EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 9780100257061
Author: BEER
Publisher: YUZU
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Chapter 11.5, Problem 57P

A block of weight W is dropped from a height h onto the horizontal beam AB and hits point D. (a) Denoting by ym the exact value of the maximum deflection at D and by y’m the value obtained by neglecting the effect of this deflection on the change in potential energy of the block, show that the absolute value of the relative error is (y’mym)/ym, never exceeding ym /2h. (b) Check the result obtained in part a by solving part a of Prob. 11.52 without taking ym into account when determining the change in potential energy of the load, and comparing the answer obtained in this way with the exact answer to that problem.

11.52 The 2-kg block D is dropped from the position shown onto the end of a 16-mm-diameter rod. Knowing that E = 200 GPa, determine (a) the maximum deflection of end A, (b) the maximum bending moment in the rod, (c) the maximum normal stress in the rod.

Chapter 11.5, Problem 57P, A block of weight W is dropped from a height h onto the horizontal beam AB and hits point D. (a)

Fig. P11.52

(a)

Expert Solution
Check Mark
To determine

The absolute value of the relative error ymymymym2h

Answer to Problem 57P

The absolute value of the relative error ymymymym2h_ is proved.

Explanation of Solution

Calculation:

Sketch the loading diagram as shown in Figure 1.

EBK MECHANICS OF MATERIALS, Chapter 11.5, Problem 57P , additional homework tip  1

Refer to Figure 1.

Apply the spring constant k for the load applied at point D.

The load Pm=kym

Calculate the maximum strain energy (Um) as shown below.

Um=12Pmym

Substitute kym for Pm.

Um=12kym×ym=12kym2

Calculate the work of the block exactly as shown below.

Work=W(h+ym)

Calculate the work of the block approximately as shown below.

Work=Wh

Equating work and strain energy as shown below.

Work=StrainEnergy(Um) (1)

Equating work and strain energy exactly as shown below

Substitute 12kym2 for Um and W(h+ym) for Work in Equation (1).

12kym2=W(h+ym)12kym2=Wh+Wym (2)

Equating work and strain energy exactly as shown below

Substitute 12kym2 for Um and Wh for Work in Equation (1).

12kym2=Wh (3)

Here, ym is the approximate value of ym.

ym2h=2Wkym2=2Whkym=2Whk (4)

Subtracting Equation (3) from Equation (2) as shown below.

12kym212kym2=Wh+WymWh12k(ym2ym2)=Wym12k(ymym)(ym+ym)=Wymymymym=2Wk(ym+ym)

Here, ymymym is the relative error

Substitute ym2h for 2Wk.

ymymym=1(ym+ym)ym2h=ym2h(ym+ym)<ym2h

Hence, the absolute value of the relative error ymymymym2h_ is proved.

(b)

Expert Solution
Check Mark
To determine

The relative error of the block.

Answer to Problem 57P

The relative error of the block is 0.152_.

Explanation of Solution

Given information:

The mass of the block is m=2kg.

The modulus of elasticity is E=200GPa .

The diameter of the rod is d=16mm.

The length of the beam is L=0.6m.

The dropping height is h=40mm.

Calculation:

Refer to part (a).

The relative error is ymymym (5)

Consider the acceleration due to gravity as g=9.81m/s2.

Calculate the weight of the block as shown below.

W=mg

Substitute 2kg for m and 9.81m/s2 for g.

W=2kg×9.81m/s2=19.62kgm/s2×1N1kgm/s2=19.62N

Calculate the moment of inertia (I) of the rod as shown below.

I=πd464

Substitute 16mm for d.

I=π×16464=3,217mm4×(1m1,000mm)4=3.217×109m4

Calculate the moment of inertia (I) of the rod as shown below.

I=πd464

Substitute 16mm for d.

I=π×16464=3,217mm4×(1m1,000mm)4=3.217×109m4

Calculate the centroid (c) of the rod as shown below.

c=d2

Substitute 16mm for d.

c=162=8mm

Sketch the deformation diagram as shown in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 11.5, Problem 57P , additional homework tip  2

Refer to Figure 2.

Refer to Appendix D “Beam Deflections and Slope” in the text book,

Calculate the maximum deflection (ym) as shown below.

ym=PmL33EI

Substitute 200GPa for E, 0.6m for L, and 3.217×109m4 for I.

ym=Pm×(0.6m)33×200GPa×109N/m21GPa×3.217×109m4ym=1.119×104PmPm=8,936.6ym

Calculate the maximum strain energy (Um) as shown below.

Um=12Pmym

Substitute 8,936.6ym for Pm.

Um=12×8,936.6ym×ym=4,468.3ym2

Calculate the work of the block as shown below.

Work=W(h+ym)

Substitute 19.62N for W and 40mm for h.

Work=19.62N(40mm×1m1,000mm+ym)=19.62(0.04+ym)=0.7848+19.62ym

Calculate the maximum deflection (ym) as shown below.

Work=StrainEnergy(Um)

Substitute 4,468.3ym2 for (Um) and 0.7848+19.62ym for Work.

4,468.3ym2=0.7848+19.62ym4,468.3ym219.62ym0.7848=0ym=0.015628m×1,000mm1mym=15.63mm

Calculate the spring constant (k) as shown below.

k=3EIL3

Substitute 200GPa for E, 0.6m for L, and 3.217×109m4 for I.

k=3×200GPa×109N/m21GPa×3.217×109m4(0.6m)3=8.936×103N/m

Calculate the approximate value of ym (ym) as shown below.

Substitute 19.62N for W, 8.936×103N/m for k, and 40mm for h in Equation (4).

ym=2×19.62N×40mm×1m1,000mm8.936×103N/m=1.7565×104=0.01325m×1,000mm1m=13.25mm

Calculate the relative error as shown below.

Substitute 15.63mm for ym and 13.25mm for ym in Equation (5).

Relativeerror=15.6313.2515.63=0.152

To check:

ym2h=13.252×40=0.166>Relativeerror=0.152

Therefore, the relative error is 0.152_.

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Chapter 11 Solutions

EBK MECHANICS OF MATERIALS

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