Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card
Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card
9th Edition
ISBN: 9781337594301
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 8, Problem 8.5.34P

A compound beam ABCD has a cable with force P anchored at C The cable passes over a pulley at D, and force P acts in the —x direction, There is a moment release just left of B. Neglect the self-weight of the beam and cable. Cable force P = 450 N and dimension variable L = 0.25 m. The beam has a rectangular cross section (b = 20 mm, it = 50 mm).

(a) Calculate the maximum normal stresses and maximum in-plane shear stress on the bottom surface of the beam at support A.

(b) Repeat part (a) for a plane stress element located at mid-height of the beam at A.

(c) If the maximum tensile stress and maximum in-plane shear stress at point A are limited to 90 MPa and 42 MPa, respectively, what is the largest permissible value of the cable force P?

  Chapter 8, Problem 8.5.34P, A compound beam ABCD has a cable with force P anchored at C The cable passes over a pulley at D, and

(a)

Expert Solution
Check Mark
To determine

The maximum normal stresses and maximum in-plane shear stress on the bottom of the beam at fixed support A.

Answer to Problem 8.5.34P

Maximum normal stresses, σ1=0 , σ2=σx=108.4ΜΡa

Maximum in-plane shear stress τmax=54.2ΜΡa

Explanation of Solution

Given information: A compound beam ABCD has a cable having force P anchored at C as shown in the figure below:

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card, Chapter 8, Problem 8.5.34P , additional homework tip  1

Cable force P=450N

Dimension variable L=0.25m

The cross section of the beam b=20mm and h=50mm

To calculate the maximum normal stress and in-plane shear stress first taken the cross-sectional properties of the beam.

Area of the rectangular beam A=bh

  A=20×50A=1000mm2

Moment of inertia of the rectangular beam is I=bh312

  I=20×(50)31220×50I=208333.333mm4

Now reactions at fixed support A

Horizontal force HA=P

  HA=450Ν

Shear force VA=4L3LP

  VA=43×450VA=600Ν

And moment at point A, MA=8PL

  MA=8×450×0.25MA=900Νm

Now, maximum normal stresses on the bottom of the beam at A is,

  σx=HAAMA(h2)I

  σx=4501000900×25×1000208333.33σx=0.45108σx=108.45ΜΡa

  σy=0

And at principle plane where there is maximum normal stress the shear stress is zero. So,

  τxy=0

Principle stresses:

  σ1=σx+σy2+( σx σy 2 )2+τxy2σ1=108.452+( ( 108.45 )02 )2+( 0)2σ1=54.225+( 54.225)2+0σ1=54.225+54.225σ1=0

  σ2=σx+σy2( σx σy 2 )2+τxy2σ2=108.452( ( 108.45 )02 )2+( 0)2σ2=54.225( 54.225)2+0σ2=54.22554.225σ2=108.45ΜΡa

Maximum in-plane shear stress

  τmax=( σx σy 2 )2+τxy2τmax=σx2τmax=54.225ΜΡa

(b)

Expert Solution
Check Mark
To determine

Maximum stresses located at the mid-height of the beam at A.

Answer to Problem 8.5.34P

Maximum normal stresses, σ1=0.703ΜΡa , σ2=1.153ΜΡa

Maximum in-plane shear stress τmax=0.928ΜΡa

Explanation of Solution

Given information: A compound beam ABCD has a cable having force P anchored at C as shown in the figure below:

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card, Chapter 8, Problem 8.5.34P , additional homework tip  2

Cable force P=450N

Dimension variable L=0.25m

The cross section of the beam b=20mm and h=50mm

Area of the rectangular beam A=bh

  A=20×50A=1000mm2

Moment of inertia of the rectangular beam is I=bh312

  I=20×(50)31220×50I=208333.333mm4

Now reactions at fixed support A

Horizontal force HA=P

  HA=450Ν

Shear force VA=4L3LP

  VA=43×450VA=600Ν

The maximum normal stresses at the mid-point of the beam at A is,

  σx=HAA

  σx=4501000σx=0.45ΜΡa

  σy=0

And, τxy=32VAA

  τxy=32×6001000τxy=0.9ΜΡa

Principle stresses:

  σ1=σx+σy2+( σx σy 2 )2+τxy2σ1=0.452+( ( 0.45 )02 )2+( 0.9)2σ1=0.225+( 0.225)2+( 0.9)2σ1=0.225+0.927698766σ1=0.70260.703ΜΡa

  σ2=σx+σy2( σx σy 2 )2+τxy2σ2=0.452( ( 0.45 )02 )2+( 0.9)2σ2=0.225( 0.225)2+( 0.9)2σ2=0.2250.927698766σ2=1.15261.153ΜΡa

Maximum in-plane shear stress

  τmax=( σx σy 2 )2+τxy2τmax=( ( 0.45 )02 )2+( 0.9)2τmax=( 0.225)2+( 0.9)2τmax=0.92760.928ΜΡa

(c)

Expert Solution
Check Mark
To determine

The maximum permissible value of the cable force P.

Answer to Problem 8.5.34P

Maximum permissible value of P Pmax=348.5Ν

Explanation of Solution

Given information: A compound beam ABCD has a cable having force P anchored at C as shown in the figure below:

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card, Chapter 8, Problem 8.5.34P , additional homework tip  3

Dimension variable L=0.25m

Maximum tensile stress at point A, σa=90ΜΡa

Maximum shear stress at point A, τa=42ΜΡa

The cross section of the beam b=20mm and h=50mm

Area of the rectangular beam A=bh

Moment of inertia of the rectangular beam is I=bh312

And moment at point A, MA=8PL

In the above figure the maximum cable force P is controlled by the tensile and maximum shear force at the bottom of the beam at point A.

Hence, tensile stress σx=σa (given)

Then, σx=PAMA(h2)I

  σa=Pbh+8PL( h 2)( bh3 12)σa=Pbh+48PLbh2or,P=σa( 1 bh+ 48L bh2 )

By substituting the values we get,

  P=90( 1 1000+ 48×0.25×1000 20 ( 50 )2 )P=90( 1 1000+ 48×250 20 ( 50 )2 )P=90( 1 1000+ 12000 50000)P=90(0.001+0.24)P=900.241Pmax=373.44Ν

And,

  τmax=σx2

Then, σx=σa=Pbh+48PLbh2

  τmax=σx2

  τa=P2bh+24PLbh2or,Pmax=τa( 1 2bh+ 24L bh2 )Pmax=42( 1 2×20×50+ 24( 0.25×1000 ) 20 ( 50 )2 )Pmax=42( 1 2000+ 6000 50000)Pmax=42(0.0005+0.12)Pmax=420.1205Pmax=348.5Ν

Conclusion:

Hence from the obtained values the maximum permissible value of the force P is

  Pmax=348.5Ν

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Chapter 8 Solutions

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card

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