In this tutorial we will look at how to calculate the bending stress of a beam using a bending stress formula that relates the longitudinal stress distribution in a beam to the internal bending moment acting on the beamâs cross section. A bending moment is the reaction induced in a structural element when an external force or moment is applied to the element causing the element to bend. The most common or simplest structural element subjected to bending moments is the beam. The diagram shows a beam which is simply supported at both ends. Chapter 04 - Shear and Moment in Beams. Definition of a Beam A beam is a bar subject to forces or couples that lie in a plane containing the longitudinal section of the bar. According to determinacy, a beam may be determinate or indeterminate. Loads applied to the beam may consist of a concentrated load (load applied at a point), uniform.
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Shear and Bending moment diagram for a simply supported beam with a concentrated load at mid-span.
Shear and bending moment diagrams are analytical tools used in conjunction with structural analysis to help perform structural design by determining the value of shear force and bending moment at a given point of a structural element such as a beam. These diagrams can be used to easily determine the type, size, and material of a member in a structure so that a given set of loads can be supported without structural failure. Another application of shear and moment diagrams is that the deflection of a beam can be easily determined using either the moment area method or the conjugate beam method.
Convention[edit]
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.
Normal convention[edit]
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 remember 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.[1]
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.
Alternative drawing convention[edit]
In structural engineering and in particular concrete design the positive moment is drawn on the tension side of the member. This convention puts the positive moment below the beam described above. A convention of placing moment diagram on the tension side allows for frames to be dealt with more easily and clearly. Additionally placing the moment on the tension side of the member shows the general shape of the deformation and indicates on which side of a concrete member rebar should be placed, as concrete is weak in tension.[2]
Calculating shear force and bending moment[edit]
Loaded beam
Moment Applied To Beam A Frame
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 beam.
The example below includes a point load, a distributed load, and an applied moment. The supports include both hinged supports and a fixed end support. The first drawing shows the beam with the applied forces and displacement constraints. The second drawing is the loading diagram with the reaction values given without the calculations shown or what most people call a free body diagram. 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 are the stepwise functions for the shear force and bending moment with the functions expanded to show the effects of each load on the shear and bending functions.
The example is illustrated using United States customary units. Point loads are expressed in kips (1 kip = 1000 lbf = 4.45 kN), distributed loads are expressed in k/ft (1 k/ft = 1 kip/ft = 14.6 kN/m), moments are expressed in ft-k (1 ft-k = 1 ft-kip = 1.356 kNm), and lengths are in ft (1 ft = 0.3048 m).
Step 1: Compute the reaction forces and moments[edit]
Free-body diagram of whole beam
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 of the entire beam.
The beam has three reaction forces, Ra, Rb at the two supports and Rc at the clamped end. The clamped end also has a reaction couple Mc. These four quantities have to be determined using two equations, the balance of forces in the beam and the balance of moments in the beam. Four unknowns cannot be found given two independent equations in these unknown variables and hence the beam is statically indeterminate. One way of solving this problem is to use the principle of linear superposition and break the problem up into the superposition of a number of statically determinate problems. The extra boundary conditions at the supports have to be incorporated into the superposed solution so that the deformation of the entire beam is compatible.
From the free-body diagram of the entire beam we have the two balance equations
Summing the forces, we have
and summing the moments around the free end (A) we have
We can solve these equations for Rb and Rc in terms of Ra and Mc :
and
If we sum moments about the first support from the left of the beam we have
If we plug in the expressions for Rb and Rc we get the trivial identity 0 = 0 which indicates that this equation is not independent of the previous two. Similarly, if we take moments around the second support, we have
Once again we find that this equation is not independent of the first two equations. We could also try to compute moments around the clamped end of the beam to get
This equation also turns out not to be linearly independent from the other two equations. Therefore, the beam is statically indeterminate and we will have to find the bending moments in segments of the beam as functions of Ra and Mc.
Step 2: Break beam into segments[edit]
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.
Step 3: Compute shear forces and moments - first piece[edit]
Free-body diagram of segment 1
Let V1 and M1 be the shear force and bending moment respectively in a cross-section of the first beam segment. As the section of the beam moves towards the point of application of the external force the magnitudes of the shear force and moment may change. This makes the shear force and bending moment a function of the position of cross-section (in this example x).
By summing the forces along this segment and summing the moments, the equations for the shear force and bending moment are obtained. These equations are:
and
Therefore,
Step 4: Compute shear forces and moments - second piece[edit]
Free-body diagram of segment 2
Taking the second segment, ending anywhere before the second internal force, we have
and
Therefore,
Notice that because the shear force is in terms 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.
Alternatively, we can take moments about the cross-section to get
Again, in this case,
Step 5: Compute shear forces and moments - third piece[edit]
Free-body diagram of segment 3
Taking the third segment, and summing forces, we have
and summing moments about the cross-section, we get
Therefore,
and
Notice that the distributed force can now be considered one force of 15 kips acting in the middle of where it is positioned.
Step 6: Compute shear forces and moments - fourth piece[edit]
Free-body diagram of segment 4
Taking the fourth and final segment, a balance of forces gives
and a balance of moments around the cross-section leads to
Solving for V4 and M4, we have
and
By plotting each of these equations on their intended intervals, you get the bending moment and shear force diagrams for this beam. In particular, at the clamped end of the beam, x = 50 and we have
Step 7: Compute deflections of the four segments[edit]
We now use the Euler-Bernoulli beam theory to compute the deflections of the four segments. The differential equation that relates the beam deflection (w) to the bending moment (M) is
Moment Applied To Beam Lyrics
where E is the Young's modulus and I is the area moment of inertia of the beam cross-section.
Substituting the expressions for M1, M2, M3, M4 into the beam equation and solving for the deflection gives us
Step 8: Apply boundary conditions[edit]
Now we will apply displacement boundary conditions for the four segments to determine the integration constants.
For the fourth segment of the beam, we consider the boundary conditions at the clamped end where w4 = dw/dx = 0 at x = 50. Solving for C7 and C8 gives
Therefore, we can express w4 as
Now, w4 = w3 at x = 37.5 (the point of application of the external couple). Also, the slopes of the deflection curves at this point are the same, i.e., dw4/dx = dw3/dx. Using these boundary conditions and solving for C5 and C6, we get Free english learning software download.
Substitution of these constants into the expression for w3 gives us
Similarly, at the support between segments 2 and 3 where x = 25, w3 = w2 and dw3/dx = dw2/dx. Using these and solving for C3 and C4 gives
Therefore,
At the support between segments 1 and 2, x = 10 and w1 = w2 and dw1/dx = dw2/dx. These boundary conditions give us
Therefore,
Step 9: Solve for Mc and Ra[edit]
Because w2 = 0 at x = 25, we can solve for Mc in terms of Ra to get
Also, since w1 = 0 at x = 10, expressing the deflection in terms of Ra (after eliminating Mc) and solving for Ra, gives
Step 10: Plot bending moment and shear force diagrams[edit]
Free-body diagram
Shear force diagram
Bending moment diagram
We can now calculate the reactions Rb and Rc, the bending moments M1, M2, M3, M4, and the shear forces V1, V2, V3, V4. These expressions can then be plotted as a function of length for each segment.
Relationship between shear force and bending moment[edit]
It is important to note the relationship between the two diagrams. The moment diagram is a visual representation of the area under the shear force 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 terms of x (linear). If the shear force is linear over an interval, the moment equation will be quadratic (parabolic).
Another note on the shear force 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 values on the graphs represent the max forces and moments that this beam will have under these circumstances.
Relationships between load, shear, and moment diagrams[edit]
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, described by Schwedler's theorem, between distributed load and shear force magnitude is:[3]
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:[4]
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.
Practical considerations[edit]![]()
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.[5]
See also[edit]References[edit]
Further reading[edit]
External links[edit]
Retrieved from 'https://en.wikipedia.org/w/index.php?title=Shear_and_moment_diagram&oldid=910400110'
Shear and moment diagram for a simply supported beam with a concentrated load at mid-span.
A bending moment is the reaction induced in a structural element when an external force or moment is applied to the element causing the element to bend.[1][2] The most common or simplest structural element subjected to bending moments is the beam. The diagram shows a beam which is simply supported at both ends. Simply supported means that each end of the beam can rotate; therefore each end support has no bending moment. The ends can only react to the shear loads. Other beams can have both ends fixed; therefore each end support has both bending moment and shear reaction loads. Beams can also have one end fixed and one end simply supported. The simplest type of beam is the cantilever, which is fixed at one end and is free at the other end (neither simple or fixed). In reality, beam supports are usually neither absolutely fixed nor absolutely rotating freely.
The internal reaction loads in a cross-section of the structural element can be resolved into a resultant force and a resultant couple. For equilibrium, the moment created by external forces (and external moments) must be balanced by the couple induced by the internal loads. The resultant internal couple is called the bending moment while the resultant internal force is called the shear force (if it is transverse to the plane of element) or the normal force (if it is along the plane of the element).
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. 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 of contraflexureâthat is the point of transition from hogging to sagging or vice versa.
Moments and torques are measured as a force multiplied by a distance so they have as unit newton-metres (N·m), or pound-foot (lbf·ft). The concept of bending moment is very important in engineering (particularly in civil and mechanical engineering) and physics.
Background[edit]
Tensile and compressive stresses increase proportionally with bending moment, but are also dependent on the second moment of area of the cross-section of a beam (that is, the shape of the cross-section, such as a circle, square or I-beam being common structural shapes). 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. In structural analysis, this bending failure is called a plastic hinge, since the full load carrying ability of the structural element is not reached until the full cross-section is past the yield stress. 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.
Moments are calculated by multiplying the external vectorforces (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-joints' 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.
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.
Draw your own house plans. Engineering descriptions of the computation of bending moments can be confusing because of unexplained sign conventions and implicit assumptions. The descriptions below use vector mechanics to compute moments of force and bending moments in an attempt to explain, from first principles, why particular sign conventions are chosen.
Computing the moment of force[edit]
Computing the moment of force in a beam.
An important part of determining bending moments in practical problems is the computation of moments of force.Let F{displaystyle mathbf {F} } be a force vector acting at a point A in a body. The moment of this force about a reference point (O) is defined as[2]
where M{displaystyle mathbf {M} } is the moment vector and r{displaystyle mathbf {r} } is the position vector from the reference point (O) to the point of application of the force (A). The Ã{displaystyle times } symbol indicates the vector cross product. For many problems, it is more convenient to compute the moment of force about an axis that passes through the reference point O. If the unit vector along the axis is e{displaystyle mathbf {e} }, the moment of force about the axis is defined as
where â
{displaystyle cdot } indicates the vector dot product.
Example[edit]
The adjacent figure shows a beam that is acted upon by a force F{displaystyle F}. If the coordinate system is defined by the three unit vectors ex,ey,ez{displaystyle mathbf {e} _{x},mathbf {e} _{y},mathbf {e} _{z}}, we have the following
Therefore,
The moment about the axis ez{displaystyle mathbf {e} _{z}} is then
Sign conventions[edit]
The negative value suggests that a moment that tends to rotate a body clockwise around an axis should have a negative sign. However, the actual sign depends on the choice of the three axes ex,ey,ez{displaystyle mathbf {e} _{x},mathbf {e} _{y},mathbf {e} _{z}}. For instance, if we choose another right handed coordinate system with Ex=ex,Ey=âez,Ez=ey{displaystyle mathbf {E} _{x}=mathbf {e} _{x},mathbf {E} _{y}=-mathbf {e} _{z},mathbf {E} _{z}=mathbf {e} _{y}}, we have
Then,
For this new choice of axes, a positive moment tends to rotate body clockwise around an axis.
Computing the bending moment[edit]
In a rigid body or in an unconstrained deformable body, the application of a moment of force causes a pure rotation. But if a deformable body is constrained, it develops internal forces in response to the external force so that equilibrium is maintained. An example is shown in the figure below. These internal forces will cause local deformations in the body.
For equilibrium, the sum of the internal force vectors is equal to the applied external force and the sum of the moment vectors created by the internal forces is equal to the moment of the external force. The internal force and moment vectors are oriented in such a way that the total force (internal + external) and moment (external + internal) of the system is zero. The internal moment vector is called the bending moment.[1]
Though bending moments have been used to determine the stress states in arbitrary shaped structures, the physical interpretation of the computed stresses is problematic. However, physical interpretations of bending moments in beams and plates have a straightforward interpretation as the stress resultants in a cross-section of the structural element. For example, in a beam in the figure, the bending moment vector due to stresses in the cross-section A perpendicular to the x-axis is given by
Expanding this expression we have,
We define the bending moment components as
The internal moments are computed about an origin that is at the neutral axis of the beam or plate and the integration is through the thickness (h{displaystyle h})
Example[edit]
Computing the bending moment in a beam.
Beam Moment Table
In the beam shown in the adjacent figure, the external forces are the applied force at point A (âFey{displaystyle -Fmathbf {e} _{y}}) and the reactions at the two support points O and B (RO=ROey{displaystyle mathbf {R} _{O}=R_{O}mathbf {e} _{y}} and RB=RBey{displaystyle mathbf {R} _{B}=R_{B}mathbf {e} _{y}}). The reactions can be computed using balances of forces and moments about point A, i.e.,
If L{displaystyle L} is the length of the beam, we have
If we solve for the reactions we have
Looking at the free body diagram of the part of the beam to the left of point X, the total moment of the external forces about the point X is
If we compute the cross products, we have
For this situation, the only non-zero component of the bending moment is
For the sum of the moments at X about the axis ez{displaystyle mathbf {e} _{z}} to be zero, we require
Beam Moment Formula
At x=xA{displaystyle x=x_{A}}, we have Mxz=FxA(LâxA)/L{displaystyle M_{xz}=Fx_{A}(L-x_{A})/L}.
Sign convention[edit]
In the above discussion, it is implicitly assumed that the bending moment is positive when the top of the beam is compressed. That can be seen if we consider a linear distribution of stress in the beam and find the resulting bending moment. Let the top of the beam be in compression with a stress âÏ0{displaystyle -sigma _{0}} and let the bottom of the beam have a stress Ï0{displaystyle sigma _{0}}. Then the stress distribution in the beam is Ïxx(y)=âyÏ0{displaystyle sigma _{xx}(y)=-ysigma _{0}}. The bending moment due to these stresses is
where I{displaystyle I} is the area moment of inertia of the cross-section of the beam. Therefore the bending moment is positive when the top of the beam is in compression.
Many authors follow a different convention in which the stress resultant Mxz{displaystyle M_{xz}} is defined as
Moment Applied To Beam
In that case, positive bending moments imply that the top of the beam is in tension. Of course, the definition of top depends on the coordinate system being used. In the examples above, the top is the location with the largest y{displaystyle y}-coordinate.
See also[edit]Moment Applied To Beam A CeilingReferences[edit]
External links[edit]
Moment Applied To End Of Cantilever Beam
Retrieved from 'https://en.wikipedia.org/w/index.php?title=Bending_moment&oldid=919369991'
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