Shear And Bending Moment Diagrams.ppt 133w4y

It is important to know how the shear forces and bending moments vary along the length of a beam that is being designed. Graphs are used to describe the change of shear forces and moments. These graphs are called shear and moment diagrams. Employing these diagrams, the maximum and minimum shear and moment are easily identified and located.

Constructing shear and moment diagrams is similar to finding the shear and moment at a particular point on a beam structure. However, instead of using an exact location, the location is a variable distance 'x'. This allows the shear and moment to be a function of the distance, x. In theory, this appears to be simple, but the equations can be complex, especially with distributed loads that are also a function of the location, x. Also, if there are multiple loads or s, more than one function must be developed, i.e. one shear and moment function for each section or span of the beam.

Tensile stresses and compressive stresses increase proportionally with bending moment, but are also dependent on the second moment of area of the cross-section of the structural element. Failure in bending will occur when the bending moment is sufficient to induce tensile stresses greater than the yield stress of the material throughout the entire cross-section. It is possible that failure of a structural element in shear may occur before failure in bending, however the mechanics of failure in shear and in bending are different. The bending moment at a section through a structural element may be defined as "the sum of the moments about that section of all external forces acting to one side of that section". The forces and moments on either side of the section must be equal in order to counteract each other and maintain a state of equilibrium so the same bending moment will result from summing the moments, regardless of which side of the section is selected.

Moments are calculated by multiplying the external vector forces (loads or reactions) by the vector distance at which they are applied. When analysing an entire element, it is sensible to calculate moments at both ends of the element, at the beginning, centre and end of any uniformly distributed loads, and directly underneath any point loads. Of course any "pin-ts" within a structure allow free rotation, and so zero moment occurs at these points as there is no way of transmitting turning forces from one side to the other. If clockwise bending moments are taken as negative, then a negative bending moment within an element will cause "sagging", and a positive moment will cause "hogging". It is therefore clear that a point of zero bending moment within a beam is a point ofcontraflexure—that is the point of transition from hogging to sagging or vice versa.

It is more common to use the convention that a clockwise bending moment to the left of the point under consideration is taken as positive. This then corresponds to the second derivative of a function which, when positive, indicates a curvature that is 'lower at the centre' i.e. sagging. When defining moments and curvatures in this way calculus can be more readily used to find slopes and deflections. Critical values within the beam are most commonly annotated using a bending moment diagram, where negative moments are plotted to scale above a horizontal line and positive below. Bending moment varies linearly over unloaded sections, and parabolically over uniformly loaded sections.

exists in a structural element when a moment is applied to the element so that the element bends. Moments and torques are measured as a force multiplied by a distance so they have as unit newtonmetres (N·m), or pound-foot or foot-pound (ft·lb). The concept of bending moment is very important in engineering (particularly in civil andmechanical engineering) and physics.

are analytical tools used in conjunction withstructural analysis to help perform structural design by determining the value of shear forceand bending moment at a given point of an element. Using these diagrams the type and size of a member of a given material can be easily determined. Another application of shear and moment diagrams is that the deflection can be easily determined using either the moment area method or the conjugate beam method.

Although these conventions are relative and any convention can be used if stated explicitly, practicing engineers have adopted a standard convention used in design practices.

The normal convention used in most engineering applications is to label a positive shear force one that spins an element clockwise (up on the left, and down on the right). Likewise the normal convention for a positive bending moment is to warp the element in a "u" shape manner (Clockwise on the left, and counterclockwise on the right). Another way to this is if the moment is bending the beam into a "smile" then the moment is positive, with compression at the top of the beam and tension on the bottom.

Normal positive shear force convention (left) and normal bending moment convention (right).

This convention was selected to simplify the analysis of beams. Since a horizontal member is usually analyzed from left to right and positive in the vertical direction is normally taken to be up, the positive shear convention was chosen to be up from the left, and to make all drawings consistent down from the right. The positive bending convention was chosen such that a positive shear force would tend to create a positive moment. A clear understanding of most beams that are analyzed can be found here. This shows most of the conventions, both in forces and s that we use to understand how beams are loaded.

The general steps for shear and moment diagrams are as follows: 1.Identify all information and draw diagram (similar to all static or mechanics problems). 2.Construct a free-body-diagram (FBD) and solve for all unknown reaction forces. 3.Cut the beam (one cut for each beam segment), draw a FBD, and solve for the unknown M and V at the cut in of x. 4.Plot the moment and shear equations developed in step 3. It should be noted that this example is fairly simple with one load type and simple s. More complex examples are given below with multiple beam segments and loads. Also, the sign convention is important and is reviewed below.

An exception to using the normal convention is used when deg concrete structures. Since concrete is weak in tension the most important part of deg a member with. From the service loads and the structural configuration the reaction loads can be determined using one several structural analysis methods including finite element method and static analysis. Once the reaction loads have been determined the loading diagram can be drawn

When constructing shear and moment diagrams, the sign convention is important so viewers will know what direction the beam is bending or shearing. Generally, it is assumed that a positive moment causes a beam to bend downward as shown in the diagram. A positive shear will skew the beam with the left side going up and the right going down, as shown. Each textbook can have different conventions, so it is important that you check what directions are positive or negative. Notice, all internal moments and shears need to cancel out if the cut section is reassembled. This means the direction is opposite on the right and left faces of the cut.

One confusing aspect of developing moment and shear diagrams is the need to have separate moment and shear functions for each beam segment. This is because a single function cannot model the moment (or shear) change over a load or (it is a discontinuous function). Each new beam section will have its own moment and shear equations as a function of the location, x. The diagram at the left gives various beams and typical loadings. After each section is cut, then a FBD is drawn for either side of the beam, just like a single section beam discussed above. Then the M and V is determined for that cut and plotted.

Moment and shear diagrams are best understood by examining an example. The simple beam at the left has a single point load between the s. The first step is to draw a FBD of the whole beam and solve for the reactions. ΣMA = 0 (10 ft) RB - (6 ft)(120 lb) = 0 RB = 72 lb ΣFy = 0 RA - 120 lb + 72 lb = 0 RA = 48 lb

With the loading diagram drawn the next step is to find the value of the shear force and moment at any given point along the element. For a horizontal beam one way to perform this is at any point to "chop off" the right end of the b The example below includes a point load, a distributed load, and an applied moment. The s include both hinged s and a fixed end . The first drawing is the situation of the element or what most people call a free body diagram

Determine Beam Section Section (1), to the left of the applied load, will have an expression for the shear force and bending moment that will differ from the section (2), to the right of the applied load. Therefore, the sections must be evaluated separately and each will have their own moment and shear equations. Section (1) (0 ≤ x ≤ 6 ft) First, cut section (1) a distance x from the left side and form a FBD, as shown. , the discarded right beam section needs to be replaced by unknown an internal shear force and moment, V1 an M1. The left beam section must stay in static equilibrium. ΣFy = 0 48 lb - V1 = 0 V1 = 48 lb ΣMcut = 0 M1 - x (48 lb) = 0 M1 = 48x ft-lb

Section (2) (6≤ x ≤ 10 ft) Now the next and last section can be cut. The left or right section can be discarded. For consistency with section (1), the left side will be examined. Again, form a FBD, as shown. The internal loads are labeled, V2 an M2 to distinguish them from section (1) shear and moment equations. Again, applying the static equilibrium equations gives, ΣFy = 0 48 lb - 120 lb - V2 = 0 V2 = -72 lb ΣMcut = 0 M2 - x (48 lb) + (x - 6)(120 lb) = 0 M1 = 720 - 72x ft-lb Plot Shear and Moment Diagrams The functions for V and M for both beam sections can be plotted to give the shear and moment over the length of the beam. The plots are given at the left. The location for maximum and minimum shear force and bending moment are easily found and evaluated.

The second drawing is the loading diagram with the reaction values given without the calculations shown. The third drawing is the shear force diagram and the fourth drawing is the bending moment diagram. For the bending moment diagram the normal sign convention was used. Below the moment diagram is the stepwise functions for the shear force and bending moment with the functions expanded to show the effects of each loading on the shear and bending functions. The example is illustrated using imperial units.

The first step obtaining the bending moment and shear force equations is to determine the reaction forces. This is done using a free body diagram.

Solving this system of linear equations, R_a=25.3 R_B= -3.5 R_C =3.3 and Mc=-14.6 [This set of equations cannot be solved, write the y = Ax form and "A" is singular - there are more unknowns than static equilibrium equations. Multiple moment equations are not independent] After the reaction forces are found, you then break the beam into pieces. The location and number of external forces on the member determine the number and location of these pieces. The first piece always starts from one end and ends anywhere before the first external force.

V is defined to be the shear force at a particular instant. As the section of the beam moves towards the external force, the location of the shear force changes. This makes the shear force in of the location of the bar ( in this example ‘x’). M is defined as the moment and is a function of x. By summing the forces along this segment and summing the moments, the equations for the shear force and bending moment are obtained. Taking the second segment, ending anywhere before the second internal force:

Notice that because the shear force is in of x, the moment equation is squared. This is due to the fact that the moment is the integral of the shear force. The tricky part of this moment is the distributed force. Since the force changes with the length of the segment, the force will be multiplied by the distance after 10 ft. i.e (x-10) the moment location is defined in the middle of the distributed force, which is also changing. This is where (x+10)/2 is derived from. Taking the third section:

Notice that the distributed force can now be considered one force of 15 kips acting in the middle of where it is positioned. Taking the fourth and final segment:

By plotting each of these equations on their intended intervals, you get the bending moment and shear force diagrams for this beam. It is important to note the relationship between the two diagrams. The moment diagram is a visual representation of the area under the shear moment diagram. That is, the moment is the integral of the shear force. If the shear force is constant over an interval, the moment equation will be in of x. If the shear force is linear over an interval, the moment equation will be quadratic. Another note on the shear moment diagrams is that they show where external force and moments are applied. With no external forces, the piecewise functions should attach and show no discontinuity. The discontinuities on the graphs are the exact magnitude of either the external force or external moments that are applied. For example, at x = 10 on the shear force diagram, there is a gap between the two equations. This gap goes from -10 to 15.3. The length of this gap is 25.3, the exact magnitude of the external force at that point. At section 3 on the moment diagram, there is a discontinuity of 50. This is from the applied moment of 50 on the structure. The maximum and minimum vales on the graphs represent the max forces and moments that this beam will have under these circumstances.

Since this method can easily become unnecessarily complicated with relatively simple problems, it can be quite helpful to understand different relations between the loading, shear, and moment diagram. The first of these is the relationship between a distributed load on the loading diagram and the shear diagram. Since a distributed load varies the shear load according to its magnitude it can be derived that the slope of the shear diagram is equal to the magnitude of the distributed load. The relationship between distributed load and shear force magnitude is:

V = dM/dx

Some direct results of this is that a shear diagram will have a point change in magnitude if a point load is applied to a member, and a linearly varying shear magnitude as a result of a constant distributed load. Similarly it can be shown that the slope of the moment diagram at a given point is equal to the magnitude of the shear diagram at that distance. The relationship between distributed shear force and bending moment is

This relationship may be helpful in determining both the moment and shear diagram without calculating the actual function. Derivation of this relationship is done in the Integration of Load Equation.

A direct result of this is that at every point the shear diagram crosses zero the moment diagram will have a local maximum or minimum. Also if the shear diagram is zero over a length of the member, the moment diagram will have a constant value over that length. By calculus it can be shown that a point load will lead to a linearly varying moment diagram, and a constant distributed load will lead to a quadratic moment diagram.

In practical applications the entire stepwise function is rarely written out. The only parts of the stepwise function that would be written out are the moment equations in a nonlinear portion of the moment diagram; this occurs whenever a distributed load is applied to the member. For constant portions the value of the shear and/or moment diagram is written right on the diagram, and for linearly varying portions of a member the beginning value, end value, and slope or the portion of the member are all that are required