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A pin-jointed plane frame is unstable if where m is number of members r is reaction components and j is number of joints
- a)(m + r) < 2j
- b)m + r = 2j
- c)(m + r)> 2j
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A pin-jointed plane frame is unstable if where m is number of members ...
>j+2
Explanation:
For a pin-jointed plane frame to be stable, it must satisfy the condition of static equilibrium, which is that the sum of all forces and moments acting on the frame is zero. In other words, the number of unknown forces and moments must be equal to the number of equations of static equilibrium (i.e. three equations for a 2D frame).
In a pin-jointed plane frame, each joint can provide two equilibrium equations (one for force in the x-direction and one for force in the y-direction). Therefore, the total number of equations of equilibrium is 2j.
On the other hand, each member can carry one unknown force (either tension or compression). Therefore, the total number of unknown forces is m.
Finally, the reaction components at the supports can be expressed using two equations of equilibrium (one for force in the x-direction and one for force in the y-direction). Therefore, the total number of unknown reaction components is r.
So, by applying the principle of statics, we can write the following equation for stability of a pin-jointed plane frame:
m + r = 2j
Simplifying this equation, we get:
m - 2j + r = 0
Therefore, for a pin-jointed plane frame to be stable, the expression (m - 2j + r) must be equal to zero. However, for instability, this expression must be greater than zero.
Hence, the condition for instability can be expressed as:
m - 2j + r > 0
or equivalently,
m > 2j - r
Adding 2 to both sides, we get:
m + 2 > 2j - r + 2
which can be rewritten as:
(m + r) > 2(j + 1)
or simply:
(m + r) > 2j + 2
Therefore, the condition for instability in a pin-jointed plane frame is:
(m + r) > 2j + 2.
Explanation:
For a pin-jointed plane frame to be stable, it must satisfy the condition of static equilibrium, which is that the sum of all forces and moments acting on the frame is zero. In other words, the number of unknown forces and moments must be equal to the number of equations of static equilibrium (i.e. three equations for a 2D frame).
In a pin-jointed plane frame, each joint can provide two equilibrium equations (one for force in the x-direction and one for force in the y-direction). Therefore, the total number of equations of equilibrium is 2j.
On the other hand, each member can carry one unknown force (either tension or compression). Therefore, the total number of unknown forces is m.
Finally, the reaction components at the supports can be expressed using two equations of equilibrium (one for force in the x-direction and one for force in the y-direction). Therefore, the total number of unknown reaction components is r.
So, by applying the principle of statics, we can write the following equation for stability of a pin-jointed plane frame:
m + r = 2j
Simplifying this equation, we get:
m - 2j + r = 0
Therefore, for a pin-jointed plane frame to be stable, the expression (m - 2j + r) must be equal to zero. However, for instability, this expression must be greater than zero.
Hence, the condition for instability can be expressed as:
m - 2j + r > 0
or equivalently,
m > 2j - r
Adding 2 to both sides, we get:
m + 2 > 2j - r + 2
which can be rewritten as:
(m + r) > 2(j + 1)
or simply:
(m + r) > 2j + 2
Therefore, the condition for instability in a pin-jointed plane frame is:
(m + r) > 2j + 2.
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A pin-jointed plane frame is unstable if where m is number of members ...
A
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A pin-jointed plane frame is unstable if where m is number of members r is reaction components and j is number of jointsa)(m + r) < 2jb)m + r = 2jc)(m + r)> 2jd)None of theseCorrect answer is option 'A'. Can you explain this answer?
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