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6th Edition

KASSIMALI + 1 other

Publisher: Cengage,

ISBN: 9781337630931

Chapter 12, Problem 21P

Textbook Problem

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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 cantilever method.

FIG. P12.13, P12.21

To determine

Find the approximate axial forces, shears, and moments for the all members of the frames using cantilever 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 *KL*

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*

Take the counterclockwise moment is positive and clockwise moment is negative.

The axial force in horizontal direction, towards right is positive and towards left side is negative.

The axial force in vertical direction, towards upward is positive and towards downward is negative.

**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 axial forces of story of the frame, pass an imaginary section *aa* through the internal hinges at the midheights of columns *JM* and *KN*, pass an imaginary section *bb* through the internal hinges at the midheights of columns *EI*, *FJ*, *GK*, and *HL*, and pass an imaginary section *cc* through the internal hinges at the midheights of columns *AE*, *BF*, *CG*, and *DH*.

Draw the free body diagram of the frame portion with the passed imaginary lines as in Figure (2).

**Column axial forces:**

**Above section aa:**

Draw the free body diagram of the frame portion above the section *aa* as in Figure (3).

Determine the location of the centroid from support *A* using the relation.

Here, *A* is the area of the column sections, *B* from support *A*, *C* from support *A*, and *D* from support *A*.

Substitute 0 ft for

The location of the centroid from the column member *JM* is 10 ft.

The given lateral load is acting on the frame to the right, therefore the axial force in column *JM* located to the left of the centroid, must be tensile whereas the axial force in column *KN* placed to the right of the centroid, must be compressive.

Determine the axial force in the column members *JM* and *KN* using equilibrium conditions.

Take moment about point *O*.

Substitute 20 ft for

The axial force in the column members *JM* and *KN* is

Draw the free body diagram of the frame portion above the section *aa* with the axial forces in the column members as in Figure (4).

Draw the free body diagram of the frame portion above the section *bb* as in Figure (5).

Consider the axial forces in the columns are to be linearly proportional to their distances from centroid.

The given lateral load is acting on the frame to the right, therefore the axial force in column *EI* and *FJ* located to the left of the centroid, must be tensile whereas the axial force in column *GK* and *HL* placed to the right of the centroid, must be compressive.

Refer Figure (5).

Determine the relationship in column axial force between the member *EI* and *FJ* using the relation.

Substitute 40 ft for

Determine the relationship in column axial force between the member *EI* and *GK* using the relation.

Substitute 30 ft for

Determine the axial force in the column members *EI*, *FJ*, *GK*, and *HL* using equilibrium conditions.

Take moment about point *P*.

Substitute

Determine the axial force in the column members *FJ*.

Substitute 3.53 k for

Determine the axial force in the column members *GK*.

Substitute 3.53 k for

Determine the axial force in the column members *HL*.

Substitute 3.53 k for

Draw the free body diagram of the frame portion above the section *bb* with the axial forces in the column members as in Figure (6).

Draw the free body diagram of the frame portion above the section *cc* as in Figure (7).

The given lateral load is acting on the frame to the right, therefore the axial force in column *AE* and *BF* located to the left of the centroid, must be tensile, whereas the axial force in column *CG* and *DH* placed to the right of the centroid, must be compressive.

Determine the relationship in column axial force between the member *AE* and *BF* using the relation.

Substitute 40 ft for

Determine the relationship in column axial force between the member *AE* and *CG* using the relation.

Substitute 30 ft for

Determine the axial force in the column members *AE*, *BF*, *CG*, and *DH* using equilibrium conditions.

Take moment about point *Q*.

Substitute

Determine the axial force in the column members *BF*.

Substitute 9.18 k for

Determine the axial force in the column members *CG*.

Substitute 9.18 k for

Determine the axial force in the column members *DH*.

Substitute 2.29 k for

Draw the free body diagram of the frame portion above the section *bb* with the axial forces in the column members as in Figure (8).

**Girder shear and moments:**

**Consider girder MN**.

Determine the shear at upper left end joint *M* using equilibrium equation.

Substitute 3 k for

Determine the shear at upper right end joint *N* using equilibrium equation.

Substitute 3 k for

Determine the moment at left end of the girder *MN* using equilibrium equations.

Substitute 3 k for

Determine the moment at right end of the girder *NM* using equilibrium equations.

Take moment about point *M*.

Substitute

**Consider girder IJ**.

Determine the shear at upper left end joint *I* using equilibrium equation.

Substitute 3.53 k for

Determine the shear at upper right end joint *J* using equilibrium equation.

Substitute 3.53 k for

Determine the moment at left end of the girder *IJ* using equilibrium equations.

Substitute 3.53 k for

Determine the moment at right end of the girder *IJ* using equilibrium equations.

Take moment about point *I*.

Substitute

**Consider girder JK.**

Determine the shear at left end joint *J* using equilibrium equation.

Substitute 3.53 k for

Determine the shear at right end joint *K* using equilibrium equation.

Substitute 1.41 k for

Determine the moment at left end of the girder *JK* using equilibrium equations.

Substitute 1.41 k for

Determine the moment at right end of the girder *JK* using equilibrium equations.

Take moment about point *J*.

Substitute

**Consider girder KL.**

Determine the shear at left end joint *K* using equilibrium equation.

Substitute 1.41 k for

Determine the shear at right end joint *L* using equilibrium equation.

Substitute 3.53 k for

Determine the moment at left end of the girder *KL* using equilibrium equations.

Substitute 3.53 k for

Determine the moment at right end of the girder *JK* using equilibrium equations.

Take moment about point *K*.

Substitute

**Consider girder EF**.

Determine the shear at left end joint *E* using equilibrium equation.

Substitute 3.53 k for

Determine the shear at right end joint *F* using equilibrium equation.

Substitute 5.65 k for

Determine the moment at left end of the girder *EF* using equilibrium equations.

Substitute 5.65 k for

Determine the moment at right end of the girder *EF* using equilibrium equations.

Take moment about point *E*.

Substitute

**Consider girder FG.**

Determine the shear at left end joint *F* using equilibrium equation.

Substitute 5.65 k for

Determine the shear at right end joint *G* using equilibrium equation.

Substitute 7.06 k for

Determine the moment at left end of the girder *FG* using equilibrium equations.

Substitute 7.06 k for

Determine the moment at right end of the girder *FG* using equilibrium equations.

Take moment about point *F*.

Substitute

**Consider girder GH.**

Determine the shear at left end joint *G* using equilibrium equation.

Substitute 7.06 k for

Determine the shear at right end joint *H* using equilibrium equation.

Substitute 5.65 k for

Determine the moment at left end of the girder *GH* using equilibrium equations.

Substitute 5.65 k for

Determine the moment at right end of the girder *GH* using equilibrium equations

Structural Analysis

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