Find the slope θ B & θ D and deflection Δ B & Δ D at point B and D of the given beam using the moment-area method.

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Answer 1

Question

Chapter 6, Problem 33P

To determine

Find the slope $θB&θD$ and deflection $ΔB&ΔD$ at point B and D of the given beam using the moment-area method.

Expert Solution & Answer

Answer to Problem 33P

The slope $θB$ at point B of the given beam using the moment-area method is $0.0099 rad(Clockwise)_$ .

The deflection $ΔB$ at point B of the given beam using the moment-area method is $0.86 in.(↓)_$ .

The slope $θD$ at point D of the given beam using the moment-area method is $0.0084 rad(Counterclockwise)_$ .

The deflection $ΔD$ at point D of the given beam using the moment-area method is $1.44 in.(↓)_$ .

Explanation of Solution

Given information:

The Young’s modulus (E) is 29,000 ksi.

The moment of inertia (I) is $6,000 in.4$ .

Calculation:

Consider flexural rigidity EI of the beam is constant.

Show the free body diagram of the given beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 1

Refer Figure (1),

Consider upward is force is positive and downward force is negative.

Consider clockwise moment is negative and counterclockwise moment is positive.

Split the given beam into two sections such as AC and CE.

Consider the portion CE:

Draw the free body diagram of the portion CE as in Figure (2).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 2

Refer Figure (2),

Consider a reaction at C and take moment about point C.

Determine the reaction at E;

$RE×(24)−(40×12)+250=0RE=23024RE=9.58 kips$

Determine the reaction at support A;

$∑V=0RA+RE−75−40=0RA=115−9.58RA=105.42 kips$

Show the reaction of the given beam as in Figure (3).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 3

Refer Figure (3),

Determine the moment at A:

$MA=(9.58×48)+250−(40×36)−(75×12)=709.84−2,340≃−1,630 kips-ft$

Determine the bending moment at B;

$MB=(9.58×36)+250−(40×24)=344.88+250−960≃−365 kips-ft$

Determine the bending moment at C;

$MC=(9.58×24)+250−(40×12)=479.92−480≃0$

Determine the bending moment at D;

$MD=(9.58×12)+250≃365 kips-ft$

Determine the bending moment at E;

$ME=250 kips-ft$

Show the $M/EI$ diagram for the given beam as in Figure (4).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 4

Show the elastic curve diagram as in Figure (5).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 5

Refer Figure (4),

Determine the slope at B;

$θB=θBA=Area of the MEI diagram between A and B=−(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $1,630EI$ for $h1$ , 12 ft for $b2$ , and $365EI$ for $h2$ .

$θB=θBA=−[(12×12×1,630EI)+(12×12×365EI)]=−1EI(1,630+3652)12=−11,970 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θB=−11,970 kips-ft2×(12 in.1 ft)229,000×6,000=−0.0099 rad=0.0099 (Clockwise)$

Hence, the slope at B is $0.0099 rad(Clockwise)_$ .

Determine the deflection between A and B using the relation;

$ΔB=ΔBA=Moment of the area of the M/EI diagram between B and A about A=−[(b1h1)(b12)+(12b2h2)(23×b2)]$

Here, $b1$ is the width of rectangle, $h1$ is the height of the rectangle, $b2$ is the width of the triangle, and $h2$ is the height of the triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $(1,630EI−365EI)$ for $h2$ .

$ΔB=ΔBA=−[(12×365EI)(122)+12×12×(1,630EI−365EI)(23×12)]=−87,000 kips⋅ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔB=ΔBA=−87,000 kips⋅ft3×(12 in.1 ft)329,000×6,000=−0.864 in.≃0.86 in.(↓)$

Hence, the deflection at B is $0.86 in.(↓)_$ .

To determine the slope at point E, it is necessary to determine the deflection at point C and the deflection between C and E.

Determine the deflection at C and A using the relation;

$ΔC=ΔCA=Moment of the area of the M/EI diagram between C and A about A=−[(b1h1)(12+b12)+(12b2h2)(12+23×b2)+(12b3h3)(23×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $(1,630EI−365EI)$ for $h2$ , 12 for $b3$ , and $365EI$ for $h3$ .

$ΔC=ΔCA=−[(12×365EI)(12+122)+(12×12×(1,630EI−365EI))(12+23×12)+(12×12×365EI)(23×12)]=−248,160 kips-ft3EI$

Determine the deflection between C and E using the relation;

$ΔCE=Moment of the area of the M/EI diagram between C and E about E=−[(12b1h1)(23×b1)+(b2h2)(12+b22)+(12b3h3)(12+13×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $250EI$ for $h2$ , 12 ft for $b3$ , and $(365EI−250EI)$ for $h3$ .

$ΔCE=−[(12×12×365EI)(23×12)+(12×250EI)(12+122)+(12×12×(365EI−250EI))(12+13×12)]=−82,560 kips-ft3EI$

Determine the slope at E using the relation;

$θE=ΔC+ΔCELCE$

Here, $LCE$ is the length between C and E.

Substitute $−248,160 kips-ft3EI$ for $ΔC$ , $−82,560 kips-ft3EI$ for $ΔCE$ , and 24 ft for $LCE$ .

$θE=−248,160 kips-ft3EI+(−82,560 kips-ft3EI)24=−13,780 kips-ft2EI$

Determine the slope between D and E using the relation;

$θDE=Area of the MEI diagram between D and E=(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $250EI$ for $h2$ .

$θDE=−[(12×12×365EI)+(12×12×250EI)]=−1EI(365+2502)12=−3,690 kips-ft2EI$

Determine the slope at D using the relation;

$θD=θE−θDE$

Substitute $−3,690 kips-ft2EI$ for $θE$ and $−3,690 kips-ft2EI$ for $θDE$ .

$θD=−13,780 kips-ft2EI−(−3,690 kips-ft2EI)=−10,090 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θD=−10,090 kips-ft2×(12 in.1 ft)229,000×6,000=−0.00835 rad≃0.0084 rad(Counterclockwise)$

Hence, the slope at D is $0.0084 rad(Counterclockwise)_$ .

Determine the deflection between D and E using the relation;

$ΔDE=Moment of the area of the M/EI diagram between D and E about E=−[(b1h1)(b12)+(12b2h2)(13×b2)]$

Here, b is the width and h is the height of respective rectangle and triangle.

Substitute 12 ft for $b1$ , $250EI$ for $h1$ , 12 for $b2$ , and $(365EI−250EI)$ for $h2$ .

$ΔDE=−[(12×250EI)(122)+(12×12×(365EI−250EI))(13×12)]=−20,760 kips-ft3EI$

Determine the deflection at D using the relation;

$ΔD=LDEθE−ΔDE$

Substitute 12 ft for $LDE$ , $−13,780 kips-ft2EI$ for $θE$ , and $−20,760 kips-ft3EI$ for $ΔDE$ .

$ΔD=12(−13,780 kips-ft2EI)−(−20,760 kips-ft3EI)=−165,360+20,760EI=−144,600 kips-ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔD=−144,600 kips-ft3×(12 in.1 ft)329,000×6,000=−1.44 in.=1.44 in.(↓)$

Hence, the deflection at point D is $1.44 in.(↓)_$ .

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Answer 2

Question

Chapter 6, Problem 33P

To determine

Find the slope $θB&θD$ and deflection $ΔB&ΔD$ at point B and D of the given beam using the moment-area method.

Expert Solution & Answer

Answer to Problem 33P

The slope $θB$ at point B of the given beam using the moment-area method is $0.0099 rad(Clockwise)_$ .

The deflection $ΔB$ at point B of the given beam using the moment-area method is $0.86 in.(↓)_$ .

The slope $θD$ at point D of the given beam using the moment-area method is $0.0084 rad(Counterclockwise)_$ .

The deflection $ΔD$ at point D of the given beam using the moment-area method is $1.44 in.(↓)_$ .

Explanation of Solution

Given information:

The Young’s modulus (E) is 29,000 ksi.

The moment of inertia (I) is $6,000 in.4$ .

Calculation:

Consider flexural rigidity EI of the beam is constant.

Show the free body diagram of the given beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 1

Refer Figure (1),

Consider upward is force is positive and downward force is negative.

Consider clockwise moment is negative and counterclockwise moment is positive.

Split the given beam into two sections such as AC and CE.

Consider the portion CE:

Draw the free body diagram of the portion CE as in Figure (2).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 2

Refer Figure (2),

Consider a reaction at C and take moment about point C.

Determine the reaction at E;

$RE×(24)−(40×12)+250=0RE=23024RE=9.58 kips$

Determine the reaction at support A;

$∑V=0RA+RE−75−40=0RA=115−9.58RA=105.42 kips$

Show the reaction of the given beam as in Figure (3).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 3

Refer Figure (3),

Determine the moment at A:

$MA=(9.58×48)+250−(40×36)−(75×12)=709.84−2,340≃−1,630 kips-ft$

Determine the bending moment at B;

$MB=(9.58×36)+250−(40×24)=344.88+250−960≃−365 kips-ft$

Determine the bending moment at C;

$MC=(9.58×24)+250−(40×12)=479.92−480≃0$

Determine the bending moment at D;

$MD=(9.58×12)+250≃365 kips-ft$

Determine the bending moment at E;

$ME=250 kips-ft$

Show the $M/EI$ diagram for the given beam as in Figure (4).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 4

Show the elastic curve diagram as in Figure (5).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 5

Refer Figure (4),

Determine the slope at B;

$θB=θBA=Area of the MEI diagram between A and B=−(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $1,630EI$ for $h1$ , 12 ft for $b2$ , and $365EI$ for $h2$ .

$θB=θBA=−[(12×12×1,630EI)+(12×12×365EI)]=−1EI(1,630+3652)12=−11,970 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θB=−11,970 kips-ft2×(12 in.1 ft)229,000×6,000=−0.0099 rad=0.0099 (Clockwise)$

Hence, the slope at B is $0.0099 rad(Clockwise)_$ .

Determine the deflection between A and B using the relation;

$ΔB=ΔBA=Moment of the area of the M/EI diagram between B and A about A=−[(b1h1)(b12)+(12b2h2)(23×b2)]$

Here, $b1$ is the width of rectangle, $h1$ is the height of the rectangle, $b2$ is the width of the triangle, and $h2$ is the height of the triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $(1,630EI−365EI)$ for $h2$ .

$ΔB=ΔBA=−[(12×365EI)(122)+12×12×(1,630EI−365EI)(23×12)]=−87,000 kips⋅ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔB=ΔBA=−87,000 kips⋅ft3×(12 in.1 ft)329,000×6,000=−0.864 in.≃0.86 in.(↓)$

Hence, the deflection at B is $0.86 in.(↓)_$ .

To determine the slope at point E, it is necessary to determine the deflection at point C and the deflection between C and E.

Determine the deflection at C and A using the relation;

$ΔC=ΔCA=Moment of the area of the M/EI diagram between C and A about A=−[(b1h1)(12+b12)+(12b2h2)(12+23×b2)+(12b3h3)(23×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $(1,630EI−365EI)$ for $h2$ , 12 for $b3$ , and $365EI$ for $h3$ .

$ΔC=ΔCA=−[(12×365EI)(12+122)+(12×12×(1,630EI−365EI))(12+23×12)+(12×12×365EI)(23×12)]=−248,160 kips-ft3EI$

Determine the deflection between C and E using the relation;

$ΔCE=Moment of the area of the M/EI diagram between C and E about E=−[(12b1h1)(23×b1)+(b2h2)(12+b22)+(12b3h3)(12+13×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $250EI$ for $h2$ , 12 ft for $b3$ , and $(365EI−250EI)$ for $h3$ .

$ΔCE=−[(12×12×365EI)(23×12)+(12×250EI)(12+122)+(12×12×(365EI−250EI))(12+13×12)]=−82,560 kips-ft3EI$

Determine the slope at E using the relation;

$θE=ΔC+ΔCELCE$

Here, $LCE$ is the length between C and E.

Substitute $−248,160 kips-ft3EI$ for $ΔC$ , $−82,560 kips-ft3EI$ for $ΔCE$ , and 24 ft for $LCE$ .

$θE=−248,160 kips-ft3EI+(−82,560 kips-ft3EI)24=−13,780 kips-ft2EI$

Determine the slope between D and E using the relation;

$θDE=Area of the MEI diagram between D and E=(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $250EI$ for $h2$ .

$θDE=−[(12×12×365EI)+(12×12×250EI)]=−1EI(365+2502)12=−3,690 kips-ft2EI$

Determine the slope at D using the relation;

$θD=θE−θDE$

Substitute $−3,690 kips-ft2EI$ for $θE$ and $−3,690 kips-ft2EI$ for $θDE$ .

$θD=−13,780 kips-ft2EI−(−3,690 kips-ft2EI)=−10,090 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θD=−10,090 kips-ft2×(12 in.1 ft)229,000×6,000=−0.00835 rad≃0.0084 rad(Counterclockwise)$

Hence, the slope at D is $0.0084 rad(Counterclockwise)_$ .

Determine the deflection between D and E using the relation;

$ΔDE=Moment of the area of the M/EI diagram between D and E about E=−[(b1h1)(b12)+(12b2h2)(13×b2)]$

Here, b is the width and h is the height of respective rectangle and triangle.

Substitute 12 ft for $b1$ , $250EI$ for $h1$ , 12 for $b2$ , and $(365EI−250EI)$ for $h2$ .

$ΔDE=−[(12×250EI)(122)+(12×12×(365EI−250EI))(13×12)]=−20,760 kips-ft3EI$

Determine the deflection at D using the relation;

$ΔD=LDEθE−ΔDE$

Substitute 12 ft for $LDE$ , $−13,780 kips-ft2EI$ for $θE$ , and $−20,760 kips-ft3EI$ for $ΔDE$ .

$ΔD=12(−13,780 kips-ft2EI)−(−20,760 kips-ft3EI)=−165,360+20,760EI=−144,600 kips-ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔD=−144,600 kips-ft3×(12 in.1 ft)329,000×6,000=−1.44 in.=1.44 in.(↓)$

Hence, the deflection at point D is $1.44 in.(↓)_$ .

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Answer 3

Question

Chapter 6, Problem 33P

To determine

Find the slope $θB&θD$ and deflection $ΔB&ΔD$ at point B and D of the given beam using the moment-area method.

Expert Solution & Answer

Answer to Problem 33P

The slope $θB$ at point B of the given beam using the moment-area method is $0.0099 rad(Clockwise)_$ .

The deflection $ΔB$ at point B of the given beam using the moment-area method is $0.86 in.(↓)_$ .

The slope $θD$ at point D of the given beam using the moment-area method is $0.0084 rad(Counterclockwise)_$ .

The deflection $ΔD$ at point D of the given beam using the moment-area method is $1.44 in.(↓)_$ .

Explanation of Solution

Given information:

The Young’s modulus (E) is 29,000 ksi.

The moment of inertia (I) is $6,000 in.4$ .

Calculation:

Consider flexural rigidity EI of the beam is constant.

Show the free body diagram of the given beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 1

Refer Figure (1),

Consider upward is force is positive and downward force is negative.

Consider clockwise moment is negative and counterclockwise moment is positive.

Split the given beam into two sections such as AC and CE.

Consider the portion CE:

Draw the free body diagram of the portion CE as in Figure (2).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 2

Refer Figure (2),

Consider a reaction at C and take moment about point C.

Determine the reaction at E;

$RE×(24)−(40×12)+250=0RE=23024RE=9.58 kips$

Determine the reaction at support A;

$∑V=0RA+RE−75−40=0RA=115−9.58RA=105.42 kips$

Show the reaction of the given beam as in Figure (3).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 3

Refer Figure (3),

Determine the moment at A:

$MA=(9.58×48)+250−(40×36)−(75×12)=709.84−2,340≃−1,630 kips-ft$

Determine the bending moment at B;

$MB=(9.58×36)+250−(40×24)=344.88+250−960≃−365 kips-ft$

Determine the bending moment at C;

$MC=(9.58×24)+250−(40×12)=479.92−480≃0$

Determine the bending moment at D;

$MD=(9.58×12)+250≃365 kips-ft$

Determine the bending moment at E;

$ME=250 kips-ft$

Show the $M/EI$ diagram for the given beam as in Figure (4).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 4

Show the elastic curve diagram as in Figure (5).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 5

Refer Figure (4),

Determine the slope at B;

$θB=θBA=Area of the MEI diagram between A and B=−(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $1,630EI$ for $h1$ , 12 ft for $b2$ , and $365EI$ for $h2$ .

$θB=θBA=−[(12×12×1,630EI)+(12×12×365EI)]=−1EI(1,630+3652)12=−11,970 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θB=−11,970 kips-ft2×(12 in.1 ft)229,000×6,000=−0.0099 rad=0.0099 (Clockwise)$

Hence, the slope at B is $0.0099 rad(Clockwise)_$ .

Determine the deflection between A and B using the relation;

$ΔB=ΔBA=Moment of the area of the M/EI diagram between B and A about A=−[(b1h1)(b12)+(12b2h2)(23×b2)]$

Here, $b1$ is the width of rectangle, $h1$ is the height of the rectangle, $b2$ is the width of the triangle, and $h2$ is the height of the triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $(1,630EI−365EI)$ for $h2$ .

$ΔB=ΔBA=−[(12×365EI)(122)+12×12×(1,630EI−365EI)(23×12)]=−87,000 kips⋅ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔB=ΔBA=−87,000 kips⋅ft3×(12 in.1 ft)329,000×6,000=−0.864 in.≃0.86 in.(↓)$

Hence, the deflection at B is $0.86 in.(↓)_$ .

To determine the slope at point E, it is necessary to determine the deflection at point C and the deflection between C and E.

Determine the deflection at C and A using the relation;

$ΔC=ΔCA=Moment of the area of the M/EI diagram between C and A about A=−[(b1h1)(12+b12)+(12b2h2)(12+23×b2)+(12b3h3)(23×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $(1,630EI−365EI)$ for $h2$ , 12 for $b3$ , and $365EI$ for $h3$ .

$ΔC=ΔCA=−[(12×365EI)(12+122)+(12×12×(1,630EI−365EI))(12+23×12)+(12×12×365EI)(23×12)]=−248,160 kips-ft3EI$

Determine the deflection between C and E using the relation;

$ΔCE=Moment of the area of the M/EI diagram between C and E about E=−[(12b1h1)(23×b1)+(b2h2)(12+b22)+(12b3h3)(12+13×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $250EI$ for $h2$ , 12 ft for $b3$ , and $(365EI−250EI)$ for $h3$ .

$ΔCE=−[(12×12×365EI)(23×12)+(12×250EI)(12+122)+(12×12×(365EI−250EI))(12+13×12)]=−82,560 kips-ft3EI$

Determine the slope at E using the relation;

$θE=ΔC+ΔCELCE$

Here, $LCE$ is the length between C and E.

Substitute $−248,160 kips-ft3EI$ for $ΔC$ , $−82,560 kips-ft3EI$ for $ΔCE$ , and 24 ft for $LCE$ .

$θE=−248,160 kips-ft3EI+(−82,560 kips-ft3EI)24=−13,780 kips-ft2EI$

Determine the slope between D and E using the relation;

$θDE=Area of the MEI diagram between D and E=(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $250EI$ for $h2$ .

$θDE=−[(12×12×365EI)+(12×12×250EI)]=−1EI(365+2502)12=−3,690 kips-ft2EI$

Determine the slope at D using the relation;

$θD=θE−θDE$

Substitute $−3,690 kips-ft2EI$ for $θE$ and $−3,690 kips-ft2EI$ for $θDE$ .

$θD=−13,780 kips-ft2EI−(−3,690 kips-ft2EI)=−10,090 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θD=−10,090 kips-ft2×(12 in.1 ft)229,000×6,000=−0.00835 rad≃0.0084 rad(Counterclockwise)$

Hence, the slope at D is $0.0084 rad(Counterclockwise)_$ .

Determine the deflection between D and E using the relation;

$ΔDE=Moment of the area of the M/EI diagram between D and E about E=−[(b1h1)(b12)+(12b2h2)(13×b2)]$

Here, b is the width and h is the height of respective rectangle and triangle.

Substitute 12 ft for $b1$ , $250EI$ for $h1$ , 12 for $b2$ , and $(365EI−250EI)$ for $h2$ .

$ΔDE=−[(12×250EI)(122)+(12×12×(365EI−250EI))(13×12)]=−20,760 kips-ft3EI$

Determine the deflection at D using the relation;

$ΔD=LDEθE−ΔDE$

Substitute 12 ft for $LDE$ , $−13,780 kips-ft2EI$ for $θE$ , and $−20,760 kips-ft3EI$ for $ΔDE$ .

$ΔD=12(−13,780 kips-ft2EI)−(−20,760 kips-ft3EI)=−165,360+20,760EI=−144,600 kips-ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔD=−144,600 kips-ft3×(12 in.1 ft)329,000×6,000=−1.44 in.=1.44 in.(↓)$

Hence, the deflection at point D is $1.44 in.(↓)_$ .

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Answer 4

Question

Chapter 6, Problem 33P

To determine

Find the slope $θB&θD$ and deflection $ΔB&ΔD$ at point B and D of the given beam using the moment-area method.

Expert Solution & Answer

Answer to Problem 33P

The slope $θB$ at point B of the given beam using the moment-area method is $0.0099 rad(Clockwise)_$ .

The deflection $ΔB$ at point B of the given beam using the moment-area method is $0.86 in.(↓)_$ .

The slope $θD$ at point D of the given beam using the moment-area method is $0.0084 rad(Counterclockwise)_$ .

The deflection $ΔD$ at point D of the given beam using the moment-area method is $1.44 in.(↓)_$ .

Explanation of Solution

Given information:

The Young’s modulus (E) is 29,000 ksi.

The moment of inertia (I) is $6,000 in.4$ .

Calculation:

Consider flexural rigidity EI of the beam is constant.

Show the free body diagram of the given beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 1

Refer Figure (1),

Consider upward is force is positive and downward force is negative.

Consider clockwise moment is negative and counterclockwise moment is positive.

Split the given beam into two sections such as AC and CE.

Consider the portion CE:

Draw the free body diagram of the portion CE as in Figure (2).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 2

Refer Figure (2),

Consider a reaction at C and take moment about point C.

Determine the reaction at E;

$RE×(24)−(40×12)+250=0RE=23024RE=9.58 kips$

Determine the reaction at support A;

$∑V=0RA+RE−75−40=0RA=115−9.58RA=105.42 kips$

Show the reaction of the given beam as in Figure (3).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 3

Refer Figure (3),

Determine the moment at A:

$MA=(9.58×48)+250−(40×36)−(75×12)=709.84−2,340≃−1,630 kips-ft$

Determine the bending moment at B;

$MB=(9.58×36)+250−(40×24)=344.88+250−960≃−365 kips-ft$

Determine the bending moment at C;

$MC=(9.58×24)+250−(40×12)=479.92−480≃0$

Determine the bending moment at D;

$MD=(9.58×12)+250≃365 kips-ft$

Determine the bending moment at E;

$ME=250 kips-ft$

Show the $M/EI$ diagram for the given beam as in Figure (4).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 4

Show the elastic curve diagram as in Figure (5).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 5

Refer Figure (4),

Determine the slope at B;

$θB=θBA=Area of the MEI diagram between A and B=−(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $1,630EI$ for $h1$ , 12 ft for $b2$ , and $365EI$ for $h2$ .

$θB=θBA=−[(12×12×1,630EI)+(12×12×365EI)]=−1EI(1,630+3652)12=−11,970 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θB=−11,970 kips-ft2×(12 in.1 ft)229,000×6,000=−0.0099 rad=0.0099 (Clockwise)$

Hence, the slope at B is $0.0099 rad(Clockwise)_$ .

Determine the deflection between A and B using the relation;

$ΔB=ΔBA=Moment of the area of the M/EI diagram between B and A about A=−[(b1h1)(b12)+(12b2h2)(23×b2)]$

Here, $b1$ is the width of rectangle, $h1$ is the height of the rectangle, $b2$ is the width of the triangle, and $h2$ is the height of the triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $(1,630EI−365EI)$ for $h2$ .

$ΔB=ΔBA=−[(12×365EI)(122)+12×12×(1,630EI−365EI)(23×12)]=−87,000 kips⋅ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔB=ΔBA=−87,000 kips⋅ft3×(12 in.1 ft)329,000×6,000=−0.864 in.≃0.86 in.(↓)$

Hence, the deflection at B is $0.86 in.(↓)_$ .

To determine the slope at point E, it is necessary to determine the deflection at point C and the deflection between C and E.

Determine the deflection at C and A using the relation;

$ΔC=ΔCA=Moment of the area of the M/EI diagram between C and A about A=−[(b1h1)(12+b12)+(12b2h2)(12+23×b2)+(12b3h3)(23×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $(1,630EI−365EI)$ for $h2$ , 12 for $b3$ , and $365EI$ for $h3$ .

$ΔC=ΔCA=−[(12×365EI)(12+122)+(12×12×(1,630EI−365EI))(12+23×12)+(12×12×365EI)(23×12)]=−248,160 kips-ft3EI$

Determine the deflection between C and E using the relation;

$ΔCE=Moment of the area of the M/EI diagram between C and E about E=−[(12b1h1)(23×b1)+(b2h2)(12+b22)+(12b3h3)(12+13×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $250EI$ for $h2$ , 12 ft for $b3$ , and $(365EI−250EI)$ for $h3$ .

$ΔCE=−[(12×12×365EI)(23×12)+(12×250EI)(12+122)+(12×12×(365EI−250EI))(12+13×12)]=−82,560 kips-ft3EI$

Determine the slope at E using the relation;

$θE=ΔC+ΔCELCE$

Here, $LCE$ is the length between C and E.

Substitute $−248,160 kips-ft3EI$ for $ΔC$ , $−82,560 kips-ft3EI$ for $ΔCE$ , and 24 ft for $LCE$ .

$θE=−248,160 kips-ft3EI+(−82,560 kips-ft3EI)24=−13,780 kips-ft2EI$

Determine the slope between D and E using the relation;

$θDE=Area of the MEI diagram between D and E=(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $250EI$ for $h2$ .

$θDE=−[(12×12×365EI)+(12×12×250EI)]=−1EI(365+2502)12=−3,690 kips-ft2EI$

Determine the slope at D using the relation;

$θD=θE−θDE$

Substitute $−3,690 kips-ft2EI$ for $θE$ and $−3,690 kips-ft2EI$ for $θDE$ .

$θD=−13,780 kips-ft2EI−(−3,690 kips-ft2EI)=−10,090 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θD=−10,090 kips-ft2×(12 in.1 ft)229,000×6,000=−0.00835 rad≃0.0084 rad(Counterclockwise)$

Hence, the slope at D is $0.0084 rad(Counterclockwise)_$ .

Determine the deflection between D and E using the relation;

$ΔDE=Moment of the area of the M/EI diagram between D and E about E=−[(b1h1)(b12)+(12b2h2)(13×b2)]$

Here, b is the width and h is the height of respective rectangle and triangle.

Substitute 12 ft for $b1$ , $250EI$ for $h1$ , 12 for $b2$ , and $(365EI−250EI)$ for $h2$ .

$ΔDE=−[(12×250EI)(122)+(12×12×(365EI−250EI))(13×12)]=−20,760 kips-ft3EI$

Determine the deflection at D using the relation;

$ΔD=LDEθE−ΔDE$

Substitute 12 ft for $LDE$ , $−13,780 kips-ft2EI$ for $θE$ , and $−20,760 kips-ft3EI$ for $ΔDE$ .

$ΔD=12(−13,780 kips-ft2EI)−(−20,760 kips-ft3EI)=−165,360+20,760EI=−144,600 kips-ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔD=−144,600 kips-ft3×(12 in.1 ft)329,000×6,000=−1.44 in.=1.44 in.(↓)$

Hence, the deflection at point D is $1.44 in.(↓)_$ .

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Answer 5

Chapter 6, Problem 33P

To determine

Find the slope $θB&θD$ and deflection $ΔB&ΔD$ at point B and D of the given beam using the moment-area method.

Expert Solution & Answer

Answer to Problem 33P

The slope $θB$ at point B of the given beam using the moment-area method is $0.0099 rad(Clockwise)_$ .

The deflection $ΔB$ at point B of the given beam using the moment-area method is $0.86 in.(↓)_$ .

The slope $θD$ at point D of the given beam using the moment-area method is $0.0084 rad(Counterclockwise)_$ .

The deflection $ΔD$ at point D of the given beam using the moment-area method is $1.44 in.(↓)_$ .

Explanation of Solution

Given information:

The Young’s modulus (E) is 29,000 ksi.

The moment of inertia (I) is $6,000 in.4$ .

Calculation:

Consider flexural rigidity EI of the beam is constant.

Show the free body diagram of the given beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 1

Refer Figure (1),

Consider upward is force is positive and downward force is negative.

Consider clockwise moment is negative and counterclockwise moment is positive.

Split the given beam into two sections such as AC and CE.

Consider the portion CE:

Draw the free body diagram of the portion CE as in Figure (2).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 2

Refer Figure (2),

Consider a reaction at C and take moment about point C.

Determine the reaction at E;

$RE×(24)−(40×12)+250=0RE=23024RE=9.58 kips$

Determine the reaction at support A;

$∑V=0RA+RE−75−40=0RA=115−9.58RA=105.42 kips$

Show the reaction of the given beam as in Figure (3).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 3

Refer Figure (3),

Determine the moment at A:

$MA=(9.58×48)+250−(40×36)−(75×12)=709.84−2,340≃−1,630 kips-ft$

Determine the bending moment at B;

$MB=(9.58×36)+250−(40×24)=344.88+250−960≃−365 kips-ft$

Determine the bending moment at C;

$MC=(9.58×24)+250−(40×12)=479.92−480≃0$

Determine the bending moment at D;

$MD=(9.58×12)+250≃365 kips-ft$

Determine the bending moment at E;

$ME=250 kips-ft$

Show the $M/EI$ diagram for the given beam as in Figure (4).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 4

Show the elastic curve diagram as in Figure (5).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 5

Refer Figure (4),

Determine the slope at B;

$θB=θBA=Area of the MEI diagram between A and B=−(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $1,630EI$ for $h1$ , 12 ft for $b2$ , and $365EI$ for $h2$ .

$θB=θBA=−[(12×12×1,630EI)+(12×12×365EI)]=−1EI(1,630+3652)12=−11,970 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θB=−11,970 kips-ft2×(12 in.1 ft)229,000×6,000=−0.0099 rad=0.0099 (Clockwise)$

Hence, the slope at B is $0.0099 rad(Clockwise)_$ .

Determine the deflection between A and B using the relation;

$ΔB=ΔBA=Moment of the area of the M/EI diagram between B and A about A=−[(b1h1)(b12)+(12b2h2)(23×b2)]$

Here, $b1$ is the width of rectangle, $h1$ is the height of the rectangle, $b2$ is the width of the triangle, and $h2$ is the height of the triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $(1,630EI−365EI)$ for $h2$ .

$ΔB=ΔBA=−[(12×365EI)(122)+12×12×(1,630EI−365EI)(23×12)]=−87,000 kips⋅ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔB=ΔBA=−87,000 kips⋅ft3×(12 in.1 ft)329,000×6,000=−0.864 in.≃0.86 in.(↓)$

Hence, the deflection at B is $0.86 in.(↓)_$ .

To determine the slope at point E, it is necessary to determine the deflection at point C and the deflection between C and E.

Determine the deflection at C and A using the relation;

$ΔC=ΔCA=Moment of the area of the M/EI diagram between C and A about A=−[(b1h1)(12+b12)+(12b2h2)(12+23×b2)+(12b3h3)(23×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $(1,630EI−365EI)$ for $h2$ , 12 for $b3$ , and $365EI$ for $h3$ .

$ΔC=ΔCA=−[(12×365EI)(12+122)+(12×12×(1,630EI−365EI))(12+23×12)+(12×12×365EI)(23×12)]=−248,160 kips-ft3EI$

Determine the deflection between C and E using the relation;

$ΔCE=Moment of the area of the M/EI diagram between C and E about E=−[(12b1h1)(23×b1)+(b2h2)(12+b22)+(12b3h3)(12+13×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $250EI$ for $h2$ , 12 ft for $b3$ , and $(365EI−250EI)$ for $h3$ .

$ΔCE=−[(12×12×365EI)(23×12)+(12×250EI)(12+122)+(12×12×(365EI−250EI))(12+13×12)]=−82,560 kips-ft3EI$

Determine the slope at E using the relation;

$θE=ΔC+ΔCELCE$

Here, $LCE$ is the length between C and E.

Substitute $−248,160 kips-ft3EI$ for $ΔC$ , $−82,560 kips-ft3EI$ for $ΔCE$ , and 24 ft for $LCE$ .

$θE=−248,160 kips-ft3EI+(−82,560 kips-ft3EI)24=−13,780 kips-ft2EI$

Determine the slope between D and E using the relation;

$θDE=Area of the MEI diagram between D and E=(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $250EI$ for $h2$ .

$θDE=−[(12×12×365EI)+(12×12×250EI)]=−1EI(365+2502)12=−3,690 kips-ft2EI$

Determine the slope at D using the relation;

$θD=θE−θDE$

Substitute $−3,690 kips-ft2EI$ for $θE$ and $−3,690 kips-ft2EI$ for $θDE$ .

$θD=−13,780 kips-ft2EI−(−3,690 kips-ft2EI)=−10,090 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θD=−10,090 kips-ft2×(12 in.1 ft)229,000×6,000=−0.00835 rad≃0.0084 rad(Counterclockwise)$

Hence, the slope at D is $0.0084 rad(Counterclockwise)_$ .

Determine the deflection between D and E using the relation;

$ΔDE=Moment of the area of the M/EI diagram between D and E about E=−[(b1h1)(b12)+(12b2h2)(13×b2)]$

Here, b is the width and h is the height of respective rectangle and triangle.

Substitute 12 ft for $b1$ , $250EI$ for $h1$ , 12 for $b2$ , and $(365EI−250EI)$ for $h2$ .

$ΔDE=−[(12×250EI)(122)+(12×12×(365EI−250EI))(13×12)]=−20,760 kips-ft3EI$

Determine the deflection at D using the relation;

$ΔD=LDEθE−ΔDE$

Substitute 12 ft for $LDE$ , $−13,780 kips-ft2EI$ for $θE$ , and $−20,760 kips-ft3EI$ for $ΔDE$ .

$ΔD=12(−13,780 kips-ft2EI)−(−20,760 kips-ft3EI)=−165,360+20,760EI=−144,600 kips-ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔD=−144,600 kips-ft3×(12 in.1 ft)329,000×6,000=−1.44 in.=1.44 in.(↓)$

Hence, the deflection at point D is $1.44 in.(↓)_$ .

Answer 6

Chapter 6, Problem 33P

To determine

Find the slope $θB&θD$ and deflection $ΔB&ΔD$ at point B and D of the given beam using the moment-area method.

Expert Solution & Answer

Answer to Problem 33P

The slope $θB$ at point B of the given beam using the moment-area method is $0.0099 rad(Clockwise)_$ .

The deflection $ΔB$ at point B of the given beam using the moment-area method is $0.86 in.(↓)_$ .

The slope $θD$ at point D of the given beam using the moment-area method is $0.0084 rad(Counterclockwise)_$ .

The deflection $ΔD$ at point D of the given beam using the moment-area method is $1.44 in.(↓)_$ .

Explanation of Solution

Given information:

The Young’s modulus (E) is 29,000 ksi.

The moment of inertia (I) is $6,000 in.4$ .

Calculation:

Consider flexural rigidity EI of the beam is constant.

Show the free body diagram of the given beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 1

Refer Figure (1),

Consider upward is force is positive and downward force is negative.

Consider clockwise moment is negative and counterclockwise moment is positive.

Split the given beam into two sections such as AC and CE.

Consider the portion CE:

Draw the free body diagram of the portion CE as in Figure (2).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 2

Refer Figure (2),

Consider a reaction at C and take moment about point C.

Determine the reaction at E;

$RE×(24)−(40×12)+250=0RE=23024RE=9.58 kips$

Determine the reaction at support A;

$∑V=0RA+RE−75−40=0RA=115−9.58RA=105.42 kips$

Show the reaction of the given beam as in Figure (3).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 3

Refer Figure (3),

Determine the moment at A:

$MA=(9.58×48)+250−(40×36)−(75×12)=709.84−2,340≃−1,630 kips-ft$

Determine the bending moment at B;

$MB=(9.58×36)+250−(40×24)=344.88+250−960≃−365 kips-ft$

Determine the bending moment at C;

$MC=(9.58×24)+250−(40×12)=479.92−480≃0$

Determine the bending moment at D;

$MD=(9.58×12)+250≃365 kips-ft$

Determine the bending moment at E;

$ME=250 kips-ft$

Show the $M/EI$ diagram for the given beam as in Figure (4).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 4

Show the elastic curve diagram as in Figure (5).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 5

Refer Figure (4),

Determine the slope at B;

$θB=θBA=Area of the MEI diagram between A and B=−(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $1,630EI$ for $h1$ , 12 ft for $b2$ , and $365EI$ for $h2$ .

$θB=θBA=−[(12×12×1,630EI)+(12×12×365EI)]=−1EI(1,630+3652)12=−11,970 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θB=−11,970 kips-ft2×(12 in.1 ft)229,000×6,000=−0.0099 rad=0.0099 (Clockwise)$

Hence, the slope at B is $0.0099 rad(Clockwise)_$ .

Determine the deflection between A and B using the relation;

$ΔB=ΔBA=Moment of the area of the M/EI diagram between B and A about A=−[(b1h1)(b12)+(12b2h2)(23×b2)]$

Here, $b1$ is the width of rectangle, $h1$ is the height of the rectangle, $b2$ is the width of the triangle, and $h2$ is the height of the triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $(1,630EI−365EI)$ for $h2$ .

$ΔB=ΔBA=−[(12×365EI)(122)+12×12×(1,630EI−365EI)(23×12)]=−87,000 kips⋅ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔB=ΔBA=−87,000 kips⋅ft3×(12 in.1 ft)329,000×6,000=−0.864 in.≃0.86 in.(↓)$

Hence, the deflection at B is $0.86 in.(↓)_$ .

To determine the slope at point E, it is necessary to determine the deflection at point C and the deflection between C and E.

Determine the deflection at C and A using the relation;

$ΔC=ΔCA=Moment of the area of the M/EI diagram between C and A about A=−[(b1h1)(12+b12)+(12b2h2)(12+23×b2)+(12b3h3)(23×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $(1,630EI−365EI)$ for $h2$ , 12 for $b3$ , and $365EI$ for $h3$ .

$ΔC=ΔCA=−[(12×365EI)(12+122)+(12×12×(1,630EI−365EI))(12+23×12)+(12×12×365EI)(23×12)]=−248,160 kips-ft3EI$

Determine the deflection between C and E using the relation;

$ΔCE=Moment of the area of the M/EI diagram between C and E about E=−[(12b1h1)(23×b1)+(b2h2)(12+b22)+(12b3h3)(12+13×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $250EI$ for $h2$ , 12 ft for $b3$ , and $(365EI−250EI)$ for $h3$ .

$ΔCE=−[(12×12×365EI)(23×12)+(12×250EI)(12+122)+(12×12×(365EI−250EI))(12+13×12)]=−82,560 kips-ft3EI$

Determine the slope at E using the relation;

$θE=ΔC+ΔCELCE$

Here, $LCE$ is the length between C and E.

Substitute $−248,160 kips-ft3EI$ for $ΔC$ , $−82,560 kips-ft3EI$ for $ΔCE$ , and 24 ft for $LCE$ .

$θE=−248,160 kips-ft3EI+(−82,560 kips-ft3EI)24=−13,780 kips-ft2EI$

Determine the slope between D and E using the relation;

$θDE=Area of the MEI diagram between D and E=(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $250EI$ for $h2$ .

$θDE=−[(12×12×365EI)+(12×12×250EI)]=−1EI(365+2502)12=−3,690 kips-ft2EI$

Determine the slope at D using the relation;

$θD=θE−θDE$

Substitute $−3,690 kips-ft2EI$ for $θE$ and $−3,690 kips-ft2EI$ for $θDE$ .

$θD=−13,780 kips-ft2EI−(−3,690 kips-ft2EI)=−10,090 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θD=−10,090 kips-ft2×(12 in.1 ft)229,000×6,000=−0.00835 rad≃0.0084 rad(Counterclockwise)$

Hence, the slope at D is $0.0084 rad(Counterclockwise)_$ .

Determine the deflection between D and E using the relation;

$ΔDE=Moment of the area of the M/EI diagram between D and E about E=−[(b1h1)(b12)+(12b2h2)(13×b2)]$

Here, b is the width and h is the height of respective rectangle and triangle.

Substitute 12 ft for $b1$ , $250EI$ for $h1$ , 12 for $b2$ , and $(365EI−250EI)$ for $h2$ .

$ΔDE=−[(12×250EI)(122)+(12×12×(365EI−250EI))(13×12)]=−20,760 kips-ft3EI$

Determine the deflection at D using the relation;

$ΔD=LDEθE−ΔDE$

Substitute 12 ft for $LDE$ , $−13,780 kips-ft2EI$ for $θE$ , and $−20,760 kips-ft3EI$ for $ΔDE$ .

$ΔD=12(−13,780 kips-ft2EI)−(−20,760 kips-ft3EI)=−165,360+20,760EI=−144,600 kips-ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔD=−144,600 kips-ft3×(12 in.1 ft)329,000×6,000=−1.44 in.=1.44 in.(↓)$

Hence, the deflection at point D is $1.44 in.(↓)_$ .

Answer 7

Chapter 6, Problem 33P

To determine

Find the slope $θB&θD$ and deflection $ΔB&ΔD$ at point B and D of the given beam using the moment-area method.

Answer 8

Expert Solution & Answer

Answer to Problem 33P

The slope $θB$ at point B of the given beam using the moment-area method is $0.0099 rad(Clockwise)_$ .

The deflection $ΔB$ at point B of the given beam using the moment-area method is $0.86 in.(↓)_$ .

The slope $θD$ at point D of the given beam using the moment-area method is $0.0084 rad(Counterclockwise)_$ .

The deflection $ΔD$ at point D of the given beam using the moment-area method is $1.44 in.(↓)_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The Young’s modulus (E) is 29,000 ksi.

The moment of inertia (I) is $6,000 in.4$ .

Calculation:

Consider flexural rigidity EI of the beam is constant.

Show the free body diagram of the given beam as in Figure (1).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 1

Refer Figure (1),

Consider upward is force is positive and downward force is negative.

Consider clockwise moment is negative and counterclockwise moment is positive.

Split the given beam into two sections such as AC and CE.

Consider the portion CE:

Draw the free body diagram of the portion CE as in Figure (2).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 2

Refer Figure (2),

Consider a reaction at C and take moment about point C.

Determine the reaction at E;

$RE×(24)−(40×12)+250=0RE=23024RE=9.58 kips$

Determine the reaction at support A;

$∑V=0RA+RE−75−40=0RA=115−9.58RA=105.42 kips$

Show the reaction of the given beam as in Figure (3).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 3

Refer Figure (3),

Determine the moment at A:

$MA=(9.58×48)+250−(40×36)−(75×12)=709.84−2,340≃−1,630 kips-ft$

Determine the bending moment at B;

$MB=(9.58×36)+250−(40×24)=344.88+250−960≃−365 kips-ft$

Determine the bending moment at C;

$MC=(9.58×24)+250−(40×12)=479.92−480≃0$

Determine the bending moment at D;

$MD=(9.58×12)+250≃365 kips-ft$

Determine the bending moment at E;

$ME=250 kips-ft$

Show the $M/EI$ diagram for the given beam as in Figure (4).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 4

Show the elastic curve diagram as in Figure (5).

Structural Analysis, SI Edition, Chapter 6, Problem 33P , additional homework tip 5

Refer Figure (4),

Determine the slope at B;

$θB=θBA=Area of the MEI diagram between A and B=−(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $1,630EI$ for $h1$ , 12 ft for $b2$ , and $365EI$ for $h2$ .

$θB=θBA=−[(12×12×1,630EI)+(12×12×365EI)]=−1EI(1,630+3652)12=−11,970 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θB=−11,970 kips-ft2×(12 in.1 ft)229,000×6,000=−0.0099 rad=0.0099 (Clockwise)$

Hence, the slope at B is $0.0099 rad(Clockwise)_$ .

Determine the deflection between A and B using the relation;

$ΔB=ΔBA=Moment of the area of the M/EI diagram between B and A about A=−[(b1h1)(b12)+(12b2h2)(23×b2)]$

Here, $b1$ is the width of rectangle, $h1$ is the height of the rectangle, $b2$ is the width of the triangle, and $h2$ is the height of the triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $(1,630EI−365EI)$ for $h2$ .

$ΔB=ΔBA=−[(12×365EI)(122)+12×12×(1,630EI−365EI)(23×12)]=−87,000 kips⋅ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔB=ΔBA=−87,000 kips⋅ft3×(12 in.1 ft)329,000×6,000=−0.864 in.≃0.86 in.(↓)$

Hence, the deflection at B is $0.86 in.(↓)_$ .

To determine the slope at point E, it is necessary to determine the deflection at point C and the deflection between C and E.

Determine the deflection at C and A using the relation;

$ΔC=ΔCA=Moment of the area of the M/EI diagram between C and A about A=−[(b1h1)(12+b12)+(12b2h2)(12+23×b2)+(12b3h3)(23×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $(1,630EI−365EI)$ for $h2$ , 12 for $b3$ , and $365EI$ for $h3$ .

$ΔC=ΔCA=−[(12×365EI)(12+122)+(12×12×(1,630EI−365EI))(12+23×12)+(12×12×365EI)(23×12)]=−248,160 kips-ft3EI$

Determine the deflection between C and E using the relation;

$ΔCE=Moment of the area of the M/EI diagram between C and E about E=−[(12b1h1)(23×b1)+(b2h2)(12+b22)+(12b3h3)(12+13×b3)]$

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , $250EI$ for $h2$ , 12 ft for $b3$ , and $(365EI−250EI)$ for $h3$ .

$ΔCE=−[(12×12×365EI)(23×12)+(12×250EI)(12+122)+(12×12×(365EI−250EI))(12+13×12)]=−82,560 kips-ft3EI$

Determine the slope at E using the relation;

$θE=ΔC+ΔCELCE$

Here, $LCE$ is the length between C and E.

Substitute $−248,160 kips-ft3EI$ for $ΔC$ , $−82,560 kips-ft3EI$ for $ΔCE$ , and 24 ft for $LCE$ .

$θE=−248,160 kips-ft3EI+(−82,560 kips-ft3EI)24=−13,780 kips-ft2EI$

Determine the slope between D and E using the relation;

$θDE=Area of the MEI diagram between D and E=(12×b1×h1+12×b2×h2)$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b1$ , $365EI$ for $h1$ , 12 ft for $b2$ , and $250EI$ for $h2$ .

$θDE=−[(12×12×365EI)+(12×12×250EI)]=−1EI(365+2502)12=−3,690 kips-ft2EI$

Determine the slope at D using the relation;

$θD=θE−θDE$

Substitute $−3,690 kips-ft2EI$ for $θE$ and $−3,690 kips-ft2EI$ for $θDE$ .

$θD=−13,780 kips-ft2EI−(−3,690 kips-ft2EI)=−10,090 kips-ft2EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$θD=−10,090 kips-ft2×(12 in.1 ft)229,000×6,000=−0.00835 rad≃0.0084 rad(Counterclockwise)$

Hence, the slope at D is $0.0084 rad(Counterclockwise)_$ .

Determine the deflection between D and E using the relation;

$ΔDE=Moment of the area of the M/EI diagram between D and E about E=−[(b1h1)(b12)+(12b2h2)(13×b2)]$

Here, b is the width and h is the height of respective rectangle and triangle.

Substitute 12 ft for $b1$ , $250EI$ for $h1$ , 12 for $b2$ , and $(365EI−250EI)$ for $h2$ .

$ΔDE=−[(12×250EI)(122)+(12×12×(365EI−250EI))(13×12)]=−20,760 kips-ft3EI$

Determine the deflection at D using the relation;

$ΔD=LDEθE−ΔDE$

Substitute 12 ft for $LDE$ , $−13,780 kips-ft2EI$ for $θE$ , and $−20,760 kips-ft3EI$ for $ΔDE$ .

$ΔD=12(−13,780 kips-ft2EI)−(−20,760 kips-ft3EI)=−165,360+20,760EI=−144,600 kips-ft3EI$

Substitute 29,000 ksi for E and $6,000 in.4$ for I.

$ΔD=−144,600 kips-ft3×(12 in.1 ft)329,000×6,000=−1.44 in.=1.44 in.(↓)$

Hence, the deflection at point D is $1.44 in.(↓)_$ .

Answer 12

Ch. 6 - Prob. 1P Ch. 6 - Prob. 2P Ch. 6 - Prob. 3P Ch. 6 - Prob. 4P Ch. 6 - Prob. 5P Ch. 6 - Prob. 6P Ch. 6 - Prob. 7P Ch. 6 - Prob. 8P Ch. 6 - Prob. 9P Ch. 6 - Prob. 10P

Find the slope θ B & θ D and deflection Δ B & Δ D at point B and D of the given beam using the moment-area method.

Concept explainers

Find the slope $θB&θD$ and deflection $ΔB&ΔD$ at point B and D of the given beam using the moment-area method.

Answer to Problem 33P

Explanation of Solution

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Chapter 6 Solutions

Find the slope θ B &amp; θ D and deflection Δ B &amp; Δ D at point B and D of the given beam using the moment-area method.

Concept explainers

Answer to Problem 33P

Explanation of Solution

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Chapter 6 Solutions

Find the slope θ B & θ D and deflection Δ B & Δ D at point B and D of the given beam using the moment-area method.