = 40 10

k-ft

Calculate the

(i) horizontal deflection at C (

C,h

(ii) vertical deflection at D (

D,v

(k) v

(k) m

(k )

For horizontal deflection at C (

C,h

D,v

= (x

x

/EA) dS + (v

v

/GA

) dS + (m

m

/EI) dS

= 0.0844 ft

For vertical deflection at D (

D,v

= (x

x

/EA) dS + (v

v

/GA

) dS + (m

m

/EI) dS

= 0.125 ft

B C

D

A

10 k

Approximate Lateral Load Analysis by Portal Method, Exercises of Statics

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