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6th Edition

KASSIMALI + 1 other

Publisher: Cengage,

ISBN: 9781337630931

Chapter 12, Problem 13P

Textbook Problem

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12.6 through 12.13 Determine the approximate axial forces, shears, and moments for all the members of the frames shown in Figs. P12.6 through P12.13 by using the portal method.

FIG. P12.13, P12.21

To determine

Find the approximate axial forces, shears, and moments for all members of the frame using portal method.

**Given information:**

The axial force acting at point *M*

The axial force acting at point *I*

The axial force acting at point *E*

The vertical distance of the member *JM* and *K*

The vertical distance of the member *EI*, *FJ*, *GK*, and *HL*

The vertical distance of the member *AE*, *BF*, *CG*, and *DH*

The horizontal distance of the members *AB*, *EF*, and *IJ*

The horizontal distance of the members *BC*, *FG*, and *JK*

The horizontal distance of the members *CD*, *GH*, and *KL*

**Calculation:**

Insert the internal hinges at the midpoints of all the members of the given frame to obtain the simplified frame for approximate analysis.

Draw the simplified frame as in Figure (1).

For the calculation of column shears in the columns of the frame, pass an imaginary section *cc*, *bb,* and *aa* through the columns just above the support level.

Draw the free body diagram of the frame portion above section *cc* as in Figure (2).

**Column shears:**

Consider the shear in the interior column *BF* and *CG* is twice as much as in the exterior columns *AE* and *CG*.

Determine the column shears at the lower end using equilibrium equations.

Substitute 15 k for

The shear force at the lower end of the columns

Determine the shear force at the interior columns.

Substitute –6.25 k for

Draw the free body diagram of frame with the column shear value at the lower end column as in Figure (3).

Draw the free body diagram of the frame portion above section *bb* as in Figure (4).

Consider the shear in the interior column *FJ* and *GK* is twice as much as in the exterior columns *EI* and *HL*.

Determine the column shears at the second top lower end of the members *EFGH* using equilibrium equations.

Substitute 15 k for

The shear force at the end of the column members

Determine the shear force at the interior column members *FJ* and *GK*.

Substitute –3.75 k for

Draw the free body diagram of frame with the column shear value at the end of the column member *EFGH* as in Figure (5).

Draw the free body diagram of the frame portion above section *cc* as in Figure (6).

Determine the column shears at the top lower end of the members *JK* using equilibrium equations.

Substitute 7.5 k for

Draw the free body diagram of frame with the column shear value at the end of the column member *JK* as in Figure (7).

**Column moments:**

Consider thecolumn members *JM* and *KN*.

Determine the moment at the column member *JM*.

Substitute 3.75 k for

The moment at the column *JM* is equal to moment at the column *KN*. Therefore, the moment at the column members *JM* and *KN* is

The moment at the internal hinge of the horizontal member *MN* has the equal maginitude and opposite direction of the moment at the vertical column members *JM* and *KN*.

Consider column members *EI*, *FJ*, *GK*, and *HL*.

Consider the anticlockwise moment is positive and clockwise moment is negative.

Determine the moment at the column member *EI*.

Substitute 3.75 k for

The moment at the column *EI* is equal to moment at the column *HL*. Therefore, the moment at the column member *EI* and *HL* is

Determine the moment at the column members *FJ* and *GK*.

Substitute

The moment at the internal hinge of the horizontal member *IJ* and *KL* have the equal maginitude and opposite direction of the moment at the vertical column members *EI* and *HL*. The moment at the internal hinge of the horizontal member *JK* has the equal maginitude and opposite direction of the moment at the vertical column member *FJ* or *GK*.

Consider lower end column members.

Consider the anticlockwise moment is positive and clockwise moment is negative.

Determine the moment at the column member *AE*.

Substitute 6.25 k for

The moment at the column *AE* is equal to moment at the column *DH*. Therefore, the moment at the column member *AE* and *DH* is

Determine the moment at the column members *BF* and *CG*.

Substitute

The moment at the internal hinge of the horizontal member *EF*, *FG*, and *GH* has the equal maginitude and opposite direction of the summazation of moment at the vertical column member (*AE* and *EI*) and (*DH* and *HL*).

Determine the moment at the internal hinge of the horizontal member *EF*.

Substitute

Determine the moment at the internal hinge of the horizontal member *GH*.

Substitute

Draw the free body diagram of frame with the column moments for frame portion *JMNK* as in Figure (8).

**Girder axial forces, moments, and shears:**

**Consider girder MN.**

Determine the girder end action at the upper left end joint *M* using the relation.

Substitute 7.5 k for

The girder end action at the horizontal member *MN* is

Determine the vertical shear at the point *M*.

Take moment about hinge at the horizontal member *MN.*

Substitute

Determine the vertical shear at point *J* using the relation.

Substitute 3 k for

The vertical end action at column members *JM* and *KN* is

**Consider girder IJ.**

Determine the girder end action at the upper left end joint *I* using the relation.

Substitute 15 k for

The girder end action at the horizontal member *IJ* is

Determine the vertical shear at the point *I*.

Take moment about hinge at the horizontal member *IJ*.

Substitute

Determine the vertical shear at point *J* using the relation.

Substitute 2 k for

The vertical end action at joint *I* and *E* is

**Consider girder JK.**

Determine the girder end action at the joint *J* using the relation.

Substitute 11.25 k for

The girder end action at the horizontal member *JK* is

Structural Analysis

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