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"FRAME" --- PORTAL AND GABLE RIGID PLANE FRAME ANALYSIS Program Description: "FRAME" is a spreadsheet program program written in MS-Excel for the purpose of plane frame analysis of portal and gable rigid plane frames subjected to various various types of loading. Specifically, the "stiffness matrix" method of analysis analysis is used to determine the unknown joint displacements, support support reactions, and member end forces. Individual frame members are also analyzed to determine the shears and intermediate moments. Plots of both the shear and moment diagrams are also produced. Also, the frame is drawn for visual confimation of geometry/configuration. geometry/configuration. This program is a workbook consisting of three (3) worksheets, described as follows:

Worksheet Name

Description

Doc Portal Frame Gable Frame

This documentation sheet Portal rigid plane frame analysis Gable rigid plane frame analysis

Program Assumptions and Limitations: 1. This program uses uses the "stiffness matrix" method of analysis and four four (4) following basic basic analysis assumptions: assumptions: a. Members must be of constant cross cross section (E and I are constant for entire entire length). b. Deflections must not significantly alter the geometry of the problem. c. Stress must remain within the "elastic" region. d. Since this is a "first-order", linear analysis, the effects of "P-delta" and shear deformation are not included. (However, significant effects due to shear deformation are limited to very short and deep members.) 2. Additional assumptions and and features are as follows: a. Frame support joints may each be either fixed or pinned. pinned. b. Frame support joints may be at different levels (elevations). c. Columns must be vertical (cannot be sloped). c. For a portal frame, the top (roof) member may be flat or sloped in either direction. 3. A vertical load, horizontal horizontal load, and and externally moment may be applied applied to any of the joints of the frame. frame. These joint loads are are to be applied in "global" axes directions. directions. Note: Joint loads applied applied directly at supports are are merely added directly to support reactions and are not reflected i n member end force values.

Procedure for Stiffness Method of Frame Analysis: 1. 2. 3. 4. 5.

6. 7.

8. 9.

Identifiy members members and joints in frame Specify near (start) joint and and far (end) joint joint for each member in frame Establish global coordinate system Calculate fixed-end moments (FEM's) (FEM's) and shears for each member member due to applied member loads Specify x, y, and z coding components components (3 in all) at each joint as follows: a. Use lowest numbers to identify unknown joint displacements (for partioning overall matrix) b. Use remaining numbers to indentify known displacements From the problem, establish the known displacements, Dk, and known external forces and reactions, Qk Determine 6x6 stiffness matrix, k', for each of the member expressed expressed in global coordinates coordinates As an example, a member numbered with 1,2, and 3 at start and 4,5, 4,5, and 6 at end would be: 1 2 3 4 5 6 A B -C -A -B -C 1 B D E -B -D E 2 -C E F C -E G 3 -A -B C A B C 4 -B -D -E B D -E 5 -C E G C -E F 6 Note: l l l l A = A*E/L* x^2+12*E*I/L^3* y^2 x and y are the "direction cosines" where: lx = (xj-xi)/L and ly = (yj-yi)/L B = (A*E/L-12*E*I/L^3)* (A*E/L-12*E*I/L^3)*lx*ly l C = 6*E*I/L^2* y (xi,yi) = joint start coordinates D = A*E/L*ly^2+12*E*I/L^3* lx^2 (xj,yj) = joint end coordinates E = 6*E*I/L^2*lx L = SQRT((xj-xi)^2+(yj-yi)^2) F = 4*E*I/L G = 2*E*I/L Merge individual individual member stiffness matrices into stiffness matrix, K, for entire entire frame Partition the structure stiffness matrix, matrix, K, as follows: End Forces Vector Partitioned Stiff. Matrix, K Displacements Vector Du Qk K11 K12 = * Dk Qu K21 K22 Expansion then leads to: Qk = K11 * Du + K12 * Dk (Eqn. 1)

Formulas Used to Determine FEM's and Shear and Moment in Individual Frame Members For Uniform or Distributed Loads: Loading functions for each uniform or distributed load evaluated at distance x = L from left end of member: FvL = -wb*(L-b-(L-e)) -wb*(L-b-(L-e)) + -1/2*(we-wb)/(e-b)*(( -1/2*(we-wb)/(e-b)*((L-b)^2-(L L-b)^2-(L-e)^2)+( -e)^2)+(we-wb) we-wb)*(L-e) *(L-e) FmL = -wb/2*((L-b)^2-(L-wb/2*((L-b)^2-(L-e)^2) e)^2) + -1/6*(we-wb)/(e-b)*(( -1/6*(we-wb)/(e-b)*((L-b)^3-( L-b)^3-(L-e)^3)+( L-e)^3)+(we-wb) we-wb)/2*(L-e)^2 /2*(L-e)^2 FqL = -wb/(6*E*I)*((L-b)^3-(L-e)^3) + -1/(24*E*I)*(we-wb)/(e-b)*((L-b)^4-(L-e)^4)+(we-wb)/(6*E*I)*(L-e)^3 FDL = -wb/(24*E*I)*((L-b)^4-(L-e)^4) + -1/(120*E*I)*(we-wb)/(e-b)*((L-b)^5-(L-e)^5)+(we-wb)/(24*E*I)*(L-e)^4 Loading functions for each uniform or distributed load evaluated at distance = x from left end of member: If x >= e: Fvx = -wb*(x-b-(x-e)) -wb*(x-b-(x-e)) + -1/2*(we-wb)/(e-b)*(( -1/2*(we-wb)/(e-b)*((x-b)^2-( x-b)^2-(x-e)^2)+ x-e)^2)+(we-wb (we-wb)*(x-e) )*(x-e) Fmx = -wb/2*((x-b)^2-(x -wb/2*((x-b)^2-(x-e)^2) -e)^2) + -1/6*(we-wb)/(e-b)*(( -1/6*(we-wb)/(e-b)*((x-b)^3-( x-b)^3-(x-e)^3)+ x-e)^3)+(we-wb (we-wb)/2*(x-e)^ )/2*(x-e)^2 2 Fqx = -wb/(6*E*I)*((x-b)^3-(x-e)^3) + -1/(24*E*I)*(we-wb)/(e-b)*((x-b)^4-(x-e)^4)+(we-wb)/(6*E*I)*(x-e)^3 FDx = -wb/(24*E*I)*((x-b)^4-(x-e)^4) + -1/(120*E*I)*(we-wb)/(e-b)*((x-b)^5-(x-e)^5)+(we-wb)/(24*E*I)*(x-e)^4 else if x >= b: Fvx = -w -wb*(x-b) + -1/2*(we-wb)/(e-b)*(x-b)^2 else: Fvx = 0 Fmx = -wb/2*(x-b)^2 + -1/6*(we-wb)/(e-b)*(x-b)^3-(x-e)^3 else: Fmx = 0 Fqx = -wb/(6*E*I)*(x-b)^3 + -1/(24*E*I)*(we-wb)/(e-b)*(x-b)^4 Fqx = 0 else: FDx = -wb/(24*E*I)*(x-b)^4 + -1/(120*E*I)*(we-wb)/(e-b)*(x-b)^5 FDx = 0 else: For Point Loads: Loading functions for each point load evaluated at distance x = L from left end of member: FvL FvL = -P FmL = -P*(L -P*(L-a) -a) FqL = -P*(L-a)^2/(2*E*I) FDL = P*(L-a)^3/(6*E*I) Loading functions for each point load evaluated at distance = x from left end of member: If x > a:

Formulas Used to Determine FEM's and Shear and Moment in Individual Frame Members (continued) Initial summation values at left end (x = 0) for shear, moment, slope, and deflection: Fixed beam: Vo = Mo = qo = Do = Simple beam: Vo = Mo = qo = Do =

(for determining FEM's in frame frame members) -12*E*I/L^3*S(FDL)-6*E*I/L^2*S(FqL) 6*E*I/L^2*S(FDL)+2*E*I/L*S(FqL) 0 0

-1/L*S(FmL) 0 1/L* 1/L*S(FDL)+L/(6*E*I)*S(FmL) 0

Summations of shear, moment, slope, and deflection at distance = x from left end of member: Shear: Moment: Slope: Deflection:

Vx = Mx = qx = Dx =

Vo+S(Fvx) Mo+Vo*x+ S(Fmx) qo+Mo*x/(E*I)+Vo*x^2/(2*E*I)+ S(Fqx) -(Do-qo*x-Mo*x^2/(2*E*I)-Vo*x^3/(6*E*I)+ S(FDx)

Reference: "Modern Formulas for Statics and Dynamics, A Stress-and-Strain Stress-and-Strain Approach" by Walter D. Pilkey and Pin Yu Chang, McGraw-Hill McGraw-Hill Book Company (1978)

"FRAME.xls" Program Version 1.0

PORTAL RIGID PLANE FRAME ANALYSIS For Fixed or Pinned Bases Job Name: Job Number:

Subject: Originator:

Checker:

Results:

Input Data:

Support Reactions: 16.0

Joint Coordinates:

Y

Joint No. 1 4

14.0

Joint No. 1 2 3 4

x (ft.) 0.0000 0.0000 26.0000 26.0000

y (ft.) 0.0000 13.0000 15.0000 0.0000

12.0

2

10.0 ) . t f ( s 8.0 i x a Y 6.0

3

2

1

Support Constraints:

Memb Member er No. No. Joint Joint No. No. 1 1 2 2 2 3 4 3 3

3

2.0

Condition Fixed Fixed

0.0 0.0

10.0

15.0

20.0

25.0

30.0

Plot of Portal Frame E (ksi) 29000 29000 29000

A (in.^2) 2 0. 0 1 1. 8 2 0. 0

I (in.^4) 723.0 612.0 723.0

L (ft.) 13.0000 26.0768 15.0000

x

Px (kips)

4

X

0.0000 0.9971 0.0000

Portal Frame Nomenclature

1.0000 0.0767 1.0000

Py (kips)

Member No. 1

b/L a/L P wb

Mz (ft-k)

M

2

we

3

x or y

Mz (ft-k) -2.29 38.27

Axial (k) 12.80 -12.80 7.16 -5.16 13.28 -13.28

Shear (k) Moment (ft-k) -2.95 -2.29 2.95 -36.04 12.28 36.04 13.72 -54.71 6.20 38.27 -6.20 54.71

Member Maximum Moments:

c/L e/L

y

Joint Loads: Joint No. 1 2 3 4

1

X-axis (ft.)

Member Properties and Data: Member No. 1 2 3

5.0

Ry (kips) 12.80 13.28

Member End Forces:

4.0

Joint No. 1 4

Rx (kips) 2.95 -6.20

+M or -M +M(max) -M(max) +M(max) -M(max) +M(max) -M(max)

M (ft-k) 2.29 -36.04 39.63 -54.71 54.71 -38.27

x or y (ft.) 0.00 13.00 12.32 26.08 15.00 0.00

L

Joint Displacements: 3.25

Member Load Nomenclature Joint No. 1 2 3 4

0.0000 0.0730 0.0674 0.0000

0.0000 -0.0034 -0.0041 0.0000

Distributed Load #4 wb (k/ft.) e/L

we (k/ft.)

b/L

N o t e : Point

loads or moments at member ends must be input as joint loads.

Member Loads:

x (in.)

y (in.)

z (rad.)

0.0000 -0.0015 0.0008 0.0000

Distributed Loads:

No. 1 2 2 3

Member Load Direct. X-Global Y-Global X-Projected X-Global

b/L 0.0000

Distributed Load #1 wb (k/ft.) e/L -1.0000

1.0000

we (k/ft.)

b/L

Distributed Load #2 wb (k/ft.) e/L

we (k/ft.)

b/L

Distributed Load #3 wb (k/ft.) e/L

we (k/ft.)

b/L

Distributed Load #5 wb (k/ft.) e/L

we (k/ft.)

-1.0000

Point Loads:

No. 1 2 2 3

Member Load Direct.

Point Load #1 P (kips) a/L

Point Load #2 P (kips) a/L

Point Load #3 P (kips) a/L

Point Load #4 P (kips) a/L

Moment #1 M (ft-kips) c/L

Moment #2 M (ft-kips) c/L

Moment #3 M (ft-kips) c/L

Moment #4 M (ft-kips) c/L

Point Load #5 P (kips) a/L

Point Load #6 P (kips) a/L

Point Load #7 P (kips) a/L

Point Load #8 P (kips) a/L

Point Load #9 P (kips) a/L

Point Load #10 P (kips) a/L

X-Global Y-Global X-Global X-Global

Applied Moments: Member No. 1 2 3

5 of 6

11/3/2013 7:48 AM

"FRAME.xls" Program Version 1.0

GABLE RIGID PLANE FRAME ANALYSIS For Fixed or Pinned Bases Job Name: Job Number:

Results:

Subject: Originator:

Support Reactions: Checker: Joint No. 1 5

Input Data: 16.0

Joint Coordinates:

Rx (kips) 1.61 -4.86

Ry (kips) 11.53 14.78

Mz (ft-k) 0.00 0.00

Y 14.0

Joint No. 1 2 3 4 5

x (ft.) 0.0000 0.0000 13.0000 26.0000 26.0000

y (ft.) 0.0000 13.0000 15.0000 13.0000 0.0000

Member End Forces:

3

12.0

2

2

) . 10.0 t f ( s 8.0 i x a Y 6.0

3

1

Memb Member er No. No. Joint Joint No. No. 1 1 2 2 2 3 3 3 4 5 4 4

4

4

4.0

Support Constraints:

2.0

Joint No. 1 5

Condition Pinned Pinned

0.0

0.0

5.0

10.0

15.0

20.0

25.0

30.0

1

5

Member Maximum Moments:

Plot of Gable Frame

Gable Frame Nomenclature Member No.

Member Properties and Data: E (ksi) 29000 29000 29000 29000

A (in.^2) 20.0 11.8 11.8 20.0

I (in.^4) 723.0 612.0 612.0 723.0

Joint No. 1 2 3 4 5

Px (kips)

Py (kips)

Mz (ft-k)

L (ft.) 13.0000 13.1529 13.1529 13.0000

x

0.0000 0.9884 0.9884 0.0000

Shear (k) Moment (ft-k) -1.61 0.00 1.61 -20.89 10.66 20.89 2.34 33.77 -0.87 -33.77 13.87 -63.14 4.86 0.00 -4.86 63.14

X

X-axis (ft.)

Member No. 1 2 3 4

Axial (k) 11.53 -11.53 6.55 -4.55 5.05 -7.05 14.78 -14.78

c/L e/L

y

1.0000 0.1521 -0.1521 1.0000

1

b/L a/L P wb

2 M

we

3

Joint Loads: x or y

4

L

Member Load Nomenclature

+M or -M +M(max) -M(max) +M(max) -M(max) +M(max) -M(max) +M(max) -M(max)

M (ft-k) 0.00 -20.89 36.55 -20.89 33.77 -63.14 63.14 0.00

x or y (ft.) 0.00 13.00 10.78 0.00 0.00 13.15 13.00 0.00

Joint Displacements:

3.25 Joint No. 1 2 3 4 5

0.0000 0.1893 0.2159 0.2423 0.0000

0.0000 -0.0031 -0.1934 -0.0040 0.0000

Distributed Load #4 wb (k/ft.) e/L

we (k/ft.)

b/L

N o t e : Point

loads or moments at member ends must be input as joint loads.

Member Loads:

x (in.)

y (in.)

z (rad.)

-0.0009 -0.0018 0.0004 0.0003 -0.0025

Distributed Loads:

No. 1 2 2 3 3 4

Member Load Direct. X-Global Y-Global X-Projected Y-Global X-Projected X-Global

b/L

Distributed Load #1 wb (k/ft.) e/L

we (k/ft.)

0.0000

-1.0000

1.0000

-1.0000

0.0000

-1.0000

1.0000

-1.0000

b/L

Distributed Load #2 wb (k/ft.) e/L

we (k/ft.)

b/L

Distributed Load #3 wb (k/ft.) e/L

we (k/ft.)

b/L

Distributed Load #5 wb (k/ft.) e/L

we (k/ft.)

Point Loads:

No. 1 2 2 3 3 4

Member Load Direct.

Point Load #1 P (kips) a/L

Point Load #2 P (kips) a/L

Point Load #3 P (kips) a/L

Point Load #4 P (kips) a/L

Moment #1 M (ft-kips) c/L

Moment #2 M (ft-kips) c/L

Moment #3 M (ft-kips) c/L

Moment #4 M (ft-kips) c/L

Point Load #5 P (kips) a/L

Point Load #6 P (kips) a/L

Point Load #7 P (kips) a/L

Point Load #8 P (kips) a/L

Point Load #9 P (kips) a/L

Point Load #10 P (kips) a/L

X-Global Y-Global X-Global Y-Global X-Global X-Global

Applied Moments: Member No. 1 2 3 4

6 of 6

11/3/2013 7:48 AM

View more...
Worksheet Name

Description

Doc Portal Frame Gable Frame

This documentation sheet Portal rigid plane frame analysis Gable rigid plane frame analysis

Program Assumptions and Limitations: 1. This program uses uses the "stiffness matrix" method of analysis and four four (4) following basic basic analysis assumptions: assumptions: a. Members must be of constant cross cross section (E and I are constant for entire entire length). b. Deflections must not significantly alter the geometry of the problem. c. Stress must remain within the "elastic" region. d. Since this is a "first-order", linear analysis, the effects of "P-delta" and shear deformation are not included. (However, significant effects due to shear deformation are limited to very short and deep members.) 2. Additional assumptions and and features are as follows: a. Frame support joints may each be either fixed or pinned. pinned. b. Frame support joints may be at different levels (elevations). c. Columns must be vertical (cannot be sloped). c. For a portal frame, the top (roof) member may be flat or sloped in either direction. 3. A vertical load, horizontal horizontal load, and and externally moment may be applied applied to any of the joints of the frame. frame. These joint loads are are to be applied in "global" axes directions. directions. Note: Joint loads applied applied directly at supports are are merely added directly to support reactions and are not reflected i n member end force values.

Procedure for Stiffness Method of Frame Analysis: 1. 2. 3. 4. 5.

6. 7.

8. 9.

Identifiy members members and joints in frame Specify near (start) joint and and far (end) joint joint for each member in frame Establish global coordinate system Calculate fixed-end moments (FEM's) (FEM's) and shears for each member member due to applied member loads Specify x, y, and z coding components components (3 in all) at each joint as follows: a. Use lowest numbers to identify unknown joint displacements (for partioning overall matrix) b. Use remaining numbers to indentify known displacements From the problem, establish the known displacements, Dk, and known external forces and reactions, Qk Determine 6x6 stiffness matrix, k', for each of the member expressed expressed in global coordinates coordinates As an example, a member numbered with 1,2, and 3 at start and 4,5, 4,5, and 6 at end would be: 1 2 3 4 5 6 A B -C -A -B -C 1 B D E -B -D E 2 -C E F C -E G 3 -A -B C A B C 4 -B -D -E B D -E 5 -C E G C -E F 6 Note: l l l l A = A*E/L* x^2+12*E*I/L^3* y^2 x and y are the "direction cosines" where: lx = (xj-xi)/L and ly = (yj-yi)/L B = (A*E/L-12*E*I/L^3)* (A*E/L-12*E*I/L^3)*lx*ly l C = 6*E*I/L^2* y (xi,yi) = joint start coordinates D = A*E/L*ly^2+12*E*I/L^3* lx^2 (xj,yj) = joint end coordinates E = 6*E*I/L^2*lx L = SQRT((xj-xi)^2+(yj-yi)^2) F = 4*E*I/L G = 2*E*I/L Merge individual individual member stiffness matrices into stiffness matrix, K, for entire entire frame Partition the structure stiffness matrix, matrix, K, as follows: End Forces Vector Partitioned Stiff. Matrix, K Displacements Vector Du Qk K11 K12 = * Dk Qu K21 K22 Expansion then leads to: Qk = K11 * Du + K12 * Dk (Eqn. 1)

Formulas Used to Determine FEM's and Shear and Moment in Individual Frame Members For Uniform or Distributed Loads: Loading functions for each uniform or distributed load evaluated at distance x = L from left end of member: FvL = -wb*(L-b-(L-e)) -wb*(L-b-(L-e)) + -1/2*(we-wb)/(e-b)*(( -1/2*(we-wb)/(e-b)*((L-b)^2-(L L-b)^2-(L-e)^2)+( -e)^2)+(we-wb) we-wb)*(L-e) *(L-e) FmL = -wb/2*((L-b)^2-(L-wb/2*((L-b)^2-(L-e)^2) e)^2) + -1/6*(we-wb)/(e-b)*(( -1/6*(we-wb)/(e-b)*((L-b)^3-( L-b)^3-(L-e)^3)+( L-e)^3)+(we-wb) we-wb)/2*(L-e)^2 /2*(L-e)^2 FqL = -wb/(6*E*I)*((L-b)^3-(L-e)^3) + -1/(24*E*I)*(we-wb)/(e-b)*((L-b)^4-(L-e)^4)+(we-wb)/(6*E*I)*(L-e)^3 FDL = -wb/(24*E*I)*((L-b)^4-(L-e)^4) + -1/(120*E*I)*(we-wb)/(e-b)*((L-b)^5-(L-e)^5)+(we-wb)/(24*E*I)*(L-e)^4 Loading functions for each uniform or distributed load evaluated at distance = x from left end of member: If x >= e: Fvx = -wb*(x-b-(x-e)) -wb*(x-b-(x-e)) + -1/2*(we-wb)/(e-b)*(( -1/2*(we-wb)/(e-b)*((x-b)^2-( x-b)^2-(x-e)^2)+ x-e)^2)+(we-wb (we-wb)*(x-e) )*(x-e) Fmx = -wb/2*((x-b)^2-(x -wb/2*((x-b)^2-(x-e)^2) -e)^2) + -1/6*(we-wb)/(e-b)*(( -1/6*(we-wb)/(e-b)*((x-b)^3-( x-b)^3-(x-e)^3)+ x-e)^3)+(we-wb (we-wb)/2*(x-e)^ )/2*(x-e)^2 2 Fqx = -wb/(6*E*I)*((x-b)^3-(x-e)^3) + -1/(24*E*I)*(we-wb)/(e-b)*((x-b)^4-(x-e)^4)+(we-wb)/(6*E*I)*(x-e)^3 FDx = -wb/(24*E*I)*((x-b)^4-(x-e)^4) + -1/(120*E*I)*(we-wb)/(e-b)*((x-b)^5-(x-e)^5)+(we-wb)/(24*E*I)*(x-e)^4 else if x >= b: Fvx = -w -wb*(x-b) + -1/2*(we-wb)/(e-b)*(x-b)^2 else: Fvx = 0 Fmx = -wb/2*(x-b)^2 + -1/6*(we-wb)/(e-b)*(x-b)^3-(x-e)^3 else: Fmx = 0 Fqx = -wb/(6*E*I)*(x-b)^3 + -1/(24*E*I)*(we-wb)/(e-b)*(x-b)^4 Fqx = 0 else: FDx = -wb/(24*E*I)*(x-b)^4 + -1/(120*E*I)*(we-wb)/(e-b)*(x-b)^5 FDx = 0 else: For Point Loads: Loading functions for each point load evaluated at distance x = L from left end of member: FvL FvL = -P FmL = -P*(L -P*(L-a) -a) FqL = -P*(L-a)^2/(2*E*I) FDL = P*(L-a)^3/(6*E*I) Loading functions for each point load evaluated at distance = x from left end of member: If x > a:

Formulas Used to Determine FEM's and Shear and Moment in Individual Frame Members (continued) Initial summation values at left end (x = 0) for shear, moment, slope, and deflection: Fixed beam: Vo = Mo = qo = Do = Simple beam: Vo = Mo = qo = Do =

(for determining FEM's in frame frame members) -12*E*I/L^3*S(FDL)-6*E*I/L^2*S(FqL) 6*E*I/L^2*S(FDL)+2*E*I/L*S(FqL) 0 0

-1/L*S(FmL) 0 1/L* 1/L*S(FDL)+L/(6*E*I)*S(FmL) 0

Summations of shear, moment, slope, and deflection at distance = x from left end of member: Shear: Moment: Slope: Deflection:

Vx = Mx = qx = Dx =

Vo+S(Fvx) Mo+Vo*x+ S(Fmx) qo+Mo*x/(E*I)+Vo*x^2/(2*E*I)+ S(Fqx) -(Do-qo*x-Mo*x^2/(2*E*I)-Vo*x^3/(6*E*I)+ S(FDx)

Reference: "Modern Formulas for Statics and Dynamics, A Stress-and-Strain Stress-and-Strain Approach" by Walter D. Pilkey and Pin Yu Chang, McGraw-Hill McGraw-Hill Book Company (1978)

"FRAME.xls" Program Version 1.0

PORTAL RIGID PLANE FRAME ANALYSIS For Fixed or Pinned Bases Job Name: Job Number:

Subject: Originator:

Checker:

Results:

Input Data:

Support Reactions: 16.0

Joint Coordinates:

Y

Joint No. 1 4

14.0

Joint No. 1 2 3 4

x (ft.) 0.0000 0.0000 26.0000 26.0000

y (ft.) 0.0000 13.0000 15.0000 0.0000

12.0

2

10.0 ) . t f ( s 8.0 i x a Y 6.0

3

2

1

Support Constraints:

Memb Member er No. No. Joint Joint No. No. 1 1 2 2 2 3 4 3 3

3

2.0

Condition Fixed Fixed

0.0 0.0

10.0

15.0

20.0

25.0

30.0

Plot of Portal Frame E (ksi) 29000 29000 29000

A (in.^2) 2 0. 0 1 1. 8 2 0. 0

I (in.^4) 723.0 612.0 723.0

L (ft.) 13.0000 26.0768 15.0000

x

Px (kips)

4

X

0.0000 0.9971 0.0000

Portal Frame Nomenclature

1.0000 0.0767 1.0000

Py (kips)

Member No. 1

b/L a/L P wb

Mz (ft-k)

M

2

we

3

x or y

Mz (ft-k) -2.29 38.27

Axial (k) 12.80 -12.80 7.16 -5.16 13.28 -13.28

Shear (k) Moment (ft-k) -2.95 -2.29 2.95 -36.04 12.28 36.04 13.72 -54.71 6.20 38.27 -6.20 54.71

Member Maximum Moments:

c/L e/L

y

Joint Loads: Joint No. 1 2 3 4

1

X-axis (ft.)

Member Properties and Data: Member No. 1 2 3

5.0

Ry (kips) 12.80 13.28

Member End Forces:

4.0

Joint No. 1 4

Rx (kips) 2.95 -6.20

+M or -M +M(max) -M(max) +M(max) -M(max) +M(max) -M(max)

M (ft-k) 2.29 -36.04 39.63 -54.71 54.71 -38.27

x or y (ft.) 0.00 13.00 12.32 26.08 15.00 0.00

L

Joint Displacements: 3.25

Member Load Nomenclature Joint No. 1 2 3 4

0.0000 0.0730 0.0674 0.0000

0.0000 -0.0034 -0.0041 0.0000

Distributed Load #4 wb (k/ft.) e/L

we (k/ft.)

b/L

N o t e : Point

loads or moments at member ends must be input as joint loads.

Member Loads:

x (in.)

y (in.)

z (rad.)

0.0000 -0.0015 0.0008 0.0000

Distributed Loads:

No. 1 2 2 3

Member Load Direct. X-Global Y-Global X-Projected X-Global

b/L 0.0000

Distributed Load #1 wb (k/ft.) e/L -1.0000

1.0000

we (k/ft.)

b/L

Distributed Load #2 wb (k/ft.) e/L

we (k/ft.)

b/L

Distributed Load #3 wb (k/ft.) e/L

we (k/ft.)

b/L

Distributed Load #5 wb (k/ft.) e/L

we (k/ft.)

-1.0000

Point Loads:

No. 1 2 2 3

Member Load Direct.

Point Load #1 P (kips) a/L

Point Load #2 P (kips) a/L

Point Load #3 P (kips) a/L

Point Load #4 P (kips) a/L

Moment #1 M (ft-kips) c/L

Moment #2 M (ft-kips) c/L

Moment #3 M (ft-kips) c/L

Moment #4 M (ft-kips) c/L

Point Load #5 P (kips) a/L

Point Load #6 P (kips) a/L

Point Load #7 P (kips) a/L

Point Load #8 P (kips) a/L

Point Load #9 P (kips) a/L

Point Load #10 P (kips) a/L

X-Global Y-Global X-Global X-Global

Applied Moments: Member No. 1 2 3

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"FRAME.xls" Program Version 1.0

GABLE RIGID PLANE FRAME ANALYSIS For Fixed or Pinned Bases Job Name: Job Number:

Results:

Subject: Originator:

Support Reactions: Checker: Joint No. 1 5

Input Data: 16.0

Joint Coordinates:

Rx (kips) 1.61 -4.86

Ry (kips) 11.53 14.78

Mz (ft-k) 0.00 0.00

Y 14.0

Joint No. 1 2 3 4 5

x (ft.) 0.0000 0.0000 13.0000 26.0000 26.0000

y (ft.) 0.0000 13.0000 15.0000 13.0000 0.0000

Member End Forces:

3

12.0

2

2

) . 10.0 t f ( s 8.0 i x a Y 6.0

3

1

Memb Member er No. No. Joint Joint No. No. 1 1 2 2 2 3 3 3 4 5 4 4

4

4

4.0

Support Constraints:

2.0

Joint No. 1 5

Condition Pinned Pinned

0.0

0.0

5.0

10.0

15.0

20.0

25.0

30.0

1

5

Member Maximum Moments:

Plot of Gable Frame

Gable Frame Nomenclature Member No.

Member Properties and Data: E (ksi) 29000 29000 29000 29000

A (in.^2) 20.0 11.8 11.8 20.0

I (in.^4) 723.0 612.0 612.0 723.0

Joint No. 1 2 3 4 5

Px (kips)

Py (kips)

Mz (ft-k)

L (ft.) 13.0000 13.1529 13.1529 13.0000

x

0.0000 0.9884 0.9884 0.0000

Shear (k) Moment (ft-k) -1.61 0.00 1.61 -20.89 10.66 20.89 2.34 33.77 -0.87 -33.77 13.87 -63.14 4.86 0.00 -4.86 63.14

X

X-axis (ft.)

Member No. 1 2 3 4

Axial (k) 11.53 -11.53 6.55 -4.55 5.05 -7.05 14.78 -14.78

c/L e/L

y

1.0000 0.1521 -0.1521 1.0000

1

b/L a/L P wb

2 M

we

3

Joint Loads: x or y

4

L

Member Load Nomenclature

+M or -M +M(max) -M(max) +M(max) -M(max) +M(max) -M(max) +M(max) -M(max)

M (ft-k) 0.00 -20.89 36.55 -20.89 33.77 -63.14 63.14 0.00

x or y (ft.) 0.00 13.00 10.78 0.00 0.00 13.15 13.00 0.00

Joint Displacements:

3.25 Joint No. 1 2 3 4 5

0.0000 0.1893 0.2159 0.2423 0.0000

0.0000 -0.0031 -0.1934 -0.0040 0.0000

Distributed Load #4 wb (k/ft.) e/L

we (k/ft.)

b/L

N o t e : Point

loads or moments at member ends must be input as joint loads.

Member Loads:

x (in.)

y (in.)

z (rad.)

-0.0009 -0.0018 0.0004 0.0003 -0.0025

Distributed Loads:

No. 1 2 2 3 3 4

Member Load Direct. X-Global Y-Global X-Projected Y-Global X-Projected X-Global

b/L

Distributed Load #1 wb (k/ft.) e/L

we (k/ft.)

0.0000

-1.0000

1.0000

-1.0000

0.0000

-1.0000

1.0000

-1.0000

b/L

Distributed Load #2 wb (k/ft.) e/L

we (k/ft.)

b/L

Distributed Load #3 wb (k/ft.) e/L

we (k/ft.)

b/L

Distributed Load #5 wb (k/ft.) e/L

we (k/ft.)

Point Loads:

No. 1 2 2 3 3 4

Member Load Direct.

Point Load #1 P (kips) a/L

Point Load #2 P (kips) a/L

Point Load #3 P (kips) a/L

Point Load #4 P (kips) a/L

Moment #1 M (ft-kips) c/L

Moment #2 M (ft-kips) c/L

Moment #3 M (ft-kips) c/L

Moment #4 M (ft-kips) c/L

Point Load #5 P (kips) a/L

Point Load #6 P (kips) a/L

Point Load #7 P (kips) a/L

Point Load #8 P (kips) a/L

Point Load #9 P (kips) a/L

Point Load #10 P (kips) a/L

X-Global Y-Global X-Global Y-Global X-Global X-Global

Applied Moments: Member No. 1 2 3 4

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11/3/2013 7:48 AM

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