Moment Diagram by Parts Solman

7/25/2019 Moment Diagram by Parts Solman

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Moment Diagram by PartsProblem 624

For the beam loaded as shown in Fig. P-624, compute the moment of area of the M

diagrams between the reactions about both the left and the right reaction.

Solution 624

MR2=0

6R1=400+1000(2)

R1=400N

MR1=0

6R2+400=1000(2)

R2=600N

Moment diagram by parts can be drawn in different ways; three are shown below.

(AreaAB)XA=12(2)(800)(43)+12(4)(2400)(103)12(2)(2000)(83)(AreaAB)XA=11733.33Nm3

(AreaAB)XB=12(2)(800)(143)+12(4)(2400)(83)12(2)(2000)(103)(AreaAB)XB=9866.67Nm3


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Problem 626

For the eam loaded as shown in Fig. P-626, compute the moment of area of the M diagrams

between the reactions about both the left and the right reaction.

Solution 626

By symmetryR1=R2=12(400)(3)R1=R2=600lb

and

(AreaAB)XA=(AreaAB)XB

(AreaAB)XA=12(5)(1500)(52)13(3)(450)(52) answer(AreaAB)XA=8250lbft3

Thus,(AreaAB)XB=8250lbft3 answer


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Problem 627

For the beam loaded as shown in Fig. P-627compute the moment of area of the M diagrams

between the reactions about both the left and the right reaction. (Hint: Resolve the

trapezoidal loading into a uniformly distributed load and a uniformly varying load.)

Solution 627

MR2=04R1=200(4)(2)+12(3)(400)(1)R1=550N

MR1=04R2=200(4)(2)+12(3)(400)(3)R2=850N

(AreaAB)XA=12(4)(2200)(83)13(4)(1600)(3)14(3)(600)(175)

(AreaAB)XA=3803.33Nm3 answer

(AreaAB)XB=12(4)(2200)(43)13(4)(1600)(1)14(3)(600)(35)(AreaAB)XB=3463.33Nm3 answer


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Problem 628

For the beam loaded with uniformly varying load and a couple as shown in Fig. P-628

compute the moment of area of the M diagrams between the reactions about both the left

and the right reaction.

Solution 628

MR2=0

10R1+400=12(6)(200)(2)R1=80lb

MR1=010R2=400+12(6)(200)(8)R2=520lb

(AreaAB)XA=400(8)(6)+12(10)(800)(203)14(6)(1200)(445)(AreaAB)XA=30026.67lbft3 answer

(AreaAB)XB=400(8)(4)+12(10)(800)(103)14(6)(1200)(65)(AreaAB)XB=23973.33lbft3 answer


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Problem 629

SolveProb. 628if the sense of the couple is counterclockwise instead of clockwiseas

shown in Fig. P-628.

Solution 629

MR2=010R1=400+12(6)(200)(2)R1=160lb

MR1=010R2+400=12(6)(200)(8)R2=440lb

(AreaAB)XA=12(10)(1600)(203)400(8)(6)14(6)(1200)(445)(AreaAB)XA=18293.33lbft3 answer

(AreaAB)XB=12(10)(1600)(103)400(8)(4)14(6)(1200)(65)(AreaAB)XB=11706.67lbft3 answer
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Problem 631

Determine the value of the couple M for the beam loaded as shown in Fig. P-631 so that the

moment of area about A of the M diagram between A and B will be zero. What is the

physical significance of this result?

Solution 631

MA=04R2+M=100(4)(2)R2=20014M

MB=04R1=100(4)(2)+MR1=200+14M

(AreaAB)XA=0

12(4)(800M)(43)13(4)(800)(1)=083(800M)=32003M=400lbft answer

The uniform load over span AB will cause segment ABto deflect downward. The moment load equal to 400 lbftapplied at the free end will cause the slope through B tobe horizontal making the deviation of A from the tangentthrough B equal to zero. The downward deflection therefore due to uniform load will becountered by the moment load.


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Problem 632

For the beam loaded as shown in Fig. P-632, compute the value of (AreaAB) barred(X)A.

From this result, is the tangent drawn to the elastic curve at B directed up or down to the

right? (Hint: Refer to thedeviation equations and rules of sign.)

Solution 632

MB

=0

3R1+200(1)=800(2)(2)R1=1000N

MA=03R2=200(4)+800(2)(1)R1=800N

(AreaAB)XA=12(2)(2000)(43)+12(1)(800)(73)13(2)(1600)(32)12(1)(400)(73)12(1)(200)(83)(AreaAB)XA=1266.67Nm3 answer

The value of (AreaAB) barred(X)Ais positive, therefore point A is above the tangent through B,thus the tangent through B is upward to the right. See the approximate elastic curve shownabove and refer to therules of signfor more information.
http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/area-moment-method-beam-deflectionshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/area-moment-method-beam-deflectionshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/area-moment-method-beam-deflectionshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/area-moment-method-beam-deflections#sign_ruleshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/area-moment-method-beam-deflections#sign_ruleshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/area-moment-method-beam-deflections#sign_ruleshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/area-moment-method-beam-deflections#sign_ruleshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/area-moment-method-beam-deflections


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Deflection of Cantilever Beams

Problem 636

The cantilever beam shown in Fig. P-636 has a rectangular cross-section 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

tA/B=1EI(AreaAB)XA10=110000(50h312)[12(2)(4)(103)12(4)(16)(83)](10004)10=3125000h3[2963](10004)h3=296(10004)125000(10)h=618.67mm answer


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Problem 637

For the beam loaded as shown in Fig. P-637, determine the deflection 6 ft from the wall.

Use E = 1.5 106psi and I = 40 in4.

Solution 637

RC=80(8)=640 lb

MC

=80(8)(4)=2560lb

ft

tB/C=1EI(AreaBC)XBtB/C=1EI[12(6)(3840)(2)6(2560)(3)13(6)(1440)(1.5)](123)tB/C=1EI[27360](123)tB/C=1(1.5106)(40)[27360](123)

tB/C=0.787968in

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


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Problem 638

For the cantilever beam shown in Fig. P-638, determine the value of EIat the left end. Is

this deflection upward or downward?

Solution 638

EItA/B=(AreaAB)XAEItA/B=2(2)(3)12(4)(1)(83)EItA/B=203=6.67 kNm3

EI= 6.67 kNm3upward answer


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Problem 639

The downward distributed load and an upward concentrated force act on the cantilever

beam in Fig. P-639. Find the amount the free end deflects upward or downward if E = 1.5

106psi and I = 60 in4.

Solution 639

tA/C=1EI(AreaAB)XAtA/C=1(1.5106)(60)[12(6)(5400)(6)13(8)(6400)(6)](123)

tA/C=0.09984in

The free end will move by 0.09984 inch downward. answer


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Problem 640

Compute the value of at the concentrated load inProb. 639.Is the deflection upward

downward?

Solution 640

RC=200(8)900=700 lb

MC=200(8)(4)900(6)=1000 lbft

tB/C=1EI(AreaBC)XBtB/C=1(1.5106)(60)[12(6)(4200)(2)1000(6)(3)13(6)(3600)(15)](123)tB/C=0.06912 in

= 0.06912 inch downward answer
http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-639-deflection-of-cantilever-beamshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-639-deflection-of-cantilever-beamshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-639-deflection-of-cantilever-beamshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-639-deflection-of-cantilever-beams


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Problem 641

For the cantilever beam shown in Fig. P-641, what will cause zero deflection at A?

Solution 641

1EI(AreaAC)XA=01EI[12(4)(4P)(83)2(400)(3)]=0P=112.5 N answer


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Problem 642

Find the maximum deflection for the cantilever beam loaded as shown in Figure P-642 if the

cross section is 50 mm wide by 150 mm high. Use E = 69 GPa.

Solution 642

RA=4(1)=4kNMA=4(1)(2.5)=10kNm

tB/A=1EI(AreaAB)XBtB/A=169000[50(1503)12][12(3)(12)(1)3(10)(1.5)13(1)(2)(0.25)](10004)tB/A=28mm

max= 28 mm answer


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Problem 643

Find the maximum value of EIfor the cantilever beam shown in Fig. P-643.

Solution 643

EItB/A=(AreaAB)XBEItB/A=12L(PL)(13L)PaL(12L)12(La)P(La)[13(La)]EItB/A=16PL312PL2a16P(La)3EItB/A=16PL312PL2a16P(L33L2a+3La2a3)EItB/A=16PL312PL2a16PL3+12PL2a12PLa2+16Pa3EItB/A=12PLa2+16Pa3EItB/A=16Pa2(3La)

ThereforeEImax=(1/6)Pa^2(3La) answer


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Problem 644

Determine the maximum deflection for the beam loaded as shown in Fig. P-644.

Solution 644

R=wo(12L)

R=12woL

M=wo(12L)(34L)M=38woL2

tA/B=1/EI(AreaAB)XA

tA/B=1/EI[1/2(L)(1/2woL^2)(1/3L)3/8woL^2(L)(1/2L)1/3(1/8woL^2)(1/2L)(1/8L)]tA/B=1/EI[1/12woL^43/16woL41/384woL^4]tA/B=1/EI[41/384woL^4]tA/B=(41woL^4)/(384EI)

Thereforemax=(41woL^4)/(384EI) answer


18/21

Problem 645

Compute the deflection and slope at a section 3 m from the wall for the beam shown in Fig.

P-645. Assume that E = 10 GPa and I= 30 106mm4.

Solution 645

R=1/2(4)(1200)R=2400N

M=1/2(4)(1200)(8/3)M=6400Nm

y3=12004y=900N/m

tB/A=1EI(AreaAB)XBtB/A=1EI[12(3)(7200)(1)3(6400)(1.5)14(3)(1350)(0.6)](10003)

tB/A=[1/10000(30106) ] [18607.5](1000^3)tB/A=62.025mm

Therefore:B=62.025mm answer

AB=1EI(AreaAB)AB=1EI[12(3)(7200)3(6400)14(3)(1350)](10002)AB=110000(30106)[9412.5](10002)AB=0.031375radianAB=1.798degree answer


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Problem 646For the beam shown in Fig. P-646, determine the value of I that will limit the maximumdeflection to 0.50 in. Assume that E = 1.5 106 psi.

M=12(5)(60)(2+53)=550lbftR=12(5)(60)=150lb

tA/B=1/EI(AreaAB)XA5=1EI[12(300)(2)(263)550(2)(9)14(5)(250)(7)](123) 5=1(1.5106)I(16394400) I=2.18592in4


20/21

Problem 647

Find the maximum value of EIfor the beam shown in Fig. P-647.

Solution 647

R=12(12L)(wo)=1/4woLM=12(12L)(wo)(56L)=5/24woL^2

EItA/B=(AreaAB)XAEItA/B=1/2L(1/4woL^2)(1/3L)L(5/24woL^2)(1/2L)1/4(1/2L)(1/24woL^2)(1/10L)EItA/B=1/24woL^45/48woL^41/1920woL^4EItA/B=(12/11920)woL^4

ThereforeEImax=(121/1920)woL^4 answer


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Problem 648

For the cantilever beam loaded as shown in Fig. P-648, determine the deflection at a

distance x from the support.

Solution 648

y/x=wo/Ly=(wo/L)x

M=1/2L(wo)(1/3L)=1/6woL^2R=12woL

Moments about B:Triangular force to the left of B:M1=1/2(Lx)(woy)(1/3)(Lx)M1=1/6(Lx)^2(wo(wox/L))M1=wo(Lx)^3 /6L

Triangular upward force:M2=1/2(xy)(1/3x)=1/6x^2(wox/L)M2=wox^3 /6L

Rectangle (woby x):

M3=wox(1/2x)=1/2wox^2

Reactions R and M:M4=Rx=1/2woLxM5=M=1/6woL^2

Deviation at B with the tangent line through CEItB/C=(AreaBC)XBEItB/C=1/4x(wox^3 /6L)(1/5x)+1/2x(1/2woLx)(1/3x)(1/6woL^2)x(1/2x)1/3x(1/2wox^2)(1/4x)EItB/C=(wo/120L)x^5+(woL/12)x^3(woL^2/12)x^2(wo/24)x^4EItB/C=(wox^2/120L)(x3+10L2x10L35Lx2)

Therefore,EI=(wox^2/120L)(x3+10L2x10L35Lx2)EI=(wox^2/120L)(10L310L2x+5Lx2x3) answer

Moment Diagram by Parts Solman

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