Area Moment Method from Mathalino

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Solution to Problem 636 | Deflection of Cantilever Beams Problem 636 The cantilever beam shown in Fig. P636 has a rectangular crosssection 50 mm wide by h mm high. Find the height h if the maximum deflection is not to exceed 10 mm. Use E = 10 GPa.

Solution 636 Click here to show or hide the solution

1 tA/B =

EI

¯ (AreaAB ) X A

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Solution to Problem 636 | Deflection of Cantilever Beams | Strength of Materials Review 1

−10 =

[−

3

10 000 (

50h

1 2

(2)(4)(

10 3

)−

1 2

(4)(16)(

8 3

4

)] (1000 )

)

12 3

−10 =

3

[−

296

125 000h

4

] (1000 )

3

4

3

h

−296(1000 ) = 125 000(−10)

h = 618.67 mm           answer

Tags: concentrated load cantilever beam beam deflection maximum deflection end deflection point load ‹ Deflection of Cantilever Beams | AreaMoment Method

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Matched Content Strength of Materials Chapter 01 Simple Stresses Chapter 02 Strain Chapter 03 Torsion Chapter 04 Shear and Moment in Beams Chapter 05 Stresses in Beams Chapter 06 Beam Deflections Double Integration Method | Beam Deflections Moment Diagram by Parts AreaMoment Method | Beam Deflections http://www.mathalino.com/reviewer/mechanicsandstrengthofmaterials/solutiontoproblem636deflectionofcantileverbeams

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Solution to Problem 637 | Deflection of Cantilever Beams Problem 637 For the beam loaded as shown in Fig. P637, determine the deflection 6 ft from the wall. Use E = 1.5 × 106 psi and I = 40 in4.

Solution 637 Click here to show or hide the solution RC = 80(8) = 640 lb MC = 80(8)(4) = 2560 lb ⋅ ft

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Solution to Problem 637 | Deflection of Cantilever Beams | Strength of Materials Review

tB/C =

tB/C =

1 EI 1 EI

¯ (AreaBC )X B

[

1 2

(6)(3840)(2) − 6(2560)(3) −

1 tB/C =

1 3

3

(6)(1440)(1.5) ] (12 )

3

[ 27 360 ] (12 ) EI 1

tB/C =

3

6

[ 27 360 ] (12 )

(1.5 × 10 )(40) tB/C = −0.787968 in

Thus, δB = | tB/C | = 0.787968 in           answer

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Solution to Problem 638 | Deflection of Cantilever Beams Problem 638 For the cantilever beam shown in Fig. P638, determine the value of EIδ at the left end. Is this deflection upward or downward?

Solution 638 Click here to show or hide the solution

¯ EI tA/B = (AreaAB ) X A EI tA/B = 2(2)(3) −

20

1 2

(4)(1)(

3

8 3

)

= = 6.67 kN ⋅ http://www.mathalino.com/reviewer/mechanicsandstrengthofmaterials/solutiontoproblem638deflectionofcantileverbeams

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Solution to Problem 638 | Deflection of Cantilever Beams | Strength of Materials Review

EI tA/B =

20 3

= 6.67 kN ⋅ m

3

∴   EIδ = 6.67 kN·m3 upward           answer

Tags: moment load cantilever beam beam deflection end deflection ‹ Solution to Problem 637 | Deflection of Cantilever Beams

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Matched Content Strength of Materials Chapter 01 Simple Stresses Chapter 02 Strain Chapter 03 Torsion Chapter 04 Shear and Moment in Beams Chapter 05 Stresses in Beams Chapter 06 Beam Deflections Double Integration Method | Beam Deflections Moment Diagram by Parts AreaMoment Method | Beam Deflections Deflection of Cantilever Beams | AreaMoment Method Solution to Problem 636 | Deflection of Cantilever Beams Solution to Problem 637 | Deflection of Cantilever Beams Solution to Problem 638 | Deflection of Cantilever Beams Solution to Problem 639 | Deflection of Cantilever Beams Solution to Problem 640 | Deflection of Cantilever Beams http://www.mathalino.com/reviewer/mechanicsandstrengthofmaterials/solutiontoproblem638deflectionofcantileverbeams

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Solution to Problem 639 | Deflection of Cantilever Beams Problem 639 The downward distributed load and an upward concentrated force act on the cantilever beam in Fig. P639. Find the amount the free end deflects upward or downward if E = 1.5 × 106 psi and I = 60 in4.

Solution 639 Click here to show or hide the solution

1 tA/C =

EI

¯ (AreaAB ) X A

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Solution to Problem 639 | Deflection of Cantilever Beams | Strength of Materials Review

EI 1

tA/C =

6

[

(1.5 × 10 )(60)

1 2

(6)(5400)(6) −

1 3

3

(8)(6400)(6) ] (12 )

tA/C = −0.09984 in

∴   The free end will move by 0.09984 inch downward.           answer

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Matched Content Strength of Materials Chapter 01 Simple Stresses Chapter 02 Strain Chapter 03 Torsion Chapter 04 Shear and Moment in Beams Chapter 05 Stresses in Beams Chapter 06 Beam Deflections Double Integration Method | Beam Deflections Moment Diagram by Parts AreaMoment Method | Beam Deflections Deflection of Cantilever Beams | AreaMoment Method Solution to Problem 636 | Deflection of Cantilever Beams Solution to Problem 637 | Deflection of Cantilever Beams Solution to Problem 638 | Deflection of Cantilever Beams http://www.mathalino.com/reviewer/mechanicsandstrengthofmaterials/solutiontoproblem639deflectionofcantileverbeams

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Solution to Problem 640 | Deflection of Cantilever Beams Problem 640 Compute the value of δ at the concentrated load in Prob. 639. Is the deflection upward downward?

Solution 640 Click here to show or hide the solution RC = 200(8) − 900 = 700 lb MC = 200(8)(4) − 900(6) = 1000 lb ⋅ ft

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Solution to Problem 640 | Deflection of Cantilever Beams | Strength of Materials Review

tB/C =

1 EI

¯ (AreaBC ) X B

1

tB/C =

6

(1.5 × 10 )(60)

[

1 2

(6)(4200)(2) − 1000(6)(3) −

1 3

3

(6)(3600)(15) ](12 )

tB/C = −0.06912 in

∴   δ = 0.06912 inch downward           answer

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Solution to Problem 641 | Deflection of Cantilever Beams Problem 641 For the cantilever beam shown in Fig. P641, what will cause zero deflection at A?

Solution 641 Click here to show or hide the solution

1 EI

¯ (AreaAC ) X A = 0

1 EI

[

1 2

(4)(4P )(

8 3

) − 2(400)(3) ] = 0

P = 112.5 N            answer

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Solution to Problem 642 | Deflection of Cantilever Beams Problem 642 Find the maximum deflection for the cantilever beam loaded as shown in Figure P642 if the cross section is 50 mm wide by 150 mm high. Use E = 69 GPa.

Solution 642 Click here to show or hide the solution RA = 4(1) = 4 kN MA = 4(1)(2.5) = 10 kN ⋅ m

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Solution to Problem 643 | Deflection of Cantilever Beams Problem 643 Find the maximum value of EIδ for the cantilever beam shown in Fig. P643.

Solution 643 Click here to show or hide the solution

¯ EI tB/A = (AreaAB ) X B EI tB/A =

EI tB/A =

1 2 1 6

L(P L)(

PL

3

3

−

1 3

1 2

L) − P aL( 2

PL a −

2

1 6

1 2

L) −

1 2

P (L − a)

3

(L − a)P (L − a)[

1 3

(L − a) ]

3

2

2

3

= − − ( −3 +3 − ) http://www.mathalino.com/reviewer/mechanicsandstrengthofmaterials/solutiontoproblem643deflectionofcantileverbeams

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Solution to Problem 643 | Deflection of Cantilever Beams | Strength of Materials Review

EI tB/A =

EI tB/A =

1 6 1 6

PL

PL

EI tB/A = −

EI tB/A = −

1 2 1 6

3

3

−

− 2

P La

1 2 1 2

2

PL a − 2

PL a −

+

1 6

1 6 1 6

P (L

PL

3

3

2

2

− 3L a + 3La

+

1 2

2

PL a −

1 2

3

−a ) 2

P La

+

1 6

3

Pa

3

Pa

P a (3L − a) 2

Therefore EI δmax =

1 6

P a (3L − a)           answer 2

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Matched Content Strength of Materials Chapter 01 Simple Stresses Chapter 02 Strain Chapter 03 Torsion Chapter 04 Shear and Moment in Beams Chapter 05 Stresses in Beams Chapter 06 Beam Deflections Double Integration Method | Beam Deflections Moment Diagram by Parts AreaMoment Method | Beam Deflections Deflection of Cantilever Beams | AreaMoment Method http://www.mathalino.com/reviewer/mechanicsandstrengthofmaterials/solutiontoproblem643deflectionofcantileverbeams

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Solution to Problem 644 | Deflection of Cantilever Beams Problem 644 Determine the maximum deflection for the beam loaded as shown in Fig. P644.

Solution 644 Click here to show or hide the solution R = wo ( R =

1 2

1 2

L)

w o L

M = wo (

M =

3 8

1 2

L)(

wo L

2

3 4

L)

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Solution to Problem 644 | Deflection of Cantilever Beams | Strength of Materials Review

tA/B =

tA/B =

1 EI 1

¯ (AreaAB ) X A

[

EI 1

tA/B =

[ EI

1 2

(L)(

1 12

[− EI

tA/B = −

2

wo L

1 tA/B =

1

41 384

41w o L

2

w o L )(

4

−

wo L

4

3 16

1 3

L) −

wo L

4

−

3 8

2

w o L (L)(

1 384

wo L

4

1 2

L) −

1 3

(

1 8

2

w o L )(

1 2

L)(

1 8

L) ]

]

]

4

384EI

Therefore δmax =

41w o L

4

           answer

384EI

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Solution to Problem 645 | Deflection of Cantilever Beams Problem 645 Compute the deflection and slope at a section 3 m from the wall for the beam shown in Fig. P645. Assume that E = 10 GPa and I = 30 × 106 mm4.

Solution 645 Click here to show or hide the solution R =

1 2

(4)(1200)

R = 2400 N

M =

1 2

(4)(1200)(

8 3

)

M = 6400 N ⋅ m

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Solution to Problem 645 | Deflection of Cantilever Beams | Strength of Materials Review

y 3

1200 =

4

y = 900 N/m

1 tB/A =

EI

¯ (AreaAB ) X B

1 tB/A =

[ EI

1 2

(3)(7200)(1) − 3(6400)(1.5) −

1 tB/A =

1 4

3

(3)(1350)(0.6) ](1000 )

3

[ −18607.5 ](1000 )

6

10000(30 × 10 ) tB/A = −62.025 mm

Therefore: δB = 62.025 mm           answer

1 θAB =

EI

(AreaAB )

1 θAB =

[ EI

1 2

(3)(7200) − 3(6400) −

1 θAB =

1 4

2

(3)(1350) ](1000 )

2

6

[ −9412.5 ](1000 )

10 000(30 × 10 ) θAB = −0.031375 radian θAB = 1.798 degree           answer

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Solution to Problem 646 | Deflection of Cantilever Beams Problem 646 For the beam shown in Fig. P646, determine the value of I that will limit the maximum deflection to 0.50 in. Assume that E = 1.5 × 106 psi.

Solution 646 Click here to show or hide the solution M = R =

1 2 1 2

(5)(60)(2 +

5 3

) = 550 lb ⋅ ft

(5)(60) = 150 lb

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Solution to Problem 646 | Deflection of Cantilever Beams | Strength of Materials Review

tA/B =

1 EI 1

−5 =

[ EI

¯ (AreaAB ) X A

1 2

(300)(2)(

26 3

) − 550(2)(9) −

1 4

3

(5)(250)(7) ](12 )

1 −5 =

6

(−16394400)

(1.5 × 10 )I I = 2.18592 in

4

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Matched Content Strength of Materials Chapter 01 Simple Stresses Chapter 02 Strain Chapter 03 Torsion Chapter 04 Shear and Moment in Beams Chapter 05 Stresses in Beams Chapter 06 Beam Deflections Double Integration Method | Beam Deflections Moment Diagram by Parts AreaMoment Method | Beam Deflections Deflection of Cantilever Beams | AreaMoment Method Solution to Problem 636 | Deflection of Cantilever Beams http://www.mathalino.com/reviewer/mechanicsandstrengthofmaterials/solutiontoproblem646deflectionofcantileverbeams

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Solution to Problem 646 | Deflection of Cantilever Beams | Strength of Materials Review

Solution to Problem 637 | Deflection of Cantilever Beams Solution to Problem 638 | Deflection of Cantilever Beams Solution to Problem 639 | Deflection of Cantilever Beams Solution to Problem 640 | Deflection of Cantilever Beams Solution to Problem 641 | Deflection of Cantilever Beams Solution to Problem 642 | Deflection of Cantilever Beams Solution to Problem 643 | Deflection of Cantilever Beams Solution to Problem 644 | Deflection of Cantilever Beams Solution to Problem 645 | Deflection of Cantilever Beams Solution to Problem 646 | Deflection of Cantilever Beams Solution to Problem 647 | Deflection of Cantilever Beams Solution to Problem 648 | Deflection of Cantilever Beams Deflections in Simply Supported Beams | AreaMoment Method Midspan Deflection | Deflections in Simply Supported Beams Method of Superposition | Beam Deflection Conjugate Beam Method | Beam Deflection Strain Energy Method (Castigliano’s Theorem) | Beam Deflection Chapter 07 Restrained Beams Chapter 08 Continuous Beams Chapter 09 Combined Stresses Chapter 10 Reinforced Beams

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Solution to Problem 647 | Deflection of Cantilever Beams Problem 647 Find the maximum value of EIδ for the beam shown in Fig. P647.

Solution 647 Click here to show or hide the solution R = M =

1 2

(

1 2

1 2

(

L)(w o ) =

1 2

L)(w o )(

5 6

1 4

wo L

L) =

5 24

wo L

2

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Solution to Problem 647 | Deflection of Cantilever Beams | Strength of Materials Review

¯ EI tA/B = (AreaAB ) X A EI tA/B =

EI tA/B =

1 2

L(

1 24

EI tA/B = −

1 4

2

w o L )(

wo L 121 1920

4

5

−

wo L

48 4

1 3

L) − L(

wo L

4

−

5 24

2

w o L )(

1 1920

wo L

1 2

L) −

1 4

(

1 2

L)(

1 24

1

2

w o L )(

10

L)

4

Therefore EI δmax =

121 1920

wo L

4

           answer

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Solution to Problem 648 | Deflection of Cantilever Beams Problem 648 For the cantilever beam loaded as shown in Fig. P648, determine the deflection at a distance x from the support.

Solution 648 Click here to show or hide the solution y x

wo

=

y =

L wo

x

L

M =

R =

1 2 1 2

L(w o )(

1 3

L) =

1 6

wo L

2

w o L

Moments about B: Triangular force to the left of B: M1 = −

M1 = −

M1 = −

1 2

1 6

(L − x)(w o − y)(

(L − x)

2

(w o −

w o (L − x)

1 3

)(L − x)

wo x L

)

3

6L

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Solution to Problem 648 | Deflection of Cantilever Beams | Strength of Materials Review

Triangular upward force: M2 =

1 2

(xy)(

1

1

x) =

3

6

2

x

wo x L

3

M2 =

wo x

6L

Rectangle (wo by x): M3 = −w o x(

1 2

x) = −

1 2

2

wo x

Reactions R and M: 1

M4 = Rx =

2

w o Lx

M5 = −M = −

1 6

wo L

2

Deviation at B with the tangent line through C ¯ EI tB/C = (AreaBC ) X B 3

EI tB/C =

EI tB/C =

1 4

x(

wo x

)(

6L wo

5

x

+

120L

1

x) +

5

wo L

3

x

12

−

1 2

1

x(

2

w o Lx)(

wo L

2 2

x

−

12

1 3

x) − (

wo

1 6

2

w o L ) x(

1 2

x) −

1 3

x(

1 2

2

w o x )(

1 4

x)

4

x

24

2

EI tB/C =

wo x

120L

3

(x

2

+ 10L x − 10L

3

− 5Lx ) 2

Therefore, 2

EI δ = −

wo x

120L

3

(x

2

+ 10L x − 10L

3

2

− 5Lx )

2

EI δ =

wo x

(10L

3

2

2

− 10L x + 5Lx

− x )           answer 3

120L

Tags: cantilever beam triangular load uniformly varying load beam deflection ‹ Solution to Problem 647 | Deflection of Cantilever

up

Beams

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Solution to Problem 653 | Deflections in Simply Supported Beams Problem 653 Compute the midspan value of EIδ for the beam shown in Fig. P653. (Hint: Draw the M diagram by parts, starting from midspan toward the ends. Also take advantage of symmetry to note that the tangent drawn to the elastic curve at midspan is horizontal.)

Solution 653 Click here to show or hide the solution By symmetry: R1 = R2 = 600(2) = 1200 N

1 http://www.mathalino.com/reviewer/mechanicsandstrengthofmaterials/solutiontoproblem653deflectionsinsimplysupported

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Solution to Problem 653 | Deflections in Simply Supported Beams | Strength of Materials Review

tA/B =

1 EI

¯ (AreaAB ) X A

1 tA/B =

[ EI 3350

tA/B =

1 2

(2.5)(3000)(

5 3

)+

1 3

(0.5)(75)(

19 8

)−

1 3

(2.5)(1875)(

15 8

)]

EI

From the figure δmidspan = tA/B

Thus EI δmidspan = 3350 N ⋅ m

3

           answer

Tags: moment diagram simple beam uniformly distributed load beam deflection maximum deflection midspan deflection ‹ Deflections in Simply Supported Beams | Area

up

Moment Method

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Matched Content Strength of Materials Chapter 01 Simple Stresses Chapter 02 Strain Chapter 03 Torsion Chapter 04 Shear and Moment in Beams Chapter 05 Stresses in Beams Chapter 06 Beam Deflections Double Integration Method | Beam Deflections http://www.mathalino.com/reviewer/mechanicsandstrengthofmaterials/solutiontoproblem653deflectionsinsimplysupported

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Solution to Problem 654 | Deflections in Simply Supported Beams Problem 654 For the beam in Fig. P654, find the value of EIδ at 2 ft from R2. (Hint: Draw the reference tangent to the elastic curve at R2.)

Solution 654 Click here to show or hide the solution ΣMR2 = 0 6R1 = 80(4)(4) R1 =

640 3

lb

ΣMR1 = 0 6R2 = 80(4)(2) R2 =

320 3

lb

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Solution to Problem 654 | Deflections in Simply Supported Beams | Strength of Materials Review

tA/C =

tA/C =

1 EI 1

¯ (AreaAC ) X A

[

EI

1 2

8960 tA/C =

(4)(2560/3)(

8 3

)+

1 2

(2)(

640 3

)(4 +

2 3

)−

1 3

(4)(640)(3) ]

3EI

1 tB/C =

EI

¯ (AreaBC ) X B

1 tB/C =

tB/C =

EI

[

1 2

1280

(2)(

640 3

)(

2 3

)]

9EI

By ratio and proportion: y

tA/C =

2

6 2

y =

8960 (

6

) 3EI

8960 y =

9EI

δB = y − tB/C

δB =

δB =

8960

1280 −

9EI

9EI

2560 3EI

EI δB =

2560 3

3

lb ⋅ ft

           answer

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Solution to Problem 655 | Deflections in Simply Supported Beams Problem 655 Find the value of EIδ under each concentrated load of the beam shown in Fig. P655.

Solution 655 Click here to show or hide the solution ΣMR2 = 0 8R1 = 200(5) + 400(1) R1 = 175 lb

ΣMR1 = 0 8R2 = 200(3) + 400(7) R2 = 425 lb

yC 1 7

1400 =

8

y C 1 = 1225 lb

yC 2

−1000 =

4

5

y C 2 = −800 lb

yB 3

1400 =

8

y B = 525 lb

¯ EI tD/A = (AreaAD ) X D EI tD/A =

1 2

8

(8)(1400)(

3 3

EI tD/A = 10 700 lb ⋅ ft

)−

1 2

(5)(1000)(

5 3

¯ EI tC /A = (AreaAC ) X C EI tC /A =

EI tC /A =

EI tC /A =

1 2 1 2

(7)(y C 1 )(

7 3

(7)(1225)(

47 225 6

3

lb ⋅ ft

)− 7 3

¯ EI tB/A = (AreaAB ) X B EI tC /A =

=

1 2

1

(3)(y B )(1)

(3)(525)(1)

1 2

)−

(4)(y C 2 )( 1 2

4 3

(4)(800)(

) 4 3

)

)−

1 2

(1)(400)(

1 3

)

EI tC /A = EI tC /A =

1 2

(3)(525)(1)

1575 2

3

lb ⋅ ft

By ratio and proportion: ¯ BE

¯ CF =

tD/A =

3

7

¯ BE =

3

C¯ F =

7

8

8

8

tD/A =

tD/A =

3 8 7 8

(10 700) =

(10 700) =

8025 2 18 725 2

Deflections: ¯ −t δB = BE B/A ¯ EI δB = EI BE − EI tB/A = 3

EI δB = 3225 lb ⋅ ft

8025 2

−

1575 2

            → answer

¯ δC = C F − tC /A ¯ EI δC = EI C F − EI tC /A =

EI δC =

4475 3

18 725 2 3

= 1491.67 lb ⋅ ft

−

47 225 6

           answer

Tags: moment diagram simple beam concentrated load beam deflection point load moment diagram by parts ‹ Solution to Problem 654 | Deflections in Simply

up

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Solution to Problem 656 | Deflections in Simply Supported Beams Problem 656 Find the value of EIδ at the point of application of the 200 N·m couple in Fig. P656.

Solution 656 Click here to show or hide the solution ΣMR2 = 0 4R1 = 500(3) + 200 R1 = 425 N

ΣMR1 = 0 4R2 + 200 = 500(1) R2 = 75 N

¯ EI tD/A = (AreaAD ) X D 1

EI tD/A =

(1)(75)(

2

6550

EI tD/A =

2 3

N ⋅ m

3

)+

3

1 2

(3)(1275)(2) −

1 2

(2)(1000)(

5 3

)

¯ EI tC /A = (AreaAC ) X C 1

EI tC /A =

2

(3)(1275)(1) −

7475

EI tC /A =

6

N ⋅ m

3

1 2

(2)(1000)(

2 3

)

C¯ E

tD/A =

3 ¯ CE =

4 3

6550 (

4

3275 ) =

3EI

¯ EI C E =

3275 2

2EI

N ⋅ m

3

¯ δC = C E − tC /A ¯ EI δC = EI C E − EI tC /A EI δC =

3275 2

−

7475 6

=

EI δC = 391.67 N ⋅ m

3

1175 3

           answer

Tags: simple beam concentrated load moment load beam deflection elastic curve ‹ Solution to Problem 655 | Deflections in Simply Supported Beams

up Solution to Problem 657 | Deflections in Simply Supported Beams ›

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Solution to Problem 657 | Deflections in Simply Supported Beams Problem 657 Determine the midspan value of EIδ for the beam shown in Fig. P657.

Solution 657 Click here to show or hide the solution ΣMR1 = 0 6R2 =

R2 =

1 2

(4)(600)(

800 3

4 3

)

N

¯ EI tA/B = (AreaAB ) X A EI tA/B =

1 2

(6)(1600)(2) −

EI tA/B = 8320 N ⋅ m

3

1 4

(4)(1600)(

¯ EI tM /B = (AreaM B ) X M EI tM /B =

1 2

(3)(800)(1) −

EI tM /B = 1198.75 N ⋅ m

3

1 4

(1)(25)(

1 5

)

4 5

)

By ratio and proportion: δm + tM /B

tA/B =

3 δm + tM /B =

6 1 2

tA/B 1

EI δm + EI tM /B = EI EI δm + 1198.75 = EI EI δm = 2961.25 N ⋅ m

2 1 2

3

tA/B

(8320)

           answer

Tags: simple beam triangular load uniformly varying load beam deflection elastic curve triangle load decreasing load ‹ Solution to Problem 656 | Deflections in Simply Supported Beams

up Solution to Problem 658 | Deflections in Simply Supported Beams › 43741 reads

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Matched Content Strength of Materials Chapter 01 Simple Stresses Chapter 02 Strain Chapter 03 Torsion Chapter 04 Shear and Moment in Beams Chapter 05 Stresses in Beams Chapter 06 Beam Deflections Double Integration Method | Beam Deflections Moment Diagram by Parts AreaMoment Method | Beam Deflections Deflection of Cantilever Beams | AreaMoment Method













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Solution to Problem 658 | Deflections in Simply Supported Beams Problem 658 For the beam shown in Fig. P658, find the value of EIδ at the point of application of the couple.

Solution 658 Click here to show or hide the solution y a

M =

L

y = M a/L

¯ EI tB/A = (AreaAB ) X B EI tB/A =

EI tB/A =

1 2 1 6

(ay)(

1 3

a)

2

a (M a/L) 3

EI tB/A =

Ma

6L

¯ EI tC /A = (AreaAC ) X C EI tC /A =

=

1 2

(LM )(

2

1 3

−

L) − M (L − a)[

(

−

2

1 2

(L − a) ]

EI tC /A =

1 6

ML

2

−

1 2

M (L − a)

2

By ratio and proportion: δB + tB/A

tC /A =

a

L a

δB =

L

tC /A − tB/A a

EI δB =

EI δB =

EI δB =

L

EI tC /A − EI tB/A

3

a L a

[

[

L

1 6

1 6

ML

ML

Ma EI δB =

[L

EI δB =

2

−

1 2

1 2

M (L − a)

M (L − a)

− 3(L − a)

2

2

2

2

−a

Ma ]−

−

1 6

6L 2

Ma

]

]

[L

2

− 3(L

2

2

2

− 2La + a ) − a

]

6L Ma

EI δB =

−

6L Ma

EI δB =

2

2

6L Ma 6L

EI δB = −

[L

2

− 3L

(3L

2

+ 6La − 3a

2

+ 6La − 4a

2

− 6La + 4a )

[ −3L

Ma

2

2

2

−a

]

]

2

6L

The negative sign indicates that the deflection is opposite to the direction sketched in the figure. Thus, Ma EI δB =

(3L

2

− 6La + 4a )   upward           answer 2

6L

Tags: simple beam moment load beam deflection elastic curve moment diagram by parts ‹ Solution to Problem 657 | Deflections in Simply

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Solution to Problem 659 | Deflections in Simply Supported Beams Problem 659 A simple beam supports a concentrated load placed anywhere on the span, as shown in Fig. P659. Measuring x from A, show that the maximum deflection occurs at x = √[(L2 b2)/3].

Solution 659 Click here to show or hide the solution ΣMR2 = 0 LR1 = P b R1 = P b/L

ΣMR1 = 0 LR2 = P a R2 = P a/L

y

Pb =

x

L Pb

y =

x

L

1

tA/D =

1 EI

¯ (AreaAD ) X A

1 tA/D =

1

[

1 tA/D =

1

[

tA/D =

3

EI 1

3

x]

2

1

[

2

x y]

3

EI 1

tA/D =

xy(

2

EI

Pb

2

x

Pb

(

x)] L

3

x

EI 3L

tC /D =

1 EI

¯ (AreaC D ) X C

1 tC /D =

1

[ EI 1

tC /D =

1

[

1 tC /D =

1

[ EI

2

EI

x) + L

2

x (1 −

)+ L

Pb [

2

(L − x) y −

2

Pb (P b −

P b(L − x)

6

1 tC /D =

(L − x)

6

EI

1

2

(L − x) (P b − y) +

6

6L

(L − x)

Pb

3

+

1 2

1 2

6

(L − x)

Pb

2

3

]

Pb (

x) − L

P b(L − x)

2

x (

)− L

Pb

2

(L − x) x −

2L

1

6

1 6

3

]

From the figure: tA/D = tC /D 1

Pb

1

3

x

[

EI 3L Pb

3

x

EI

3

3

3

2x

3

2x

= L

0 = L 2

3

3

Pb +

3

2

2

= L

2

−b

2

2

2

− Lb

2

2

2

− − − − − − − L2 − b 2

   (okay!)

3

−b

+ 3(L − x) x − Lb

−b

3

6

L

2

− 3x

Pb

2

− 3L x + 3Lx 2

] 6

2

3(L − x) x

2

− 3Lx

3

(L − x) x −

− 3L x + 3Lx

x = √

3

Pb

2

(L − x) x − 2L

+

= (L − x)

3

3

L

= (L

Pb +

2L

(L − x)

0 = L

3x

(L − x)

=

3

6L

6L

2x

2x

(L − x)

Pb =

3L

L

Pb

=

2

2

3

− x ) + 3(L 3

−x

2

2

2

− 2Lx + x )x − Lb 2

+ 3L x − 6Lx

3

+ 3x

− Lb

2

2

1 6

3

Pb ]

3

Pb ]

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