Trave con carrello a dx e doppio pendolo a sx, carico concentrato inclinato

Beam with trolley on the right and double pendulum on the left, concentrated inclined load. Static diagram This is a horizontal beam with a span of 7 meters, constrained by a trolley on the left and a double pendulum on the right. The axis of the trolley (perpendicular to the sliding plane) forms a 50° angle with the horizontal. The axis of the double pendulum (parallel to the two connecting rods) is vertical. The concentrated load P = 6 kN is 4 meters from the trolley and directed downward. The load P forms a 70° angle with the beam axis. Static stability check The trolley eliminates only one degree of freedom, the one along the axis. The trolley imposes a constraining reaction RA directed along its axis. RA will form a 50° angle with the horizontal axis of the beam. The double pendulum with a vertical axis prevents rotation and vertical translation, leaving the horizontal sliding component free. The double pendulum therefore provides two constraint reactions: a vertical force RBy and a reactive moment MB. The overall degree of constraint is 3 for the three degrees of freedom of the beam. Therefore, since GDV = DOF, the beam is isostatic. Determination of the loads and calculation of the carriage reaction We begin by decomposition of the concentrated load P = 6 kN, inclined at 70° with respect to the beam axis. The horizontal component is Px = Pcos(70°) = 6cos(70°) = 2.052 kN (directed to the right). The vertical component is Py = Psen(70°) = 6sin(70°) = 5.638 kN (directed downward). Since the double pendulum with a vertical axis is unable to provide any horizontal reaction, the entire axial force induced by the load must be balanced by the horizontal component of the carriage reaction RA. From the equilibrium at the horizontal translation RAX+Px=0, we obtain RAX=-Px=-2.052 kN. We then calculate the magnitude of the cart's reaction: RA=RAX/cos(50°)=-2.052/cos(50°)=-3.191 kN. The negative sign indicates that the cart is in traction. Consequently, the vertical component of the cart's reaction is RAy=RAsin(50°)=-3.191sin(50°)=-2.444 kN and is directed downward. Calculating the Reactions of the Double Pendulum Let's proceed to determine the reactions of the double pendulum: the vertical force RBy and the concentrated moment MB. We write the equilibrium equation for the vertical translation: RAy-Py+ RBy=0. Substituting the numerical values ​​including the sign, we obtain -2.444-5.638+RBy=0. Hence, the vertical reaction of the double pendulum RBy=2.444+5.638=8.082 kN (directed upward). Next, write the rotational equilibrium equation by choosing the left end of the beam as the pole and assuming counterclockwise rotation as positive. The Py component, with a swing length of 4 meters, generates a clockwise moment. The reaction RBy, with a swing length of 7 meters, generates a counterclockwise moment. The rotational equilibrium equation has this form -(5.638x4)+(8.082)+MB=0. Hence, the calculation of MB: MB=+(5.638x4)-(8.082)=-34.021 kN. The minus sign indicates that the moment acts in a clockwise direction. Study of the various inclination combinations of the constraint axes To complete the study, the overall stability of the system is analyzed as the axes of the two constraints vary, considering the fundamental combinations of Vertical (V), Horizontal (O), and Inclined (I) axes: • Inclined Carriage - Bipendulum V (Base Case): Isostatic. The constraint directions are not parallel and do not generate coincident centers of rotation. • Carriage V - Bipendulum Vvertical: Labile for horizontal loads, hyperstatic for vertical loads. The carriage with a vertical axis reacts with a vertical force. The vertical bipendulum reacts with a vertical force and a concentrated moment. • Carriage O - Bipendulum V: Isostatic. The carriage experiences a horizontal reaction RAX equal and opposite to Px: RAX = -2.052 kN. The bipendulum reacts with a vertical force RBy directed upward equal and opposite to Py and with a concentrated moment MB. Therefore, RBy = Py = 58.638 kN. The moment MB is calculated as follows: MB = -7RBy + 4Py = -7.5638 + 4.5638 = -16.914 kN.m • Carriage I - Bipendulum O: Isostatic. The bipendulum with a horizontal axis prevents horizontal translation and rotation. The inclined axis of the carriage ensures overall equilibrium. • Carriage V - Bipendulum O: Isostatic. The carriage reacts vertically with a force RAy equal and opposite to Py. The bipendulum reacts horizontally with a force RBx equal and opposite to Px and again with a moment MB equal to Py.4. Performing the calculations, we have RAy = 5.638 kN, RBy = 2.052 kN, MB = Py.4 = 22.552 kN.m • Carriage O - Bipendulum O: Labile for vertical loads and hyperstatic for horizontal loads. The cart reacts horizontally. The bipendulum reacts with a horizontal force and a moment. • Cart I - Bipendulum I: Isostatics. In this case, too, it will be sufficient to apply the three cardinal equations of statics as explained in the video lesson.