Moment Diagram by Parts Solman

Moment Diagram by Parts Problem 624

For the beam loaded as shown in Fi g. P-624, compute the moment of area of the M diagrams between the reactions about both the left and the right reaction.

Solution 624

ΣMR2=0

6R 1=400+1000(2) R 1=400N ΣMR1=0

6R 2+400=1000(2) R 2=600N

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

(AreaAB)X¯ A=12(2)(800)(43)+12(4)(2400)(103)−12(2)(2000)(83) (AreaAB)X¯ A=11733.33N⋅m3 (AreaAB)X¯ B=12(2)(800)(143)+12(4)(2400)(83)−12(2)(2000)(103) (AreaAB)X¯ B=9866.67N⋅m3

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 t he right reaction.

Solution 626

By symmetry R 1=R 2=12(400)(3) R 1=R 2=600lb and

(AreaAB)X¯ A=(AreaAB)X¯ B

(AreaAB)X¯ A=12(5)(1500)(52)−13(3)(450)(52) (AreaAB)X¯ A=8250lb⋅ft3 Thus, (AreaAB)X¯ B=8250lb⋅ft3

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Problem 627

For the beam loaded as shown in Fi g. P-627compute the moment of area of the M diagrams between the reactions about both the left and t he right reaction. (Hint: Resolve the trapezoidal loading into a uniformly distributed load and a uniformly varying load.)

Solution 627

ΣMR2=0

4R 1=200(4)(2)+12(3)(400)(1) R 1=550N ΣMR1=0

4R 2=200(4)(2)+12(3)(400)(3) R 2=850N

(AreaAB)X¯ A=12(4)(2200)(83)−13(4)(1600)(3)−14(3)(600)(175) (AreaAB)X¯ A=3803.33N⋅m3

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(AreaAB)X¯ B=12(4)(2200)(43)−13(4)(1600)(1)−14(3)(600)(35) answer (AreaAB)X¯ B=3463.33N⋅m3

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

ΣMR2=0

10R 1+400=12(6)(200)(2) R 1=80lb ΣMR1=0

10R 2=400+12(6)(200)(8) R 2=520lb

(AreaAB)X¯ A=400(8)(6)+12(10)(800)(203)−14(6)(1200)(445) (AreaAB)X¯ A=30026.67lb⋅ft3 answer  (AreaAB)X¯ B=400(8)(4)+12(10)(800)(103)−14(6)(1200)(65) answer  (AreaAB)X¯ B=23973.33lb⋅ft3

Problem 629

Solve Prob. 628 if the sense of the couple is counterclockwise instead of clockwise as shown in Fig. P-628.

Solution 629

ΣMR2=0

10R 1=400+12(6)(200)(2) R 1=160lb ΣMR1=0

10R 2+400=12(6)(200)(8) R 2=440lb

(AreaAB)X¯ A=12(10)(1600)(203)−400(8)(6)−14(6)(1200)(445) answer  (AreaAB)X¯ A=18293.33lb⋅ft3 (AreaAB)X¯ B=12(10)(1600)(103)−400(8)(4)−14(6)(1200)(65) answer (AreaAB)X¯ B=11706.67lb⋅ft3

Problem 630

For the beam loaded as shown in Fi g. P-630, compute the value of (Area AB)barred(X)A . From the result determine whether the tangent drawn to the elastic curve at B slopes up or down to the right. (Hint: Refer to the deviation equations and rules of sign.)

Solution 630

ΣMR2=0

4R 1+200(2)=12(3)(400)(1) R 1=50N ΣMR1=0

4R 2=200(6)+12(3)(400)(3) R 2=750N (AreaAB)X¯ A=12(4)(200)(83)−14(3)(600)(175) (AreaAB)X¯ A=−463.33N⋅m3 answer  The value of (AreaAB) barred(X)A is negative; therefore point A is below the tangent through B, thus the tangent through B slopes downward to the right. See the approximate elastic curve shown to the right and refer to the rules of sign for more information.

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

ΣMA=0

4R 2+M=100(4)(2) R 2=200−14M ΣMB=0

4R 1=100(4)(2)+M R 1=200+14M (AreaAB)X¯ A=0 12(4)(800−M)(43)−13(4)(800)(1)=0 83(800−M)=32003 answer  M=400lb⋅ft The uniform load over span AB will cause segment AB to deflect downward. The moment load equal to 400 lb·ft applied at the free end will cause the slope through B to  be horizontal making the deviation of A from the tangent through B equal to zero. The downward deflection therefore due to uniform load will be countered by the moment load.

Problem 632

For the beam loaded as shown in Fi g. P-632, compute the value of (Area AB) 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 the deviation equations and rules of sign.)

Solution 632

ΣMB=0

3R 1+200(1)=800(2)(2) R 1=1000N ΣMA=0

3R 2=200(4)+800(2)(1) R 1=800N

(AreaAB)X¯ A=12(2)(2000)(43)+12(1)(800)(73)−13(2)(1600)(32)−12(1)(400)(73)−12(1)(200)(83) answer  (AreaAB)X¯ A=1266.67N⋅m3 The value of (AreaAB) barred(X)A is 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 shown above and refer to the rules of sign for more information.

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)X¯ A −10=110000(50h312)[−12(2)(4)(103)−12(4)(16)(83)](10004) −10=3125000h3[−2963](10004) h3=−296(10004)125000(−10) answer  h=618.67mm

Problem 637

For the beam loaded as shown in Fi g. P-637, determine the deflection 6 ft from the wall. Use E = 1.5 × 106 psi and I = 40 in 4.

Solution 637

R C=80(8)=640 lb MC=80(8)(4)=2560lb⋅ft

tB/C=1EI(AreaBC)X¯ B tB/C=1EI[12(6)(3840)(2)−6(2560)(3)−13(6)(1440)(1.5)](123) tB/C=1EI[27360](123) tB/C=1(1.5×106)(40)[27360](123) tB/C=−0.787968 in Thus, δB = | tB/C | = 0.787968 in

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Problem 638 For the cantilever beam shown in Fig. P-638, determine the value of EI δ at the left end. Is

this deflection upward or downward?

Solution 638

EItA/B=(AreaAB)X¯ A EItA/B=2(2)(3)−12(4)(1)(83) EItA/B=203=6.67 kN⋅m3 ∴

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EIδ = 6.67 kN·m  upward

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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 × 106 psi and I = 60 in 4.

Solution 639

tA/C=1EI(AreaAB)X¯ A tA/C=1(1.5×106)(60)[12(6)(5400)(6)−13(8)(6400)(6)](123) tA/C=−0.09984 in ∴

The free end will move by 0.09984 inch downward.

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Problem 640

Compute the value of δ at the concentrated load in Prob. 639. Is the deflection upward downward?

Solution 640

R C=200(8)−900=700 lb MC=200(8)(4)−900(6)=1000 lb⋅ft

tB/C=1EI(AreaBC)X¯ B tB/C=1(1.5×106)(60)[12(6)(4200)(2)−1000(6)(3)−13(6)(3600)(15)](123) tB/C=−0.06912 in ∴

δ = 0.06912 inch downward

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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)X¯ A=0 1EI[12(4)(4P)(83)−2(400)(3)]=0 answer  P=112.5 N

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

R A=4(1)=4kN MA=4(1)(2.5)=10kN⋅m

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

δmax = 28 mm

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Problem 643

Find the maximum value of EI δ for the cantilever beam shown in Fig. P-643.

Solution 643

EItB/A=(AreaAB)X¯ B EItB/A=12L(PL)( 13L)−PaL(12L)−12(L−a)P(L−a)[13(L−a)] EItB/A=16PL3−12PL2a−16P(L−a)3 EItB/A=16PL3−12PL2a−16P(L3−3L2a+3La2−a3) EItB/A=16PL3−12PL2a−16PL3+12PL2a−12PLa2+16Pa3 EItB/A=−12PLa2+16Pa3 EItB/A=−16Pa2(3L−a) Therefore EIδmax=(1/6)Pa^2(3L−a)

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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)X¯ A 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^4−3/16woL4−1/384woL^4] tA/B=1/EI[−41/384woL^4] tA/B=(−41woL^4)/(384EI) Therefore δmax=(41woL^4)/(384EI)

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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 × 10 6 mm4.

Solution 645

R=1/2(4)(1200) R=2400N M=1/2(4)(1200)(8/3) M=6400N⋅m

y3=12004 y=900N/m tB/A=1EI(AreaAB)X¯ B tB/A=1EI[12(3)(7200)(1)−3(6400)(1.5)−14(3)(1350)(0.6)](10003) tB/A=[1/10000(30×106) ] [−18607.5](1000^3) tB/A=−62.025mm Therefore: δB=62.025mm

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θAB=1EI(AreaAB) θAB=1EI[12(3)(7200)−3(6400)−14(3)(1350)](10002) θAB=110000(30×106)[−9412.5](10002) θAB=−0.031375radian θAB=1.798degree answer

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

M=12(5)(60)(2+53)=550lb⋅ft R=12(5)(60)=150lb

tA/B=1/EI(AreaAB)X¯A −5=1EI[12(300)(2)(263)−550(2)(9)−14(5)(250)(7)](123) −5=1(1.5×106)I(−16394400)

I=2.18592in4

Problem 647 Find the maximum value of EI δ for the beam shown in Fig. P-647.

Solution 647

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

EItA/B=(AreaAB)X¯ A EItA/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^4−5/48woL^4−1/1920woL^4 EItA/B=(−12/11920)woL^4 Therefore EIδmax=(121/1920)woL^4

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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/L y=(wo/L)x M=1/2L(wo)(1/3L)=1/6woL^2 R=12woL Moments about B:

Triangular force to the left of B: M1=−1/2(L−x)(wo−y)(1/3)(L−x) M1=−1/6(L−x)^2(wo−(wox/L)) M1=−wo(L−x)^3 /6L Triangular upward force: M2=1/2(xy)(1/3x)=1/6x^2(wox/L) M2=wox^3 /6L Rectangle (wo by x): M3=−wox(1/2x)=−1/2wox^2 Reactions R and M: M4=Rx=1/2woLx M5=−M=−1/6woL^2 Deviation at B with the tangent line through C

EItB/C=(AreaBC)X¯ B EItB/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^4 EItB/C=(wox^2/120L)(x3+10L2x−10L3−5Lx2) Therefore, EIδ=(−wox^2/120L)(x3+10L2x−10L3−5Lx2) EIδ=(wox^2/120L)(10L3−10L2x+5Lx2−x3)

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