problem 1\nthe object shown consists of pin - connected members ab, bc, and cd. the object is externally…

Explanation:

Step1: Calculate total load on AB

The uniformly - distributed load on AB is $15\ kN/m$ and length $L_{AB}=10\ m$. So the total load $P_1 = 15\times10=150\ kN$ acting at the mid - point of AB (i.e., $x = 5\ m$ from A).

Step2: Calculate total load on CD

The uniformly - distributed load on CD is $16\ kN/m$ and length $L_{CD}=6\ m$. So the total load $P_2=16\times6 = 96\ kN$ acting at the mid - point of CD (i.e., $x = 3\ m$ from D).

Step3: Take moment about A

Let the vertical reaction at D be $V_D$ and horizontal reaction at D be $H_D$. The horizontal reaction at A is $H_A = 0$ (since there is no horizontal load). The moment equilibrium equation $\sum M_A=0$: [150\times5+96\times(10 + 4+3)-V_D\times(10 + 4+3+6)=0] [750+96\times17 - V_D\times23=0] [750 + 1632-23V_D=0] [23V_D=2382] [V_D=\frac{2382}{23}\approx103.57\ kN]

Step4: Calculate vertical reaction at A

Using vertical force equilibrium $\sum F_y = 0$, let the vertical reaction at A be $V_A$. [V_A+V_D-150 - 96=0] [V_A=150 + 96 - V_D] [V_A=246-103.57 = 142.43\ kN]

Step5: Analyze joint B

Consider the equilibrium of joint B. Let the force in member BC be $F_{BC}$. Resolve the forces in the vertical direction at joint B. The vertical component of $F_{BC}$, $F_{BCy}=\frac{4}{5}F_{BC}$. At joint B, considering vertical equilibrium: [V_A-150 - F_{BCy}=0] [142.43-150-\frac{4}{5}F_{BC}=0] [\frac{4}{5}F_{BC}=142.43 - 150=- 7.57] [F_{BC}=-\frac{7.57\times5}{4}=-9.4625\ kN] The horizontal component of $F_{BC}$, $F_{BCx}=\frac{3}{5}F_{BC}=\frac{3}{5}\times(-9.4625)=-5.6775\ kN$

Step6: Analyze joint C

At joint C, considering vertical equilibrium: Let the vertical reaction at C be $V_C$. [V_C - 16\times3-\frac{4}{5}F_{BC}=0] [V_C=48+\frac{4}{5}\times9.4625] [V_C=48 + 7.57=55.57\ kN] Considering horizontal equilibrium at joint C, the horizontal reaction at C is equal to the horizontal component of $F_{BC}$ in magnitude but opposite in direction. So $H_C = 5.6775\ kN$

Answer:

$H_A = 0\ kN$, $V_A=142.43\ kN$, $H_B = 5.6775\ kN$, $V_B = 7.57\ kN$ (downward), $H_C = 5.6775\ kN$, $V_C=55.57\ kN$, $H_D = 0\ kN$, $V_D = 103.57\ kN$