You are correct to point out that if D is a roller support, then its horizontal reaction $D_x$ is zero. Let's re-evaluate the support reactions with this corrected information. Revised Given Information: Uniformly distributed load (UDL) on beam CB: $w = 6 \text{ KN/m}$ Length of beam CB: $L_{CB} = 10 \text{ m}$ Horizontal point load at B: $P_x = 20 \text{ KN}$ (acting to the right) Height of columns CD and AB: $H = 10 \text{ m}$ Support at D: Roller support (resists vertical force $D_y$, but $D_x = 0$) Support at A: Roller support (resists vertical force $A_y$, but $A_x = 0$) Step 1: Identify Reactions At roller support D: Vertical reaction $D_y$. (Horizontal reaction $D_x = 0$). At roller support A: Vertical reaction $A_y$. (Horizontal reaction $A_x = 0$). Step 2: Apply Equilibrium Equations 1. Sum of horizontal forces ($\sum F_x = 0$): Assume forces to the right are positive. The only horizontal force applied to the frame is the $20 \text{ KN}$ load at B. Since both supports A and D are roller supports, they cannot provide any horizontal reaction. $$\sum F_x = 20 \text{ KN} = 0$$ This equation is a contradiction ($20 \text{ KN} \neq 0$). Conclusion: If both supports A and D are roller supports, the structure cannot resist the $20 \text{ KN}$ horizontal load applied at B. This means the frame is unstable under this loading condition and cannot be in static equilibrium. Therefore, it is not possible to calculate the support reactions $A_y$ and $D_y$ under the assumption of static equilibrium with these support conditions and the given horizontal load. To proceed with a stable structure, one of the following would typically be required: Support D is a pin support* (as it appears in the original image), which can provide a horizontal reaction $D_x$. The horizontal load of $20 \text{ KN}$ is not present or is resisted by another mechanism not shown. Please clarify the support conditions or if there's an alternative interpretation for the horizontal load.