Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 2, Problem 2.2.20P

A framework ABC consists of two rigid bars AB and BC. Each having a length b (see the first part of the figure part a). The bars have pin connections at A, B, and C and are joined by a spring of stiffness k. The spring is attached at the midpoints of the bars. The framework has a pin support at A and a roller support al C, and the bars are at an angle a to the horizontal.

When a vertical load P is applied at joint B (see the second part of the figure part a.) the roller support C moves to the right, the spring is stretched, and the angle of the bars decreases from a to the angle ??.

(a) Determine the angle 0 and the increase S in the distance between points A and C. Also find reactions at A and C. (Use the following data: b = 200 mm. ft = 3.2 kN/m. a = 45°. and

P = 50 N.)

(b) Repeat part (a) if a translational spring kt= kll is added at C and a rotational spring kr= kb-l2 is added at A (see figure pan b).

  Chapter 2, Problem 2.2.20P, A framework ABC consists of two rigid bars AB and BC. Each having a length b (see the first part of

(a)

Expert Solution
Check Mark
To determine

The angle θ and the increase δ in the distance between points A and C, also, find the reactions at A and C.

Answer to Problem 2.2.20P

The angle θ=35.1o

The distance between points A and C, δ=44.6mm

The reaction at support C RC=25N

The reaction at support A RA=25N

Explanation of Solution

Given information:

The length of the rigid bars b = 200 mm

The stiffness of the spring k = 3.2 kN/m

The angle of bars from the horizontal = α=45o

The valueof load P = 50 N

  Mechanics of Materials (MindTap Course List), Chapter 2, Problem 2.2.20P , additional homework tip  1

Figure: Initial and displaced positions of the framework.

Calculation:

Let us consider the structure in its displaced position. Let us use the free body diagrams of left-hand bars and right-hand bars.

From the free body diagram of left-hand bar,

   MB=0HAh+kδ2(h2)RA( L 2 2)+kr(αθ)=0

Reaction at support A,

   RA=2L2(k1δh+kδ2(h2))+kr(αθ)

From the free body diagram of right-hand bar,

   MB=0kδ2(h2)k1δh+Rc( L 2 2)=0

   RC=2L2(kδ2(h2)+k1δh)

From overall FBD of the beams,

   FH=0HAk1δ=0

   HA=k1δ

   FV=0RA+RC=P

   MA=0kr(αθ)PL22+RcL2=0

   Rc=1L2(PL22kr(αθ))

We have two expressions for Rc, equating both of the expressions and then substitute expressions for L2, kr, k1, h and δ .

   1L2(PL22kr(αθ))=2L2(kδ2(h2)+k1δh)

Now,

   1L2(PL22kr(αθ))=(2L2(k 2bcosθcosα2× bsinθ2)+k1(( 2b( cosθ )cosα))(bsinθ))=0

Let us substitute the numerical values of variables and calculate the values of angle θ and the increase δ in the distance.

   b=200mmk = 3.2 kN/mα=45oP=50Nk1=0kr=0

   L2=2bcos(θ)L1=2bcos(α)δ=L2L1δ=2b(cosθcosα)h=bsinθ

   12×0.2cosθ(50× 2×0.2×cosθ20×( 45θ))(1 2×0.2×cosθ( 0× 2×0.2cosθcos45 2 × bsinθ 2 )+0×( ( 2b( cosθ )cosα ))( bsinθ))=0

Solving the above equation gives,

   θ=35.1o and

   δ=44.6mm

Now, let us compute the reactions at point C and A.

The reaction at point C

   RC=2L2(kδ2×h2+k1δh)RC=1LC(P L 2 2kr( αθ))RC=25N

And the reaction at point A is

   RA=2L2(k1δh+kδ2( h 2 )+kr( αθ))RA=25N

Conclusion: Thus, the angle θ=35.1o

The distance between points A and C, δ=44.6mm.

The reaction at support C RC=25N.

The reaction at support A RA=25N.

(b)

Expert Solution
Check Mark
To determine

The angle θ and the increase δ in the distance between points A and C, also, find the reactions at A and C.

Answer to Problem 2.2.20P

The angle θ=43.3o

The distance between points A and C, δ=8.19mm

The reaction at support C RC=18.5N

The reaction at support A RA=31.5N

The moment reaction at point A is MA=1.8989N.m

Explanation of Solution

Given information:

The length of the rigid bars b = 200 mm

The stiffness of the spring k = 3.2 kN/m

The angle of bars from the horizontal = α=45o

The value of load P = 50 N

  Mechanics of Materials (MindTap Course List), Chapter 2, Problem 2.2.20P , additional homework tip  2

Figure: Initial and displaced positions of the framework.

Calculation:

Let us consider the structure in its displaced position. Let us use the free body diagrams of left-hand bars and right-hand bars.

From the free body diagram of left-hand bar,

   MB=0HAh+kδ2(h2)RA( L 2 2)+kr(αθ)=0

Reaction at support A,

   RA=2L2(k1δh+kδ2(h2))+kr(αθ)

From the free body diagram of right-hand bar,

   MB=0kδ2(h2)k1δh+Rc( L 2 2)=0

   RC=2L2(kδ2(h2)+k1δh)

From overall FBD of the beams,

   FH=0HAk1δ=0

   HA=k1δ

   FV=0RA+RC=P

   MA=0kr(αθ)PL22+RcL2=0

   Rc=1L2(PL22kr(αθ))

We have two expressions for Rc, equating both of the expressions and then substitute expressions for L2, kr, k1, h and δ .

   1L2(PL22kr(αθ))=2L2(kδ2(h2)+k1δh)

Now,

   1L2(PL22kr(αθ))=(2L2(k 2bcosθcosα2× bsinθ2)+k1(( 2b( cosθ )cosα))(bsinθ))=0

Let us substitute the numerical values of variables and calculate the values of angle θ and the increase δ in the distance.

   b=200mmk = 3.2 kN/mα=45oP=50Nk1=k2kr=k2b2

   L2=2bcos(θ)L1=2bcos(α)δ=L2L1δ=2b(cosθcosα)h=bsinθ

   12×0.2cosθ(50× 2×0.2×cosθ2 3.2× 10 3 × ( 0.2 ) 2 2×( 45θ))( 1 2×0.2×cosθ ( 3.2× 2×0.2cosθcos45 2 × 0.2×sinθ 2 ) + 3.2× 10 3 2 ×( ( 2×0.2×( cosθ )cos45 ) )( 0.2×sinθ ))=0

Solving the above equation gives,

   θ=43.3o and

   δ=8.19mm

Now, let us compute the reactions at point C and A.

The reaction at point C,

   RC=2L2(kδ2×h2+k1δh)RC=1L2(P L 2 2kr( αθ))RC=18.5N

And the reaction at point A is

   RA=2L2(k1δh+kδ2( h 2 )+kr( αθ))RA=31.5N

Now, moment reaction at point A is,

   MA=kr(αθ)MA=3.2× 1032×(0.2)2×(4543.3)×π180

   MA=1.8989N.m

Conclusion: Thus, the angle θ=43.3o

The distance between points A and C, δ=8.19mm.

The reaction at support C RC=18.5N.

The reaction at support A RA=31.5N.

The moment reaction at point A is MA=1.8989N.m.

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Chapter 2 Solutions

Mechanics of Materials (MindTap Course List)

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