Force acting on steel rod (deformation)

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foo9008
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in this problem , i could only understand that P3 = 120KN , i can't understand why P2= -60kN and why P1 = 240kN ? can someone explain ?

Homework Equations

The Attempt at a Solution

 
Attachments
  • Diagram of a stepped steel rod fixed at A with points A, B, C, D; segment areas 581 mm² and 194 mm²; loads 300 kN and 150 kN shown, with 150 kN pointing right at D; lengths 300 mm, 300 mm and 400 mm.
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  • Free-body diagrams of a stepped bar fixed at A with axial loads 300 kN, 150 kN and 120 kN at points B, C and D, with internal forces P1 = 240x10^3 N, P2 = -60x10^3 N and P3 = 120x10^3 N, and the deflection sum formula evaluated to 1.73x10^-3 m.
    2280007.jpg
    27.6 KB · Views: 554
In the same way that you can do a shear force diagram for lateral loads on a beam you can do a tension force diagram for axial loads on a shaft .

Basically plot a graph of axial tension against position along shaft . Have a try .
 
Nidum said:
In the same way that you can do a shear force diagram for lateral loads on a beam you can do a tension force diagram for axial loads on a shaft .

Basically plot a graph of axial tension against position along shaft . Have a try .
What do you mean? can you explain further?
 
Chestermiller said:
If you do a force balance on the entire rod, what do you get for the reaction force on the left end?
how to do it ? i know only the P3 ... which is 120kN , how to do for P1 and P2 ?
 
Attachments
  • Free-body diagram of a rod segment shown as a rectangle with arrows pointing outward at both ends, each labelled P3. The labels are placed below the left and right arrows.
    229.png
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foo9008 said:
how to do it ? i know only the P3 ... which is 120kN , how to do for P1 and P2 ?
You have 3 forces acting on the rod: 300 kN, 180 kN, and 120 kN. In order of the rod to be in static equilibrium, there must also be a force applied by the wall on the bar at location A. What does that force have to be?
 
Chestermiller said:
You have 3 forces acting on the rod: 300 kN, 180 kN, and 120 kN. In order of the rod to be in static equilibrium, there must also be a force applied by the wall on the bar at location A. What does that force have to be?
what i found is this ? how to find the force applied by the wall on the bar at location A. ??
btw , i am not sure which 1 is P3 and which one is P2
 
Attachments
  • Hand-drawn sketch of a stepped bar fixed at left end A, with points B, C and D marked; arrows show loads of 200kN and 180kN at BC, 60kN and 120kN at CD, and 120kN pulling right at D.
    230.jpg
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foo9008 said:
what i found is this ? how to find the force applied by the wall on the bar at location A. ??
btw , i am not sure which 1 is P3 and which one is P2
Forget about P1, P2, and P3 for now. Are you saying that you don't know how to do a 1 dimensional equilibrium force balance on a solid object?
 
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Chestermiller said:
Forget about P1, P2, and P3 for now. Are you saying that you don't know how to do a 1 dimensional equilibrium force balance on a solid object?
yes