Looking specifically at joint C, what is the force in members BC and CD?

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BlackTom AI · Accepted Answer

We begin by restating the problem setup and listing the options clearly.
Question: Looking specifically at joint C, what is the force in members BC and CD?
Answer options:
- A: BC = 0 and CD = P1 (compression)
- B: BC = (3/5) P1 (tension) and CD = (4/5) P1 (compression)
- C: BC = 0 and CD = P1 (tension)
- D: BC = P2 (compression) and CD = (4/5) P1 (tension)

Now, we analyze each option step by step with the typical truss reasoning in mind. 
Option A: BC = 0 and CD = P1 (compression).
- If BC carries zero force, joint C relies on BC to resist forces via the other connected member(s) and any external load at C. Since there is an external downward load P1 at C, the vertical reaction must be carried by the members connected to C through their geometry. For a two-member connection at C (BC and CD) with an external load, it’s possible for one member to take all the load while the other remains zero, depending on the orientation and stiffness of the members. Here BC is horizontal; if BC carries no force, CD would have to provide the vertical component necessary to balance the external load at C (through the truss’ geometry and other joint reactions). If CD is a diagonal member connected to D and B, its orientation would determine whether it can develop the required vertical force component at C. The statement “CD = P1 (compression)” claims that CD transmits the full vertical load in compression from C toward the rest of the structure. This is plausible if the geometry aligns such that the diagonal CD can push on C to counterbalance P1. However, we must verify whether the diagonal’s line of action from C through D to the rest of the truss yields a compressive force that resolves the vertical load at C. The assumption that BC is zero force is consistent only if the diagonal CD can carry the vertical load component entirely and the joint equilibrium at C (sum of forces in horizontal and vertical directions) is satisfied without BC contributing.

Option B: BC = (3/5) P1 (tension) and CD = (4/5) P1 (compression).
- This option distributes the P1 load between BC and CD with specific ratios 3/5 and 4/5 and assigns opposite senses (BC in tension, CD in compression). To justify this, one would analyze joint C with components along BC (horizontal) and CD (diagonal). The vertical external load P1 must be balanced by the vertical components of member forces, while the horizontal components must balance to zero at joint C. The chosen fractions imply a particular geometry where the diagonal CD carries a sizable portion of the load with a specific directional effect, and BC contributes a horizontal component in tension. This configuration could be correct if the diagonal CD’s angle and the joint C reactions produce exactly those force components. A detailed joint equilibrium calculation would be needed to verify or refute the exact 3/5 and 4/5 split.

Option C: BC = 0 and CD = P1 (tension).
- Similar to Option A in assuming BC carries no force, but here CD is specified as tension rather than compression. If the diagonal CD is oriented such that the force along CD acts from C toward D in tension, this would imply the external load at C creates a pulling effect along CD. However, for a downward external load at C, the diagonal’s orientation would determine whether the force along CD points into C or away from C. If CD were in tension, the member would be pulling away from joint C along its length. This could be possible if the rest of the structure provides the balancing reactions, but it must satisfy both vertical and horizontal equilibrium at C. The key pitfall is assuming the sign (tension vs compression) without checking the joint’s force components; many truss problems yield compression in diagonals under downward loads depending on the geometry.

Option D: BC = P2 (compression) and CD = (4/5) P1 (tension).
- This option introduces P2, a second external load, into the balance at joint C. In the given schematic, P2 appears at joint D, not at C. If we are strictly analyzing joint C, incorporating P2 would imply an influence of the D joint load through the truss geometry, which could be plausible but requires careful propagation of forces from the external loads to joint C via the member connections. The stated magnitudes also assert that BC is in compression and CD in tension with a fixed ratio (4/5) P1 for CD, which may conflict with the actual truss geometry unless the internal force distribution is such that CD’s axial force equals 4/5 of P1 in tension. Without performing a full signal-free joint analysis, this option seems less likely because it references P2 directly in a way that doesn’t align with a straightforward C-joint equilibrium unless a more complex transfer path is present.

In summary, each option presents a distinct interpretation of how joint C balances the external load(s) through members BC and CD. To determine which option is correct, one would perform a joint equilibrium analysis at C, taking into account the geometry (angles and lengths) of BC and CD, the direction of the external load P1 applied at C, and the reaction forces at the connected joints through the rest of the truss. The horizontal and vertical force balance at C must both be satisfied, and the axial force signs (tension vs compression) must align with the member directions and joint equilibrium. Numerical solving would reveal whether BC = 0 and CD carries P1 (compression), or whether a split like (3/5 P1 in BC and 4/5 P1 in CD with CD in compression) is required, or whether another distribution occurs (as in options B or D) once all joint equations are enforced.

MECHENG 2010 SP2025 (10190) Bonus Review Quiz Midterm 3

Looking specifically at joint C, what is the force in members BC and CD?

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