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 1, Problem 1.3.30P

Space frame A BCD is clamped at A, except it is Free to translate in the .v direction. There is also a roller support at D, which is normal to line CDE. A triangularly distributed Force with peak intensity q0 = 75 N/m acts along AB in the positive - direction. Forces Px= 60 N and Pz = = 45 N are applied at joint C, and a concentrated moment My = 120 N . m acts at the mid-span of member BC.

(a) Find reactions at supports A and I).

(b) Find internal stress resultants N. E’I T, and .11 at the mid-height of segment AB.

  Chapter 1, Problem 1.3.30P, Space frame A BCD is clamped at A, except it is Free to translate in the .v direction. There is also

(a)

Expert Solution
Check Mark
To determine

Reactions at support A,D.

Answer to Problem 1.3.30P

The correct answers are:

Ay=120 N, Az=60 N, MAx=70 Nm, MAy=142.5 Nm, MAz=180 Nm,Dx=60 N, Dy=120 N, Dz=30 N.

Explanation of Solution

Given Information:

You have following figure with all relevant information,

Mechanics of Materials (MindTap Course List), Chapter 1, Problem 1.3.30P , additional homework tip  1

and q0=75 N/m, Px=60 N,  Pz=45 N, and My=120 Nm.

Draw free body diagram of joints and use equilibrium of forces to determine the unknowns.

Calculation:

Draw free body diagram as shown in the following figure,

Mechanics of Materials (MindTap Course List), Chapter 1, Problem 1.3.30P , additional homework tip  2

The forces and corresponding position vectors are,

    Force Position vector
      A=(0,Ay,Az)   rA=(0,0,0)
      Q=(0,0,q0)=(0,0,75)   rQ=(0,43,0)
      P=(Px,0,Pz)=(60,0,45)   rC=(1.5,2,0)
      D=(Dx,Dy,Dz)=D(0.5)2+22+(0.5)2(1,2,0.5)=D(0.5,1,0.25)   rD=(1.5,2,0)0.75(0.5)2+22+(0.5)2(1,2,0.5)=(1.5,2,0)0.75(0.5,1,0.25)=(1.87,1.25,0.18)

Take equilibrium of forces vector form,

  F=0A+Q+P+D=0(0,Ay,Az)+(0,0,75)+(60,0,45)+D(0.5,1,0.25)=0

The vector equation yields three equations in components form as below,

  600.5D=0        ....(1)Ay+D=0        ....(2)Az+30+0.25D=0        ....(3)

Solve the three equations to get D=120 N, Ay=120 N, Az=60 N.

Now, calculate components of D as,

  D=(Dx,Dy,Dz)==D(0.5,1,0.25)=120(0.5,1,0.25)=(60 N,120 N,30 N)

Now take equilibrium of moments about A in vector form as,

  MA=0M+My+rQ×Q+rC×P+rD×D=0(Mox,Moy,Moz)+(0,My,0)+(0,43,0)×(0,0,75)+(1.5,2,0)×(60,0,45)+(1.87,1.25,0.18)×(60,120,30)=0

Evaluate the cross products to get,

  (Mox,Moy,Moz)+(0,My,0)+(100,0,0)+(90,67.5,120)+(60,45,300)=0[(Mox+70),(Moy+142.5),(Moz+180)]=0(Mox,Moy,Moz)=(70 Nm.,142.5 Nm,180 Nm)

Conclusion:

Therefore the forces and moments are:

Ay=120 N, Az=60 N, MAx=70 Nm, MAy=142.5 Nm, MAz=180 Nm,Dx=60 N, Dy=120 N, Dz=30 N.

(b)

Expert Solution
Check Mark
To determine

Internal stress resultants N,V,T and M at mid height of AB.

Answer to Problem 1.3.30P

The correct answers are:

N=120 N, V=41.3 N, T=142.5 Nm, M=180.7 Nm.

Explanation of Solution

Given Information:

You have following figure with all relevant information,

Mechanics of Materials (MindTap Course List), Chapter 1, Problem 1.3.30P , additional homework tip  3

and q0=75 N/m, Px=60 N,  Pz=45 N, and My=120 Nm.

Draw free body diagram of joints and use equilibrium of forces to determine the unknowns.

Calculation:

Draw free body diagram as shown in the following figure,

Mechanics of Materials (MindTap Course List), Chapter 1, Problem 1.3.30P , additional homework tip  4

The forces and corresponding position vectors are,

    Force Position vector
      A=(0,Ay,Az)   rA=(0,0,0)
      Q=(0,0,q0)=(0,0,75)   rQ=(0,43,0)
      P=(Px,0,Pz)=(60,0,45)   rC=(1.5,2,0)
      D=(Dx,Dy,Dz)=D(0.5)2+22+(0.5)2(1,2,0.5)=D(0.5,1,0.25)   rD=(1.5,2,0)0.75(0.5)2+22+(0.5)2(1,2,0.5)=(1.5,2,0)0.75(0.5,1,0.25)=(1.87,1.25,0.18)

Take equilibrium of forces vector form,

  F=0A+Q+P+D=0(0,Ay,Az)+(0,0,75)+(60,0,45)+D(0.5,1,0.25)=0

The vector equation yields three equations in components form as below,

  600.5D=0        ....(1)Ay+D=0        ....(2)Az+30+0.25D=0        ....(3)

Solve the three equations to get D=120 N, Ay=120 N, Az=60 N.

Now, calculate components of D as,

  D=(Dx,Dy,Dz)==D(0.5,1,0.25)=120(0.5,1,0.25)=(60 N,120 N,30 N)

Now take equilibrium of moments about A in vector form as,

  MA=0M+My+rQ×Q+rC×P+rD×D=0(Mox,Moy,Moz)+(0,My,0)+(0,43,0)×(0,0,75)+(1.5,2,0)×(60,0,45)+(1.87,1.25,0.18)×(60,120,30)=0

Evaluate the cross products to get,

  (Mox,Moy,Moz)+(0,My,0)+(100,0,0)+(90,67.5,120)+(60,45,300)=0[(Mox+70),(Moy+142.5),(Moz+180)]=0(Mox,Moy,Moz)=(70 Nm.,142.5 Nm,180 Nm)

Calculation internal stress resultants at mid point of AB :

Consider the following free body diagram,

Mechanics of Materials (MindTap Course List), Chapter 1, Problem 1.3.30P , additional homework tip  5

Analyze the lower part of the above free body diagram,

Take equilibrium of forces in vector form,

  F=0A+N+V+(0,0,q04)=0(0,Ay,Az)+(0,N,0)+(Vx,0,Vz)+(0,0,754)=0(0,120,60)+(0,N,0)+(Vx,0,Vz)+(0,0,18.75)=0

Solve the above equation in component form to get,

N=120 N,  Vx=0, Vz=41.3 N.

Take equilibrium of moments about A in vector form as,

  MA=0(Mox,Moy,Moz)+(Mx,0,Mz)+(0,T,0)+(0,23,0)×(0,0,18.75)+(0,1,0)×(Vx,0,Vz)+(0,1,0)×(0,N,0)=0(Mox,Moy,Moz)+(Mx,0,Mz)+(0,T,0)+(0,23,0)×(0,0,18.75)+(0,1,0)×(0,0,41.25)+(0,1,0)×(0,120,0)=0

Evaluate the cross products and solve to get,

  Mx=16.25 Nm, Mz=180 Nm, T=142.5 Nm.

Equivalent moment,

M=Mx2+Mz2=16.252+1802=180.7 Nm

Conclusion:

Thus the internal stress resultants are:

N=120 N, V=41.3 N, T=142.5 Nm, M=180.7 Nm.

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Chapter 1 Solutions

Mechanics of Materials (MindTap Course List)

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