Consider a jib crane with an external load suspended from the pulley. The dimensions of the crane members are shown in the figure. A systematic analysis of the frame structure is required to determine the reaction forces at the pin joints, assuming that the pulleys are frictionless.
![Figure 1](/files/ftp_upload/14291/14291_PT_Fig_1v2.svg)
The system has two main structural components: a two-force member BD and a multi-force member ABC. The two-force member BD refers to a straight element subjected only to forces at its two ends, B and D, with no additional forces acting along its length. These forces are equal in magnitude but opposite in direction, resulting in the member being either in pure tension or compression. On the other hand, the multi-force member ABC is subjected to more than two forces distributed along its length. These forces may include external loads, reaction forces at pin joints, and the force exerted by the cable. Due to multiple forces acting on member ABC, it experiences a more complex stress distribution compared to the simpler two-force member BD.
![Figure 2](/files/ftp_upload/14291/14291_PT_Fig_2v2.svg)
Considering the lower pulley section, the load weight balances the tension in the cables resulting in an upward tension of 10 kN for each cable. Now, considering the upper pulley section, the tension T in the vertical cable is directed downwards, while it points towards joint A for the horizontal cable. The tension in the vertical cable is also 10 kN as it is part of the same continuous cable system.
In member DB, the force FBD can be resolved into its horizontal and vertical components using a slope triangle. The moment equilibrium condition at joint A gives FBD as 50 kN.
The horizontal force equilibrium condition can be applied to joint A.
![Equation 1](/files/ftp_upload/14291/14291_PT_Equation_1v2.svg)
Substituting the values of the length AB, AC, and radius of the pulley C, the force FBD is obtained as 50 kN.
The horizontal force equilibrium condition gives the reaction force at A as 40 kN.
![Equation 2](/files/ftp_upload/14291/14291_PT_Equation_2v2.svg)
Similarly, using the vertical force equilibrium condition, the vertical reaction force at A is estimated as -20 kN.
![Equation 3](/files/ftp_upload/14291/14291_PT_Equation_3v2.svg)
The force equilibrium conditions can be applied at joint D to obtain the horizontal and vertical reaction forces at D.
![Equation 4](/files/ftp_upload/14291/14291_PT_Equation_4v2.svg)
![Equation 5](/files/ftp_upload/14291/14291_PT_Equation_5v2.svg)
The obtained results indicate that the horizontal and vertical reaction forces at point D are -30 kN and 40 kN, respectively.