Step 2: Equilibrium conditions to get the support force

Now you need to get the support force at the pin B. What is the force in kN at pin B?

Question

Step 2: Equilibrium conditions to get the support force

Now you need to get the support force at the pin B. What is the force in kN at pin B?

BlackTom AI · Accepted Answer

The problem asks for the force at pin B in a statically determinate beam system under given loads.
First, visualize the setup: a horizontal beam from A to the pin at B with a 30 kN downward load applied 2 m from A, and a support reaction at A (Ax, Ay) and a tension member at B with a force FB acting along the member.
Since pin A provides both horizontal and vertical reactions, and there is a connecting member at B forming a triangle with the beam, taking moments about A is a natural starting point to isolate FB.
Optionally, you could also sum moments about any other point, but choosing A eliminates Ax and Ay from the moment equation, simplifying the calculation.
Now, compute the moment contributions about point A:
- The 30 kN load is located 2 m from A, creating a clockwise moment of 30 kN × 2 m = 60 kN·m about A.
- The force FB acts along the member at B. The geometry of the member gives the perpendicular distance (moment arm) to A as (6 m) times a factor derived from the member’s orientation; in the provided solution, the moment arm factor is (4/5) for the vertical projection, corresponding to sin(theta) or the appropriate trigonometric relationship of the member orientation.
Thus, FB contributes a moment about A: FB × (4/5) × 6 m = FB × (24/5) m = 4.8 FB kN·m.
For rotational equilibrium, sum of moments about A must be zero: 4.8 FB − 60 = 0, which leads to FB = 60 / 4.8 = 12.5 kN.
A few alternative checks help reinforce the result:
- If you instead used the raw geometry to express the moment arm of FB, you’d still end up with the same factor that relates FB to the moment about A due to the direction of FB along the member.
- Summing moments about A without including any hinge reactions at A is valid because A is a pin and the moment contributions from Ax and Ay are zero about A.
- If you attempted to sum moments about B, you would need to know reactions at A and the lever arms accordingly; the math would be consistent but more cumbersome, and would still yield the same FB once all forces are accounted for.
In summary, applying rotational equilibrium about A with the moment contribution of FB expressed through its geometry gives FB = 12.5 kN, confirming the value.

Step 2: Equilibrium conditions to get the support force Now you need to get the support force at the pin B. What is the force in kN at pin B?数值题

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