Slope–deflection method – stiffness factor for a prismatic member For a prismatic beam member of length L and constant second moment of area I, what is the joint rotational stiffness factor (end moment per unit rotation at the near end) when the far end is fixed?

Introduction / Context:
In the slope–deflection and moment distribution methods, the rotational stiffness of a member at a joint is required to relate end moments to joint rotations. For prismatic members the stiffness factor has well-known values depending on the boundary condition at the far end.

Given Data / Assumptions:

• Prismatic (constant E and I) beam member, length L.
• Small deflection, linear elasticity.
• Near-end rotation is applied; far end is fixed against rotation and translation.

Concept / Approach:
The stiffness factor k is defined as the end moment required at the near end to produce unit rotation at that end with the far end fixed and with no joint translation (sidesway). Classical beam theory yields k = 4EI / L for this case. If the far end were simply supported, k would be 3EI / L.

Step-by-Step Solution:
Definition: k = M_required / θ, with θ = 1 rad at near end.Slope–deflection derivation gives: M_near = (4EI/L) * θ_near + (2EI/L) * θ_far.With far end fixed, θ_far = 0 ⇒ M_near = 4EI/L.

Verification / Alternative check:
Compare to standard tables used in moment distribution: joint stiffness = 4EI/L (far end fixed) and 3EI/L (far end hinged).

Why Other Options Are Wrong:

• 2EI/L is too small; corresponds to carry-over relation, not stiffness here.
• 3EI/L applies when the far end is hinged.
• 6EI/L and EI/L^2 are not the standard rotational stiffness factors for this boundary condition.

Common Pitfalls:
Confusing stiffness with carry-over factor; remember: carry-over moment = 1/2 of near-end moment for prismatic members.

Final Answer:
4EI / L