Drawning the diagrams of internal forces

Now you will know a few rules for drawing graphs of internal forces in structural elements. This is a very important stage in your learning, because it is the basis for getting through subjects such as building mechanics. I would like to point out that I am not the author of the following material. I let him use it, because I could not explain it to myself better. The author of the text is a lecturer at the University of Zielona Góra. I would like to remind you once again that this is something that you really need to devote a lot of time to.

I. RELATIONSHIPS BETWEEN CHARTS OF INTERNAL FORCE, TYPE OF EXTERNAL LOAD AND STATIC CONSTRUCTION SCHEME.

  1. If in the considered range of the bar q (x) = 0 then the transverse force is a constant function and the linear – bending moment.
  2. If in the considered range of the bar q (x) ≠ 0 then the transverse force is a function by 1 degree higher than q (x), and the bending moment is by 2 degrees higher, q (x) = const, so T (x) = ax + b, M (x) = ax2 + bx + c. The convexity of the bending moment parabola is always directed at the load return q (x).
  3. If the transverse force is positive, then the bending moment diagram drops (grows in the algebraic sense) when moving from the left side of the graph to the right.
  4. If the transverse force is negative, the bending moment diagram rises (decreases in the algebraic sense) when moving from the left side of the graph to the right.
  5. If, when passing through 0, the transverse forces function changes the sign, the bending moment function reaches extreme at this point. Maximum when changing characters from “+” to “-“; minimum when the character changes from “-” to “+”.
  6. In the cross-section of a bar in which the perpendicular force (parallel, bending moment) is applied to the axis of the bar, on the diagram of transverse forces (axial, bending moment) a jump is obtained by the value of this force (moment).
  7. If the rod load consists of many concentrated forces perpendicular to its axis, then the extreme bending moment will occur in the section in which the transverse force changes the sign. At the points of application of the remaining concentrated forces, the moment diagram collapses.
  8. On the extreme articulated support, the internal joint and the free end of the rod the bending moment function is reset (exception – see point I.6).
  9. The existence of an unloaded internal joint does not affect the distribution of transverse and axial forces.
  10. The values of support reactions do not depend on the shape of the structure, but only on the mutual positioning of the supports and the external load.

\\ HA = HA' \\ VA = VA' \\ RB = RB'

 

Most of the compounds given above result from the fact that the transverse force is the first derivative of the bending moment.

II. THE PRINCIPLE OF SUPERPOSITION (ADDING IMPACTS).

If the system is loaded with several types of external load (for example: several concentrated forces + bending moment + load), then internal force diagrams can be drawn from each load separately and then added to each other.

III. PROCESS AT DRAW “FROM THE MEMORY” OF INTERNAL FORCES.

  1. Kinematic and static analysis. If the system is geometrically unchangeable and statically determinable, you can go to point 2.
  2. Determination of the type of construction:
    a. beams and cantilever frames – point IV;
    b. beams and frames supported articulated – point V;
    c. symmetrical three-jointed frames – point VI;
    d. unsymmetrical three-joint frames – point VII;
    e. beams and composite frames – point VIII:
  • division into independent and dependent elements;
  • defining the type of each element;
  • using points IV, V, VI, VII depending on the type of element.

IV. BEAMS AND CANTILEVER FRAME.

In this type of constructions, you can immediately draw graphs of internal forces, omitting the step of calculating the reaction. The analysis should start from the free end of the system. In the case of a frame composed of several rods, each rod can be treated as a simple support fixed at the point of connection with another rod (point II.1.). The following rules will help you draw the charts:

  • intuitive sketching of the bending line of the bracket to show on which side to draw the bending moment diagram (this graph should always be drawn on the side of stretched fibers);
  • in bar connections at an angle, the torque graph must be continuous (the value at the end of one bar must be equal to the value at the beginning of the other);
  • load parallel to the axis of the rod causes in this rod a constant bending moment;
  • in the points lying at the intersection of the direction of force and the axis of the rod, the bending moment is zeroed;

Example.

From the condition of the continuity of the torque function it follows that in point B of the BC bar the torque value is also 16kNm. The uniform load is parallel to the BC bar, which results in a constant value of the moment along the entire length of the bar (the arm of the resulting load force is the same for each bar point BC). If M = const then T = 0. The resultant return of the uniform load shows that the BC bar is compressed, so N = -8kNm.

Due to the continuity of the torque function, its value at point C of the CD is also 16kNm. At the point 2m below C the torque value is 0, because through this point passes the line of the uniform load force. The value of torque at point D can be determined from the ratio (M / 4 = 16/2, so M = 32kNm). Since, moving from the left to the right of the CD rod, the moment diagram rises upwards, so the transverse force will be positive and equal to the resulting uniform load. There are no loads acting along its axis in the CD rod, so N = 0

V. BEAMS AND FRAMES – SUPPORTED ARTICULATED.

In these constructions one of the reactions (usually horizontally) can be found immediately using the equilibrium condition in the sum of views of all forces on one of the coordinate system axes (this action can be performed in memory – the response must be equal to the value of the applied load in a given direction and have the opposite phrase). This reaction, together with the corresponding external load, gives the moment (a pair of forces) that must be balanced for a moment from the remaining reactions in the other direction. On this basis, we determine the returns of these reactions and their values – the value of the moment divided by the distance between the reactions. Knowing all reactions, you can draw graphs of internal forces by treating component rods like brackets (see point V).

The frame is loaded with horizontal force at point B, the only place where the reaction to this load can arise is the point D. The external force and the horizontal reaction have the same directions and values, but the opposite turns and are shifted to each other by 3m. They are therefore a pair of forces generating a torque of 4kN × 3m = 12kNm turning the structure counter-clockwise. Because the system is to be in equilibrium, this moment must be balanced by another 12kNm moment acting clockwise and triggered by a pair of vertical forces. On this basis, the returns of vertical reactions and their values can be determined (the moment of 12kNm divided by the 4m arm gives the force value 3kN).

After determining the reaction, you can go on to draw graphs of internal forces. AB and CD rods can be treated as supports fixed at points B and C, loaded with end-focused forces. In the CD rod the moment stretches the fibers on the left side and takes values from 0 in D to 12kNm in C, the transverse force is negative (because from left to right the graph of moments rises up) and has the value of horizontal reaction (perpendicular to this bar). The vertical reaction causes compression, therefore N = -3kN. The AB bar will only be stretched by a force of 3kN.

The graph of moments in the horizontal bar can be drawn knowing the moment values at points C (12kNm) and D (0kNm) and knowing that there is no external load on the length of this rod. Therefore, it is enough to combine both values with a straight line. From left to right, the moment diagram drops, so the transverse force will be positive and will take the value of the force perpendicular to this bar, i.e. the 3kN reaction. Both horizontal forces (external and reaction) evoke compression in the BC rod – 4kN.

The verification of the correctness of the obtained graphs can be the balance test (sum of moments and the sum of the projections of forces on both axes) cut out of the construction of nodes, for example C and B.

VI. SYMMETRICAL THREE-JOINTED FRAMES.

Vertical reactions in external joints in such frames should be calculated using the condition of equilibrium in the form of the sum of projections of forces on the vertical axis and symmetry of the system (both reactions must be equal in value and have the same turns). In contrast to the articulated frames, in these constructions, even when there is no horizontal load, horizontal reactions will arise. You can find them by considering the balance of one part of the frame and calculating (in memory) the sum of the moments relative to the central joint. Horizontal reactions in the second frame will have the same values but opposite turns (symmetry). After determining the reaction, the internal force can be calculated by treating the component rods as brackets (see point V).

From the symmetry of the system, it follows that the vertical reactions at points A and D will be directed upwards and will have a value of 4kN. Then the external force 8kN should be arbitrarily assigned to one of the component frames (any) and its balance should be considered. The most convenient way to calculate in memory is the sum of moments relative to the point E. The vertical reaction of 4kN gives the torque (4kN × 2m = 8kNm) rotating the system in a clockwise direction relative to the point E. This moment must be balanced by the moment coming from horizontal reactions at points D and E. Their values are 8kNm / 4m = 2kN. It is still necessary to check the sum of views of all forces on the vertical axis. In order to ensure balance at point E, a vertical reaction of 4 kN must appear.

Using the reactions calculated for one of the component frames and the symmetry of the system and treating individual rods as brackets, you can quickly draw graphs of internal forces.

VII. UNSYMMETRICAL THREE-JOINTED FRAMES.

 

In such systems, it is quite difficult to determine the “memory” of the reaction. They can be calculated analytically using the static equilibrium conditions. After determining the internal force reaction, it is possible to calculate in the memory treating the component rods as brackets (see point V).

VIII. BEAMS AND COMPOSITE FRAMES.

Complex structures consist of 2 types of elements:

  • independent elements – they are geometrically immutable and can exist independently;
  • dependent elements – they are geometrically variable and can’t carry loads by themselves.

Calculations should start with dependent elements. After qualifying the element to a particular type of systems, you can determine reactions and internal forces (points V, VI, VII, VIII). Reactions from them are then loaded with independent elements. Here the course of proceedings is repeated. The following rules will help you draw the charts:

  • in the internal joints the bending moment value is 0;
  • on the internal articulated supports the moment diagram breaks down, and its value at this point can be found using proportions;
  • internal joints do not affect the distribution of transverse and normal forces;
  • on the internal articulated props, jumps with values equal to the reactions on these supports will be created on the cross-forces diagram.

I wish you good luck and I recommend a lot of tasks to really do these charts from memory.