Aim of this task was to study the effect of different forces on the bending moment in the beam and the result show that there is a linear relationship between bending moment and applied load. Experimental and theoretical bending moment shows perfect linear relationship with applied load with very little difference in the values of bending moment. Calculate the reactions at the supports of a beam, automatically plot the Bending Moment, Shear Force and Axial Force Diagrams Toggle navigation BEAM GURU.COM Beam calculator ONLINE.

CALCULATING BENDING MOMENT OF SIMPLY SUPPORTED BEAM 1. FOR CONCENTRATING LOAD- (End reaction X L1 ) ( where L1 is the distance between end reaction and that point where the load or external force is acted ) 2.

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.

• 1Convention
• 2Calculating shear force and bending moment

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]

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]

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]

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, R_a, R_b at the two supports and R_c at the clamped end. The clamped end also has a reaction couple M_c. 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

${\displaystyle \sum F=0,\sum M_A=0.}$

Summing the forces, we have

${\displaystyle -10-(1)(15)+R_a+R_b+R_c=0}$

and summing the moments around the free end (A) we have

${\displaystyle (R_a)(10)+(R_b)(25)+(R_c)(50)-(1)(15)(17.5)-50+M_c=0.}$

We can solve these equations for R_b and R_c in terms of R_a and M_c :

${\displaystyle R_b=37.5-1.6R_a+0.04M_c}$

and

${\displaystyle R_c=-12.5+0.6R_a-0.04M_c.}$

If we sum moments about the first support from the left of the beam we have

${\displaystyle (10)(10)-(1)(15)(7.5)+(R_b)(15)+(R_c)(40)-50+M_c=0.}$

If we plug in the expressions for R_b and R_c 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

${\displaystyle (10)(25)-(R_a)(15)+(1)(15)(7.5)+(R_c)(25)-50+M_c=0.}$

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

${\displaystyle (10)(50)-(R_a)(40)-(R_b)(25)+(1)(15)(32.5)-50+M_c=0.}$

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 R_a and M_c.

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]

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:

${\displaystyle \sum F=-10-V_1=0}$

and

${\displaystyle \sum M_A=-V_{1}x+M_1=0.}$

Therefore,

${\displaystyle V_1=-10\text{and}{M}_1=-10x.}$

Step 4: Compute shear forces and moments - second piece[edit]

Taking the second segment, ending anywhere before the second internal force, we have

${\displaystyle \sum F=-10+R_a-(1)(x-10)-V_2=0}$

and

${\displaystyle \sum M_A=R_a(10)-(1)(x-10)\frac{(x+10)}{2}-V_{2}x+M_2=0.}$

Therefore,

${\displaystyle V_2=R_a-x\text{and}{M}_2=-50+R_a(x-10)-\frac{x^2}{2}.}$

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

${\displaystyle \sum M_A=10x-R_a(x-10)+(1)(x-10)\frac{(x-10)}{2}+M_2=0.}$

Again, in this case,

${\displaystyle M_2=-50+R_a(x-10)-\frac{x^2}{2}.}$

Step 5: Compute shear forces and moments - third piece[edit]

Taking the third segment, and summing forces, we have

${\displaystyle -10+R_a+R_b-(1)(15)-V_3=0}$

and summing moments about the cross-section, we get

${\displaystyle (10)(x)-R_a(x-10)-R_b(x-25)+(1)(15)(x-17.5)+M_3=0.}$

Therefore,

${\displaystyle V_3=25-R_a-R_b=R_c}$

and

${\displaystyle M_3=262.5+R_a(x-10)+R_b(x-25)-25x=-675+R_a(30-0.6x)-M_c(1-0.04x)+12.5x.}$

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]

Taking the fourth and final segment, a balance of forces gives

${\displaystyle -10+R_a+R_b-(1)(15)-V_4=0}$

and a balance of moments around the cross-section leads to

${\displaystyle (10)(x)-R_a(x-10)-R_b(x-25)+(1)(15)(x-17.5)-50+M_4=0.}$

Solving for V₄ and M₄, we have

${\displaystyle V_4=25-R_a-R_b=R_c}$

and

${\displaystyle M_4=312.5+R_a(x-10)+R_b(x-25)-25x=-625+R_a(30-0.6x)+M_c(0.04x-1)+12.5x.}$

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

${\displaystyle M_4=M_c=-937.5+40R_a+25R_b.}$

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

${\displaystyle \frac{d^{2}w}{d{x}^2}=-\frac{M}{EI}}$

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

${\displaystyle C_7=-\frac{1250}{3EI}(-625+M_c+30R_a)\text{and}{C}_8=\frac{125}{EI}(-125+6R_a).}$

Therefore, we can express w₄ as

${\displaystyle w_4=-\frac{1}{300EI}(x-50{)}^2\left[-5(6R_a-125)(x-50)+2M_c(x+25)\right].}$

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

${\displaystyle C_5=-\frac{625}{12EI}(-5675+8M_c+240R_a)\text{and}{C}_6=\frac{250}{EI}\left(3R_a-70\right).}$

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

${\displaystyle C_3=-\frac{3125}{24EI}(-1645+4M_c+64R_a)\text{and}{C}_4=\frac{25}{12EI}\left(-40325+6M_c+120R_a\right).}$

Therefore,

At the support between segments 1 and 2, x = 10 and w₁ = w₂ and dw₁/dx = dw₂/dx. These boundary conditions give us

${\displaystyle C_1=-\frac{125}{24EI}(-40145+100M_c+1632R_a)\text{and}{C}_2=\frac{25}{4EI}(-1315+2M_c+48R_a).}$

Therefore,

${\displaystyle w_1=\frac{5}{24EI}\left[1026125-39450x+8x^3+20M_c(-125+3x)+480R_a(-85+3x)\right].}$

Step 9: Solve for M_c and R_a[edit]

Because w₂ = 0 at x = 25, we can solve for M_c in terms of R_a to get

${\displaystyle M_c=175-7.5R_a.}$

Also, since w₁ = 0 at x = 10, expressing the deflection in terms of R_a (after eliminating M_c) and solving for R_a, gives

${\displaystyle R_a=25.278⟹M_c=-14.585.}$

Step 10: Plot bending moment and shear force diagrams[edit]

Bending Moments In Beams

We can now calculate the reactions R_b and R_c, 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]

${\displaystyle \frac{dQ}{dx}=-q}$

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]

${\displaystyle \frac{dM}{dx}=Q}$

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]

1. ^Livermore C, Schmidt H, Williams J, Socrate S. '2.001 Mechanics & Materials I, Fall 2006'. Lecture 5: MIT OpenCourseWare: Massachusetts Institute of Technology. Retrieved 25 October 2013.
2. ^'Moment Diagram Sign Convention Poll'. Eng-Tips Forum. Retrieved 25 October 2013.
3. ^Emweb.unl.edu
4. ^Beer, Ferdinand P.; E. Russell Johnston; John T. DeWolf (2004). Mechanics of Materials. McGraw-Hill. pp. 322–323. ISBN0-07-298090-7.
5. ^Hibbeler, R.C (1985). Structural Analysis. Macmillan. pp. 146–148.

Further reading[edit]

• Cheng, Fa-Hwa. 'Shear Forces and Bending Moments in Beams' Statics and Strength of Materials. New York: Glencoe, McGraw-Hill, 1997. Print.
• Spotts, Merhyle Franklin, Terry E. Shoup, and Lee Emrey. Hornberger. 'Shear and Bending Moment Diagrams.' Design of Machine Elements. Upper Saddle River, NJ: Pearson/Prentice Hall, 2004. Print.

External links[edit]

• FREE Online Shear Force and Bending Moment Diagram (SFD & BMD) Calculator. (Note: only free up to 3 point loads.)

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 contraflexure—that 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.

• 2Computing the moment of force
• 3Computing the bending moment

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.

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]

An important part of determining bending moments in practical problems is the computation of moments of force.Let ${\displaystyle \mathbf{F}}$ 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]

${\displaystyle \mathbf{M}=\mathbf{r}\times \mathbf{F}}$

where ${\displaystyle \mathbf{M}}$ is the moment vector and ${\displaystyle \mathbf{r}}$ is the position vector from the reference point (O) to the point of application of the force (A). The ${\displaystyle \times }$ 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 ${\displaystyle \mathbf{e}}$, the moment of force about the axis is defined as

${\displaystyle M=\mathbf{e}\cdot \mathbf{M}=\mathbf{e}\cdot (\mathbf{r}\times \mathbf{F})}$

where ${\displaystyle \cdot }$ indicates the vector dot product.

Example[edit]

The adjacent figure shows a beam that is acted upon by a force ${\displaystyle F}$. If the coordinate system is defined by the three unit vectors ${\displaystyle {\mathbf{e}}_x,{\mathbf{e}}_y,{\mathbf{e}}_z}$, we have the following

Calculating Moments On A Beam

${\displaystyle \mathbf{F}=0{\mathbf{e}}_x-F{\mathbf{e}}_y+0{\mathbf{e}}_z\text{and}\mathbf{r}=x{\mathbf{e}}_x+0{\mathbf{e}}_y+0{\mathbf{e}}_z.}$

Moment Applied To Beam Formula

Therefore,

The moment about the axis ${\displaystyle {\mathbf{e}}_z}$ is then

${\displaystyle M_z={\mathbf{e}}_z\cdot \mathbf{M}=-Fx.}$

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 ${\displaystyle {\mathbf{e}}_x,{\mathbf{e}}_y,{\mathbf{e}}_z}$. For instance, if we choose another right handed coordinate system with ${\displaystyle {\mathbf{E}}_x={\mathbf{e}}_x,{\mathbf{E}}_y=-{\mathbf{e}}_z,{\mathbf{E}}_z={\mathbf{e}}_y}$, we have

${\displaystyle \mathbf{F}=0{\mathbf{E}}_x+0{\mathbf{E}}_y-F{\mathbf{E}}_z\text{and}\mathbf{r}=x{\mathbf{E}}_x+0{\mathbf{E}}_y+0{\mathbf{E}}_z.}$

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]

Beam With Moment Applied To Both Ends

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

${\displaystyle {\mathbf{M}}_x={\int }_A\mathbf{r}\times ({\sigma }_{xx}{\mathbf{e}}_x+{\sigma }_{xy}{\mathbf{e}}_y+{\sigma }_{xz}{\mathbf{e}}_z)dA\text{where}\mathbf{r}=y{\mathbf{e}}_y+z{\mathbf{e}}_z.}$

Expanding this expression we have,

${\displaystyle {\mathbf{M}}_x={\int }_A\left(-y{\sigma }_{xx}{\mathbf{e}}_z+y{\sigma }_{xz}{\mathbf{e}}_x+z{\sigma }_{xx}{\mathbf{e}}_y-z{\sigma }_{xy}{\mathbf{e}}_x\right)dA=:M_{xx}{\mathbf{e}}_x+M_{xy}{\mathbf{e}}_y+M_{xz}{\mathbf{e}}_z.}$

We define the bending moment components as

${\displaystyle \left[\begin{array}{c}{M}_{xx}\\ M_{xy}\\ M_{xz}\end{array}\right]:={\int }_A\left[\begin{array}{c}y{\sigma }_{xz}-z{\sigma }_{xy}\\ z{\sigma }_{xx}\\ -y{\sigma }_{xx}\end{array}\right]dA.}$

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 (${\displaystyle h}$)

Example[edit]

In the beam shown in the adjacent figure, the external forces are the applied force at point A (${\displaystyle -F{\mathbf{e}}_y}$) and the reactions at the two support points O and B (${\displaystyle {\mathbf{R}}_O=R_O{\mathbf{e}}_y}$ and ${\displaystyle {\mathbf{R}}_B=R_B{\mathbf{e}}_y}$). The reactions can be computed using balances of forces and moments about point A, i.e.,

${\displaystyle R_O+R_B-F=0\text{and}-{\mathbf{r}}_A\times {\mathbf{R}}_O+{\mathbf{r}}_B\times {\mathbf{R}}_B=\mathbf{0}.}$

If ${\displaystyle L}$ is the length of the beam, we have

${\displaystyle {\mathbf{r}}_A=x_A{\mathbf{e}}_x\text{and}{\mathbf{r}}_B=(L-x_A){\mathbf{e}}_x.}$

If we solve for the reactions we have

${\displaystyle R_O=\left(1-\frac{x_A}{L}\right)F\text{and}{R}_B=\frac{x_A}{L}F.}$

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

${\displaystyle \mathbf{M}=-({\mathbf{r}}_X-{\mathbf{r}}_A)\times \mathbf{F}-{\mathbf{r}}_X\times {\mathbf{R}}_O=\left[(x_A-x){\mathbf{e}}_x\right]\times \left(-F{\mathbf{e}}_y\right)-\left(x{\mathbf{e}}_x\right)\times \left(R_O{\mathbf{e}}_y\right).}$

If we compute the cross products, we have

For this situation, the only non-zero component of the bending moment is

${\displaystyle {\mathbf{M}}_{xz}=-\left[{\int }_z{\int }_{-h/2}^{h/2}y{\sigma }_{xx}dydz\right]{\mathbf{e}}_z.}$

For the sum of the moments at X about the axis ${\displaystyle {\mathbf{e}}_z}$ to be zero, we require

${\displaystyle \mathbf{M}+{\mathbf{M}}_{xz}=\mathbf{0}\text{or}-\frac{F{x}_A}{L}(L-x)+M_{xz}=0\text{or}{M}_{xz}=\frac{F{x}_A}{L}(L-x).}$

At ${\displaystyle x=x_A}$, we have ${\displaystyle M_{xz}=F{x}_A(L-x_A)/L}$.

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 ${\displaystyle -{\sigma }_0}$ and let the bottom of the beam have a stress ${\displaystyle {\sigma }_0}$. Then the stress distribution in the beam is ${\displaystyle {\sigma }_{xx}(y)=-y{\sigma }_0}$. The bending moment due to these stresses is

${\displaystyle M_{xz}=-\left[{\int }_z{\int }_{-h/2}^{h/2}y(-y{\sigma }_0)dydz\right]={\sigma }_{0}I}$

where ${\displaystyle I}$ 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 ${\displaystyle M_{xz}}$ is defined as

${\displaystyle {\mathbf{M}}_{xz}=\left[{\int }_z{\int }_{-h/2}^{h/2}y{\sigma }_{xx}dydz\right]{\mathbf{e}}_z.}$

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 ${\displaystyle y}$-coordinate.

See also[edit]

References[edit]

1. ^ ^abGere, J.M.; Timoshenko, S.P. (1996), Mechanics of Materials:Forth edition, Nelson Engineering, ISBN0534934293
2. ^ ^abBeer, F.; Johnston, E.R. (1984), Vector mechanics for engineers: statics, McGraw Hill, pp. 62–76

External links[edit]