Statics - Stress Resultants скачать в хорошем качестве

Statics - Stress Resultants

Hi, this is a preview of the online course "Statics" on https://learn-engineering.com​ Enroll now for 29 dollars per month! Script: In this slide we will now determine the stress resultants using a simple example. As an example we have a beam whose dead weight can be neglected. The beam is supported at point A by a simple support and at point B by a hinged support. At point D a force F is applied. The force F is inclined by the angle alpha from the beam. The distance to point A is given by l1 and the distance to point B by l2.In the next step we define our coordinate system. Now we form the Free Body Diagram. We represent the beam as a black line. The Simple Support A acts on the beam with the vertical force Az. The Hinged Support has the vertical force Bz and additionally the horizontal force Bx. Now we draw the force F. We can decompose this into a vertical component, F times sinus alpha equal to Fz, and into a horizontal component, F times cosinus alpha equal to Fx. We can then determine the forces of the supports. If we consider the sum of all forces in the x-direction, we find that Bx equals Fx. If we sum up all the forces in the z-direction and form a moment around point A or point B, we can determine the forces Az and Bz. Now we can make the first cut. To do this, we start from the left side of the beam and make sure that we are still in front of the first force, i.e. in front of force F, and make the cut there. It is always important to enter the running coordinate, in this case the x-coordinate. At the point of the Simple Support we can now directly enter the calculated force l2 divided by l1+2 times Fz. At the positive end of the beam at the intersection we now enter the three stress resultants. N for the normal component, V for the vertical component and M for the moment. All stress resultants are assumed to be positive because we are at the positive end of the beam.Now we look at our free body diagram again and mentally pass the force F and make sure that no new force comes after this and execute our cut. Now we draw the cut beam exactly as in the first cut, but this time the force F is included with its components Fz and Fx.Now, are we done with the cuts or do we need a third cut. We see in the Free Body Diagram that after our second cut there is only the support force of the hinged support, which is at the end of the beam. Since the force is at the end of the beam, we do not need another cut and can start diagramming.These diagrams are called normal-force, shear-force and bending-moment diagrams. To determine the curves, we form an equilibrium condition for each component and calculate the forces. In addition, we divide the diagrams at point D, at the point where the force D acts, into two areas in order to separate the first section and the second section.With the first section and the normal force, we can then form the sum of all forces in the x-direction. Since we only have the stress resultant N in the first section, we have N equal to zero, so we cannot enter anything but N equal to zero in the normal-force diagram. In the next step we calculate the sum of all forces in the z-direction for the first section. Here we calculate V equal to l2 divided by l1+2 times Fz. Since the stress resultant V has a positive result, we enter this in the shear-force diagram. For the bending-moment diagram we now form the moment around the intersection and calculate M equal to l2 divided by l1+2 times Fz times x. Here we see that the moment depends on the running variable x. The higher x, the higher the moment. That is, the higher x, the higher the moment. This is a linear function, so we have to draw the bending moment as an increasing line in the bending-moment diagram. At the end of the first section we can enter l1 for x and thus calculate the amount of moment at point D, which is: l1 times l2 divided by l1 plus l2 times Fz.We now carry out the same procedure for the second cut and can thus calculate the normal force as N equal to minus Fx through the sum of all forces in the x-direction. For the vertical force, we now also calculate the sum of all forces in the z-direction, which leads us to V equal to l2 divided by l1 plus 2 minus 1 times Fz. Since this force is now negative, we also enter this in the shear-force diagram. For the bending-moment diagram, we again form the moment at the end of the section and thus arrive at a linear equation in which the bending moment decreases with a higher x-portion. We can easily check this by substituting l1 plus 2 for x, which leads to l2 times Fz minus l2 times Fz, which is equal to zero.