supplement: statically indeterminate frames supplement - cantilever...supplement cantilever method...

Supplement

Cantilever Method

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Supplement: Statically Indeterminate Frames

Approximate Analysis - Cantilever Method

In this supplement, we consider another approximate method of solving statically

indeterminate frames subjected to lateral loads known as the “Cantilever Method.”

• Like the “Portal Method,” this approximate analysis provides a means to solve a

statically indeterminate problem using a simple model of the structure that is

statically determinate.

• This method is more accurate than the “Portal Method” for tall and narrow

buildings.

Statically Indeterminate Frames

Assumptions for the analysis of statically indeterminate frames by the cantilever

method include the following.

• A tall building is conceptualized as a cantilever beam as far as the axial stresses in

the columns are concerned.

• Axial stresses in the columns of a building frame vary linearly with distance from

the center of gravity of the columns.

- The linear variation is on stress, not the axial force.

• Points of inflection are located at the mid-points of all the beams (horizontal

members).

- A “point of inflection” is where the bending moment changes from positive

bending to negative bending.

- Bending moment is zero at this point.

• Points of inflection are located at the mid-heights of all the columns (vertical

members).

• The axial stresses in the columns at

each story level vary as the distances

of the columns from the center of

gravity of the columns.

• It is usually assumed that all columns

in a story are of equal area, at least

for preliminary analysis.

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Cantilever Method

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Example Problem: Statically Indeterminate Frame

Given: The 3-story frame, loaded as

shown.

Find: Analyze the frame to determine

the approximate axial forces, shear

forces, and bending moments in each

member using the cantilever method.

Comments

If we take a cut at each floor, we

expose three actions (axial force,

shear force, and bending moment) in

each of the 3 columns.

• For each cut, there are only 3 equations of equilibrium, but 9 unknowns.

• For each floor, there are 6 more unknowns than equations of equilibrium.

For the three cuts (three floors), the frame is statically indeterminate (internally)

to the 18th degree: 27 unknowns and only 9 equations of equilibrium for 3 free-

body diagrams.

• By introducing the hinges at the

mid-point of each beam and at the

mid-height of each column, we

make 15 assumptions.

• In addition, for each floor we make

assumptions regarding the

variation of the column stress

(relating 3 unknowns to a single

unknown) - an additional 6

assumptions.

• In total, we made 21 assumptions,

which allow the frame to be

analyzed using the equations of

equilibrium.

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Cantilever Method

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Solution

Top story

Find the “center of gravity (centroid)” of the columns in the top story.

• Assume in this problem that all columns in the top story have equal areas (and let

that area = A).

x = (∑xiAi )/(∑Ai ) = 0 (A) + 12(A) + 30 (A) = 14.0

3A

For this problem, because the columns are of equal area, the column axial forces will

vary linearly as well as the axial stress.

Let F be the force in the left column.

• Relative axial forces in the other columns are (2/14) F in the middle column and

(16/14) F in the right column.

Then, sum moments about the mid-point of the left column (i.e. point O).

∑MO = 0 = - 6 (12) - (2/14) F (12) + (16/14) F (30)

0 = - 72 - 1.7143 F + 34.2857 F

32.5714 F = 72

F = 2.21 kips

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Cantilever Method

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A similar procedure is carried out for the other stories.

2nd story

∑MO = 0 = - 12 (18) - 6 (12) - (2/14) F (12) + (16/14) F (30)

0 = - 216 - 72 - 1.7143 F + 34.2857 F

32.5714 F = 288

F = 8.84 kips

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1st story

∑MO = 0 = - 12 (30) - 12 (18) - 6 (12) - (2/14) F (12) + (16/14) F (30)

0 = - 360 - 216 - 72 - 1.7143 F + 34.2857 F

32.5714 F = 648

F = 19.89 kips

Summary of the axial loads

Left - (F) Middle - (2/14)F Right - (16/14)F

Top story + 2.21 + 0.32 - 2.53

Middle story + 8.84 + 1.26 - 10.10

Bottom story + 19.89 + 2.84 - 22.74

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Cantilever Method

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The final step is to separate the frame into its component free-body diagrams, making

cuts at the hinge locations to determine the rest of the internal forces.

Using the top left corner as a FBD:

∑MO = 0 = 2.21 (6) – V2 (6)

V2 (6) = 13.26

V2 = 2.21 kips ←

∑Fx = 0 = - 2.21 + 12 + N1

N1 = - 9.79

N1 = 9.79 kips ←

∑Fy = 0 = - 2.21 + V1

V1 = 2.21 kips ↑

Using the top center portion as a FBD:

∑MO = 0 = + 2.21 (15) + 0.32 (9) – V2 (6)

V2 (6) = 33.15 + 2.88 = 36.03

V2 = 6.00 kips ←

∑Fx = 0 = 9.79 – 6.00 + N1

N1 = - 3.79

N1 = 3.79 kips ←

∑Fy = 0 = - 2.21 – 0.32 + V1

V1 = 2.53

V1 = 2.53 kip ↑

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Using the top right corner as a FBD:

∑Fx = 0 = 3.79 – V1

V1 = 3.79 kip ←

The summation diagram for the top floor is shown below.

The other floors are solved in a similar manner.

A “short cut” version of this method of analysis may be used.

• The principle of equilibrium is still followed; however, separate FBDs and

equilibrium equations are not developed for each successive portion of the frame.

• Instead, portions of the frame are drawn and the equations of equilibrium for each

portion of the frame are performed mentally without writing them out.

The summation diagrams are shown on the following pages.

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Cantilever Method

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FBD of All Floors (Axial Forces and Shear Forces)

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Cantilever Method

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FBD of All Floors FBD of All Floors

(Bending Moment in the Columns) (Bending Moment in the Beams)

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Cantilever Method

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Comparison with other methods

If the frame is designed according to the forces determined from the Portal Method

or Cantilever Method, the frame will be safe but not necessarily efficient.

To compare the Portal Method and Cantilever Method with a structural analysis using

computer software (STRUDL), three cases were investigated.

• Case 1: The EI (where E is Young’s modulus and I is the moment of inertia for the

member) values for each member is proportional to the maximum bending moment

in that member according to the Portal Method.

• Case 2: The EI value in each member is proportional to the maximum bending

moment in that member according to the Cantilever Method.

• Case 3: The EI value is the same for all members (i.e. all areas are equal).

Reference for Members Relative Moment of Inertia

Values for Case 1

Relative Moment of Inertia Relative Moment of Inertia

Values for Case 2 Values for Case 3

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Cantilever Method

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Member

Axial Forces Shear Forces Bending Moments

Port

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Met

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Can

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Met

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STRUDL Analysis

Port

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Met

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Can

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Met

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STRUDL Analysis

Port

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Met

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Can

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Met

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STRUDL Analysis

Case

1

Case

2

Case

3

Case

1

Case

2

Case

3

Case 1 Case 2 Case 3

Near Far Near Far Near Far

Col

umns

1 27.0 19.89 29.01 22.26 29.00 9.0 6.63 9.59 7.24 11.57 54.0 39.78 66.18 48.90 49.92 36.93 86.34 52.47

2 -9.0 2.84 -14.53 -3.39 -15.01 18.0 18.01 18.03 18.03 14.17 108.0 108.06 127.77 88.62 128.73 87.66 96.75 73.26

3 -18.0 -22.73 -14.48 -18.87 -13.99 9.0 11.36 8.38 10.73 10.27 54.0 68.16 61.32 39.18 78.57 50.22 81.15 42.06

4 12.0 8.84 13.66 10.51 14.66 6.0 4.42 6.77 5.19 7.34 36.0 26.52 47.31 39.93 31.68 30.57 38.79 49.32

5 -4.0 1.26 -6.90 -1.68 -7.62 12.0 11.99 12.04 12.04 11.29 72.0 71.94 73.53 70.92 73.65 70.77 64.95 70.50

6 -8.0 -10.10 -6.75 -8.83 -7.04 6.0 7.59 5.19 6.78 5.37 36.0 45.54 31.77 30.57 41.52 39.81 25.71 38.73

7 3.0 2.21 3.43 2.64 4.33 3.0 2.21 3.38 2.59 3.61 18.0 13.26 20.31 20.25 15.57 15.51 15.33 27.99

8 -1.0 0.32 -1.74 -0.42 -2.19 6.0 6.00 6.02 6.02 5.95 36.0 36.00 36.18 36.03 36.18 36.00 30.06 41.34

9 -2.0 -2.53 -1.70 -2.23 -2.15 3.0 3.79 2.61 3.39 2.44 18.0 22.74 15.69 15.57 20.43 20.31 7.92 21.33

Beam

s

10 -9.0 -9.79 -9.18 -9.95 -7.78 15.0 11.05 15.35 11.75 14.34 90.0 66.30 90.21 94.05 68.61 72.33 91.26 80.85

11 -3.0 -3.77 -3.18 -3.96 -4.90 10.0 12.63 7.72 10.04 6.95 90.0 113.67 68.10 70.95 88.95 91.74 57.36 67.77

12 -9.0 -9.79 -8.61 -9.40 -8.27 9.0 6.63 10.22 7.87 10.32 54.0 39.78 60.21 62.43 46.11 48.27 64.65 59.25

13 -3.0 -3.80 -2.59 -3.38 -2.93 6.0 7.57 5.05 6.61 4.89 54.0 68.13 44.64 46.26 58.68 60.24 41.31 46.68

14 -9.0 -9.79 -8.62 -9.41 -8.39 3.0 2.21 3.44 2.65 4.33 18.0 13.26 20.25 21.00 15.51 16.26 27.99 24.00

15 -3.0 -3.79 -2.61 -3.39 -244 2.0 2.53 1.70 2.27 2.15 18.0 22.77 15.03 15.57 19.77 20.31 17.34 21.33

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Cantilever Method

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Example Problem - Statically Indeterminate Frame

Given: The 2-story frame

loaded as shown.

Find: Analyze the frame to

determine the approximate

axial forces, shear forces,

and bending moments in each

member using the cantilever

method.

Solution

Find the “center of gravity (centroid)” of the columns in the top story.

• In this example, the columns have different cross-sectional areas.

x = (∑xiAi )/(∑Ai ) = 0 (625) + 18(500) + 31.5 (375) + 54 (625)

625 + 500 + 375 + 625

x = 0 + 9,000 + 11,812.5 + 33,750 x = 25.68’

2,125

For this problem, because the columns are not of equal area, only the column axial

stresses will vary linearly.

• The column axial forces will not vary linearly.

Following is a summary of the relative axial stresses and axial loads in the columns.

• Set “f” as the stress in the left column.

Left column Second column Third column Right column

Stress f (7.68/25.68)f -(5.82/25.68)f –(28.32/25.68)f

Axial Load 625f (7.68/25.68)(500)f -(5.82/25.68)(375)f –(28.32/25.68)(625)f

Axial Load 625f 149.53f -84.99f -689.25f

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Then, sum moments about the mid-point of the left column.

∑MO = 0 = - 36 (6) - 149.53f (18) + 84.99f (31.5) + 689.25f (54)

0 = - 216 - 2,691.54f + 2,677.19f + 37,219.50f

37,205.15f = 216

f = 0.00581 ksi

The corresponding axial loads are 3.63 k in the left column, 0.87 k in the second

column, -0.49 k in the third column, and -4.00 k in the right column.

A similar procedure is carried out for the first story.

1st story

∑MO = 0 = - 36 (20) - 45 (8) - 149.53f (18) + 84.99f (31.5) + 689.25f (54)

0 = - 720 - 360 - 2,691.54f + 2,677.19f + 37,219.50f

37,205.15f = 1,080

f = 0.0290 ksi

The corresponding axial loads are 18.14 k in the left column, 4.34 k in the second

column, -2.47 k in the third column, and -20.01 k in the right column.

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Summary of the axial loads

Left column Second column Third column Right column

Top story 3.63 0.87 -0.49 -4.00

Bottom story 18.14 4.34 -2.47 -20.01

The final step is to separate the frame into its component free-body diagrams, making

cuts at the hinge locations to determine the rest of the internal forces.

The summation diagrams follow.

FBD of All Floors (Axial Forces and Shear Forces)

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Cantilever Method

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FBD of All Floors

(Bending Moment in the Columns)

FBD of All Floors

(Bending Moment in the Beams)

supplement: statically indeterminate frames supplement - cantilever...supplement cantilever method...

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