**Shear Forces & Bending Moments I**

We will now turn our attention to the forces and torque which develop in a loaded beam. Up to this point we have generally looked at only axial members – members in simple tension or compression; and have considered the forces, stresses, and deformations which occur in such members. We will now look at a particular type of non-axial member – loaded horizontal beams, and will begin the process of determining the forces, toque, stresses, and deformations which occur in these beams. And as we proceed on we will also consider the problem of beam design and/or beam selection. In this first topic, we will focus only on the **SHEAR FORCES **and **BENDING MOMENTS** (internal torque) which occur in loaded beams. These quantities are very important, as we shall see, since the axial and shear stresses which will develop in the beam depend on the values of the shear forces and bending moments in the beam.

To understand the shear forces and bending moments in a beam, we will look at a simple example. In Diagram 1, we have shown a simply supported 20 ft. beam with a load of 10,000 lb. acting downward right at the center of the beam. Due to symmetry the two support forces will be equal, with a value of 5000 lb. each. This is the static equilibrium condition for the whole beam.

Next let’s examine a section of the beam. We will cut the beam a arbitrary distance (x) between 0 and 10 feet, and apply static equilibrium conditions to the left end section as shown in Diagram 2.. We can do this since as the entire beam is in static equilibrium, then a section of the beam must also be in equilibrium.

In Diagram 2a, we have shown left section of the beam, x feet, long – where x is an arbitrary distance greater than 0 ft. and less than 10 ft. Notice if we just include the 5000 lb. external support force, the section of the beam is clearly not in equilibrium. Neither the sum of forces (translational equilibrium), nor the sum of torque (rotational equilibrium) will sum to zero – as required for equilibrium. Therefore, since we know the **beam section is in equilibrium**, there must be some forces and/or torque not accounted for.

In diagram 2b, we have shown the missing force and torque. The 10,000 lb. load which we originally applied to the beam, and the support force cause internal “shearing forces” and internal torque called “bending moments” to develop. (We have symbolically shown these in Diagram 2c.) When we cut the beam, the internal shear force and bending moment at that point then become an external force and moment (torque) acting on the section. We have shown these in Diagram 2b, and labeled them **V (shear force)** and **M (bending moment)**.

Please note that M is a moment or torque – not a force. It does not appear in the sum of forces equation when we apply static equilibrium to the section – which will be our next step.**Equilibrium Conditions:****Sum of Forces in y-direction**: **+ 5000 lb. – V = 0** , solving **V = 5000 lb.****Sum of Toque about left end**: **-V * x + M = 0 **, we next substitute the value of V from the force equation into the torque equation: **- 5000 lb. * x + M = 0** , then solving for **M = 5000x (ft-lb.)**

These are the equations for the shear force and bending moments for the section of the beam from 0 to 10 feet. Notice that the internal shear force is a constant value of 5000 lb. for the section, but that the value of the internal torque (bending moment) varies from 0 ft-lb. at x = 0, to a value of 50,000 ft-lb. at x = 10 ft.

**[We really should not put exactly 0 ft., and 10 ft. into our equation for the bending moment. The reason is that at 0 and 10 ft., there are ‘point loads/forces’ acting. That is, we have our forces acting at point – and a point has zero area, so the stress (F/A) at these points would in theory be infinite. Of course, a stress can not be infinite, and we can not apply a force at a point – it is actually applied over some area (even if the area if small). However, in ‘book’ problems we normally apply forces at a point. To deal with this difficulty, we actually skip around these points. We cut our section at 0′ < x < 6′. Still when we put values into our expressions we put in values such as x = 9.99999999 ft, and round it off (numerically) to 10 ft. This is, in effect, cheating a bit. We are putting in the value x = 10 ft., but only because the number we actually put in was rounded off to 10 ft. It all may sound confusing, but it works, and will become clear as we do several examples.]**

First, however we will finish analyzing our simple beam. So far we have found expressions for the shear force and bending moments **(V1 = 5000 lb, M1 = 5000x ft-lb)** for section 1 of the beam, between 0 and 10 ft. Now we will look at the next section of the beam. We cut the beam at distance x (ft) from the left end, where x is now greater than 10 ft. and less then 20 ft. and then look at entire section to the left of where we cut the beam (See Diagram 3). Where the beam was cut, we have an internal shear force and bending moment – which now become external. These are shown in Diagram 3 as V2 and M2. (We add the ‘2’, to indicate we are looking at section two of the beam.)

We next apply static equilibrium conditions to the beam section, and obtain:**Equilibrium Conditions:****Sum of Forces in y-direction**: **+ 5000 lb. -10,000 lb. – V = 0** , solving **V2 = -5000 lb.****Sum of Toque about left end**: **-10,000 lb * 10 (ft) -V * x (ft) + M = 0** , we next substitute the value of V from the force equation into the torque equation : -10,000 lb * 10 ft. – (-5000 lb) * x (ft) + M = 0 , then solving for **M2 = -[5000x (ft-lb.) – 100,000] ft-lb.**

The two expressions above give the value of the internal shear force and bending moment in the beam, between the distances of the 10 ft. and 20 ft. A useful way to visualize this information is to make **Shear Force and Bending Moment Diagrams** – which are really the graphs of the shear force and bending moment expressions over the length of the beam. (See Diagram 4.)

These are a quite useful way of visualizing how the shear force and bending moments vary through out the beam. We have completed our first Shear Force/Bending Moment Problem. We have determined the expressions for the shear forces and bending moments in the beam, and have made accompanying shear force and bending moment diagrams.

Now that we have the general concepts concerning shear forces and bending moments, we want to step back for a moment and become a little more specific concerning some details, such as choosing the direction of the shear forces and bending moments.

**Shear Forces and Bending Moments II**

Before continuing with a second example of determining the shear forces and bending moments in a loaded beam, we need to take a moment to discuss the sign associated with the shear force and bending moment. The signs associated with the shear force and bending moment are defined in a different manner than the signs associated with forces and moments in static equilibrium.

**The Shear Force is positive if it tends to rotate the beam section clockwise with respect to a point inside the beam section.****The Bending Moment is positive if it tends to bend the beam section concave facing upward. (Or if it tends to put the top of the beam into compression and the bottom of the beam into tension.)**

In the beam section shown in Diagram 1, we have shown the **Shear Force V** and **Bending Moment M** acting in positive directions according to the definitions above.

Notice that there is a possibility for a degree of confusion with sign notation. When summing forces, the direction of V shown in the diagram is in the negative y-direction, yet it is a positive shear force. This can lead to some confusion unless we are careful. We will deal with possible confusion by always working from the left for our beam sections, and always choosing V & M in a positive direction according to the shear force and bending moments conventions defined above. That is, we will always select the V & M directions as shown in Diagram 1. This approach will simplify the sign conventions, as we will see in the next example.

However before the next example, we will look at the causes of the internal bending moment in a little greater detail.

In Diagram 2a, we have shown a simply supported loaded beam, and have indicated in an exaggerated way the bending caused by the load. If we then cut the beam and look at a left end section, we have the Diagram 2b.

In this diagram we have, for the sake of clarity, left out the vertical shear force which develops, but have shown horizontal forces (-F_{x} and + F_{x}). These forces develop since, as the beam bends, the top region of the beam is put into compression and the bottom region of the beam is put into tension. As a result there are internal horizontal (x-direction) forces acting in the beam; however for every positive x-force, there is an equal and opposite negative x-force. **Thus the net horizontal (x-direction) internal force in the beam section is zero.** **However, even though the actual x-forces cancel each other, the torque produced by these x-forces is not zero**. Looking at Diagram 2c and mentally summing torque about the center of the beam, we see that the horizontal x-forces cause a net toque – which we call the internal bending moment, M. This is the cause of the internal bending moment (torque) inside a loaded beam.

We now continue by proceeding very slowly and carefully through a somewhat extended example(s). We will also examine an alternate method for determining the bending moments in a beam.

A loaded, simply supported beam is shown. For this beam we would like to determine expressions for the internal shear forces and bending moments in each section of the beam, and to make shear force and bending moment diagrams for the beam.

We will work very slowly and carefully, step by step, through the solution for this example.

**Solution: Part A. **We first find the support forces acting on the structure. We do this in the normal way, by applying static equilibrium conditions for the beam.**STEP 1:** Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions.

**STEP 2:** Break any forces not already in x and y direction into their x and y components.**STEP 3:** Apply the equilibrium conditions.**Sum F _{y }= -4,000 lb. – (1,000 lb./ft)(8 ft) – 6,000 lb. + B_{y} + D_{y} = 0**

**Sum T**

_{B}= (D_{y})(8 ft) – (6,000 lb.)(4 ft) + (1,000 lb./ft)(8 ft)(4 ft) + (4,000 lb.)(8 ft) = 0Solving for the unknowns:

**B**(The negative sign indicates that D

_{y}=23,000 lb.; D_{y}= -5,000 lb._{y}acts the opposite of the initial direction we chose.)

**Part B**: Now we will determine the **Shear Force** and **Bending Moment** expressions for each section of the loaded beam. For this process we will ‘cut’ the beam into sections, and then use the translational equilibrium condition for the beam section (Sum of Forces = zero) to determine the **Shear Force** expressions in each section. Determining the **Bending Moment** expression for each section of the beam may be done in two ways.

**1) By applying the rotational equilibrium condition for the beam section (Sum of Torque = zero), and solving for the bending moment.****2) By Integration. The value of the bending moment in the beam may be found from****. That is, the bending moment expression is the integral of the shear force expression for the beam section.**

We now continue with the example. We begin by starting at the left end of the beam, and cutting the beam a distance “x” from the left end – where x is a distance greater than zero and less the position where the loading of the beam changes in some way. In this problem we see that from zero to eight feet there is a uniformly distributed load of 1000 lb./ft. However this ends at eight feet (the loading changes). Thus for section 1, we will cut the beam at distance x from the left end, where x is greater than zero and less then eight feet.

**Section 1**: Cut the beam at x, where 0 < x < 8 ft., and analyze left hand section.**1.** Draw a FBD of the beam section shown and labeling all forces and toque acting – including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) (See Diagram – Section 1) Notice we have drawn the shear force and bending moment in their positive directions according to the defined sign convention discussed earlier, and have labeled them as V1 and M1, as this is section 1 of the beam.**2.** We check that we have all forces in x & y components (yes)**3.** Apply translational equilibrium conditions to determine the shear force expression.

Sum Fx = 0 (no net external x- forces)

Sum Fy = -4,000 lb. – 1,000 lb./ft *(x) ft – V1 = 0 ; and solving: **V1 = [-4,000 – 1,000x] lb.**

This expression gives us the values of the internal shear force in the beam between 0 and 8 ft. Notice as x nears zero, the shear force value in the beam goes to – 4000 lb., and as x approaches 8 ft., the shear force value becomes -12,000 lb., and that is negative everywhere between 0 and 8 ft. Let’s think for a moment what this negative sign tells us. Since we found the shear force (V) by static equilibrium conditions, the negative sign tells us that we choose the incorrect direction for the shear force – that the shear force acts in the opposite direction. However, we choose the positive direction of the shear force (by its definition) and so the negative sign also tells us we have a negative shear force.**To try to simplify a somewhat confusing sign situation we may say this: As long as we work from the left end of the beam, and choose the initial direction of the shear force and bending moment in the positive direction (by their definition), then when we solve for the shear force and bending moment, the sign which results is the correct sign as applies to the shear force and bending moment values.**

If we graph the shear force expression above, we obtain the graph shown of the internal shearing force in the beam for the first eight feet. We next will determine the bending moment expression for this first beam section.

**4.** We can find the bending moment from static equilibrium principles; summing torque about the left end of the beam.

Referring to the free body diagram for beam section 1, we can write:

Sum Torque _{left end} = **-1000 lb/ft * (x) * (x/2) – V1 (x) + M1 = 0 **

To make sure we understand this equation, let’s examine each term. The first term is the torque due to the uniformly distributed load – 1000 lb./ft * (x) ft (this is the load) times (x/2) which is the perpendicular distance, since the uniform load may be considered to act in the center, which is x/2 from the left end. Then we have the shear force V1 times x feet to the left end, and finally we have the bending moment M1 (which needs no distance since it is already a torque).

Next we substitute the expression for V1 (V1 = [-4,000 – 1,000x] lb.) from our sum of forces result above into the torque equation to get:**Sum Torque _{left end} = -1000 lb/ft * (x) * (x/2) -[-4,000 – 1,000x] (x) + M1 = 0 **;

and solving for

**M1 = [-500x**

^{2}– 4,000x] ft-lb.This is our expression for the internal torque inside the load beam for section 1, the first eight feet, which is graphed in the diagram below.

**5.** Finally, we may also obtain the expression for the bending moment by integration of the shear force expression. The integrals we will be using are basic types.

For a simple, brief review and/or introduction to basic calculus concepts,**Please Select: ****Simple Derivatives/Integrals**.

Continuing with our example:

Integration:= 1000(1/2 x^{2}) – 4000 (x) + C1; so **M1 = -500x ^{2} – 4,000x + C1**

As we the results above show, when we do an indefinite integral, the result include an arbitrary constant, in this case called C1. To determine the correct value for C1 for our problem we must apply a boundary condition: That is, we must know the value of the bending moment at some point on our interval into to find the constant.

**For simply supported beams (with no external torque applied to the beam) the value of the bending moment will be zero at the ends of the beam.**

**(There are many ways to explain why this must be so. One of the easiest explanations is to remember that the bending moment value at a point in a simply supported beam is equal to the total area under the shear force diagram up to that point. However, at the left end, as x goes to zero, the area under the shear force diagram would also go to zero, and thus so would the bending moment value.)**

So we have for our “boundary condition” that at

**x = 0, M1 = 0**. We put these values into our expression for the bending moment (M1 = -500x

^{2}– 4,000x + C1), and solve for the value of the integration constant, C1; that is:

0 = -500(0)

^{2}– 4,000(0) + C1, and solving: C1 = 0

Therefore:

**M1 = [-500x**for 0 < x < 8 ft., is our final expression for the bending moment over the first section. (Note, it is the same as found above by summing torque for the beam section.)

^{2}– 4,000x] ft-lb.We now continue with the next section of the beam. Referring to the beam diagram, we see that at a location just greater than 8 ft., there is no loading, and that this continues until 12 ft. where there is a point load of 6,000 lb. So for our second section, we cut the beam at a location “x”, where x is greater than 8 ft., and less than 12 ft – and then analyze the entire left hand section of the beam.

**Section 2**: We cut the beam at x, where 8 < x <12 ft., and analyze the entire section left of where we cut the beam.

1. Draw a FBD of the beam section shown and labeling all forces and toque acting – including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) (See Diagram – Section 2) Notice we have drawn the shear force and bending moment in their positive directions according to the defined sign convention discussed earlier, and have labeled them as V2 and M2, as this is section 2 of the beam.

2. We check that we have all forces in x & y components (yes)

3. Apply translational equilibrium conditions (forces only):

Sum Fx = 0 (no net external x- forces)

Sum Fy = -4,000 lb. – 1,000 lb./ft(8 ft) + 23,000 lb. – V2 = 0, Solving: **V2 = 11,000 lb.**

4. We may determine the bending moment expression by applying rotational equilibrium conditions, or by integration. Once more we will do it both ways for this section.**Rotational Equilibrium**:**Sum of Toque _{left end} = – (1000 lb./ft * 8 ft) * 4 ft. + 23,000 lb. * 8 ft. – V2 * x + M2 = 0**; then we substitute the value for

**V2 (V2 = 11,000 lb)**from above and obtain:

**- (1000 lb./ft * 8 ft) * 4 ft. + 23,000 lb. * 8 ft. – (11,000 lb.) * x + M2 = 0**; and then solving for M2 we find:

**M2 = [11,000x – 152,000] ft-lb.**

From integration of the shear force, we find:

**Integration:**, or

**M2 = 11,000x + C2**

We get our boundary condition from another characteristic of the bending moment expression – which is that the bending moment must be continuous. That is, the value of the bending moment at the end of the first beam section, and the value of the bending moment at the beginning of the second beam section must agree – they must be equal. We determine the value of the bending moment from our M1 equation as x approaches 8 ft. (M1 = [-500 (8)^{2} – 4,000(8)] = -64,000 ft-lb.)

Then our boundary condition to find C2 is: at** x=8 ft M=-64,000 ft-lb.** We apply our boundary condition to find C2.

Apply BC: -64,000 ft-lb. = 11,000 lb. (8) + C2, Solving: **C2 = -152,000 ft-lb**.

Therefore: **M2 = [11,000x – 152,000] ft-lb. **for 8 < x < 12

In like manner we proceed with section 3 of the beam, cutting the beam at a location greater than 12 ft. and less 16 ft., and then analyzing the entire section left of where we cut the beam.

**Section 3**: Cut the beam at x, where 12 < x < 16 ft. Analyze left hand section.

1. FBD. (See Diagram Section 3)

2. All forces in x & y components (yes)

3. Apply translational equilibrium conditions (forces only):

Sum Fx = 0 (no net external x- forces)Sum Fy = -4,000 lb. – 1,000 lb./ft(8 ft) + 23,000 lb. -6,000 lb. – V3 = 0, and Solving:**V3 = 5,000 lb.**

4. We may determine the bending moment expression by applying rotational equilibrium conditions, or by integration. Once more we will do it both ways for this section.**Rotational Equilibrium**:**Sum of Toque _{left end} = – (1000 lb./ft * 8 ft) * 4 ft. + 23,000 lb. * 8 ft. – 6,000 lb. * 12 ft -V3 * x + M3 = 0**; then we substitute the value for

**V3 (V3 = 5,000 lb)**from above and obtain:

**- (1000 lb./ft * 8 ft) * 4 ft. + 23,000 lb. * 8 ft. – 6,000 lb. * 12 ft – (5,000 lb.) * x + M3 = 0**; and then solving for M3 we find:

**M3 = [5,000x – 80,000] ft-lb.**

We will find the bending moment expression for this section using integration only.

**Integration**, or

**M3 = 5,000x + C3**

We obtain a boundary condition for section 3 by remembering that at a free end or simply supported (no external torque) end, the bending moment must go to zero, thus we have the boundary condition to find C3:

**at x = 16 ft., M = 0 ft-lb**.

Apply BC: 0 = 5,000(16) + C3, and Solving: C3 = -80,000 ft-lb.

Therefore:

**M3 = [5,000x – 80,000] ft-lb.**for 12 < x < 16

We now have our expressions for the shear forces and bending moments in each section of the loaded beam (summarized below). Additional, we have shown the shear force and bending moment diagrams for the entire beam – which is a visual representation of the internal shear forces and internal torque in the beam due to the loading.

**Part C:**Shear Force and Bending Moment Diagrams: Using the expressions found above, we can draw the shear force and bending moment diagrams for our loaded beam.

**V1 = -1,000x+4,000 **lb.; **V2 = 11,000 **lb.; **V3 = 5,000** lb**M1 =-500x ^{2}+4,000x **ft-lb.;

**M2 = 11,000x -152,000**ft-lb.;

**M3 = 5,000×-80,000**ft-lb

**Cantilever Beam – Example 3**

In this example we have a loaded, cantilever beam, as shown . For this beam we would like to determine expressions for the internal shear forces and bending moments in each section of the beam, and to draw the shear force and bending moment diagrams for the beam.

**Solution:Part A:** Our first step will be to determine the support reactions and external torque acting on the loaded beam. For a cantilevered beam (with one end embedded or rigidly fixed at the wall), the wall may exert horizontal and vertical forces and an external torque (which we will call an external moment, and label

**M**) acting on the beam – as we have shown in the free body diagram of the beam.

_{ext}**STEP 1**: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions.

**STEP 2**: Break any forces not already in x and y direction into their x and y components.**STEP 3**: Apply the static equilibrium conditions.**Sum F _{x }= A_{x} = 0**

**Sum F**_{y }= -4,000 lbs – 3,000 lbs – (2,000 lbs/ft)(6 ft) – 2,000 lbs + A_{y}= 0**Sum T**_{A}= (-4,000 lbs)(4 ft) – (3,000 lbs)(8 ft) -(2,000 lbs/ft)(6 ft)(11 ft) – (2,000 lbs)(14 ft) +M_{ext}= 0**Solving for the unknowns: A**_{y}= 21,000 lbs; M_{ext}= 200,000 ft-lbs**Part B**: Determine the Shear Force and Bending Moment expressions for each section of the loaded beam. For this process we will ‘cut’ the beam into sections, and then use static equilibrium condition – sum of forces to determine the shear force expressions, and Integration to determine the bending moment expressions in each section of the beam.

**Section 1**: Cut the beam at an arbitrary location x, where 0 < x < 4 ft. (since at 4 ft. the beam loading changes.), and analyze left hand beam section, from x to the left end of the beam.

**1.** Draw a FBD of the beam section showing and labeling all forces and toque acting – including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) (See Diagram – Section 1) Notice we have drawn the shear force and bending moment in their positive directions according to the defined sign convention for the shear force and bending moment discussed earlier, and have labeled them as V1 and M1, as this is section 1 of the beam.

**2.** Resolve all forces into x & y components (yes)

**3.** Apply translational equilibrium conditions (forces only) to determine the shear force expression:**Sum Fx = 0 (no net external x- forces)**** Sum Fy = 21,000 lbs – V1 = 0 ; **Solving for shear force:

**V1 = 21,000 lbs**

**4.**We now find the bending moment expression by summing torque about the left end. That equation would be as follows:

**Sum of Torque left end = -V1 * x + 200,000 ft-lb + M1 = 0**;

when we substitute in the value for

**V1 = 21,000 lb**. we obtain the equation:

**Sum of Torque left end = -21,000 lb * x + 200,000 ft-lb + M1 = 0**.

And solving for

**M1 = 21,000 x -200,000 ft-lb.**

[This is the same result we will find by integration. In general, particularly for non-point loads, integration is the faster method.]

Now we find the bending moment equation by integration of the shear force expression.

Integration, and then

**M1 = 21,000x + C1**

The boundary condition, which we will use to determine the integration constant C1, is different for the cantilevered end of a beam as compared with the end of a simply supported beam (or a free end). Since we have an external moment acting at the end of the cantilevered beam, **the value of the bending moment must become equal to the negative of the external moment at the cantilevered end of the beam**,** in order for the beam to be in rotational equilibrium** (sum of torque = 0). Thus our boundary condition to determine C1 is: **at x = 0, M = -200,000 ft-lb** (That is, for a cantilever beam, the value of the bending moment at the wall is equal to the negative of the external moment.)

Applying the boundary condition and solving for the integration constant:**-200,000 = 21000(0) + C1; and so C1 = -200,000**

Then the bending moment expression is: **M1 = [21,000x – 200,000] ft-lb** for 0 < x < 4 ft.**Section 2**: Since the loading changes at 8 ft, due to the point load and the beginning of the uniformly distributed load, for section 2 we cut the beam at location x, where 4 < x < 8 ft.; and analyze the left hand beam section.

**1**. Draw a FBD of the beam section showing and labeling all forces and toque acting – including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) We label them as V2 and M2, as this is section 2 of the beam.

**2**. Resolve all forces into x & y components (yes).**3**. Apply translational equilibrium conditions (forces only):**Sum Fx = 0** (no net external x- forces)**Sum Fy = 21,000 lbs – 4,000 lbs – V2 = 0 **; and solving: **V2 = 17,000 lb****4.** We now find the bending moment expression by summing torque about the left end. That equation would be as follows:**Sum of Torque left end = -4000 lb.* 4 ft. -V2 * x + 200,000 ft-lb + M2 = 0**;

when we substitute in the value for **V2 = 17,000 lb**. we obtain the equation:**Sum of Torque left end = -4000 lb.* 4 ft. -17,000 lb. * x + 200,000 ft-lb + M2 = 0**

And solving for **M2 = [ 17,000x – 184,000] ft-lb** for 4 < x < 8 ft**.**

Next we also determine the bending moment expression by integration of the shear force equation from above.

Integration: , and so **M2 = 17,000x + C2**

We obtain our boundary condition for beam section 2 by remembering that the bending moment must be continuous along the beam. This means that value of the bending moment at the end of section 1 (at x = 4 ft.) must also be the value of the bending moment at the beginning of section 2 (at x = 4 ft.). Thus our boundary condition to find the integration constant C2 is: at **x = 4 ft., M = -116,000 ft-lb**. (from equation M1).

Now applying the boundary condition and solving for the integration constant, C2, we have:**-116,000 ft-lbs = 17,000(4) + C2, **and so** C2 = -184,000 ft-lb**., and our expression for the bending moment on beam section 2 is: **M2 = [ 17,000x – 184,000] ft-lb** for 4 < x < 8 ft.

We continue in like manner for the last section of the beam

**Section 3**: Cut the beam at x, where 8 < x < 14 ft. and analyze left hand beam section, from x to the left end of the beam.

**1.** Draw a FBD of the beam section showing and labeling all forces and toque acting – including the shear force and bending moment (which act as an external force and torque at the point where we cut the beam.) We label them as V3 and M3, as this is section 3 of the beam.

**2.** Resolve all forces into x & y components (yes)**3.** Apply translational equilibrium conditions (forces only): **Sum Fx = 0** (no net external x- forces)**Sum Fy = 21,000 lbs – 4,000 lbs – 3,000 lbs – (2,000 lbs/ft)(x-8)ft – V3 = 0**

Solving for the bending moment: **V3 = [-2,000x + 30,000] lb**

**4.**We now find the bending moment expression by summing torque about the left end. That equation would be as follows:

**Sum of Torque left end = -4000 lb.* 4 ft. -3000 lb.* 8 ft. – 2000 lb/ft *(x-8′)*[(x-8)/2 + 8′] -V3 * x + 200,000 ft-lb + M3 = 0**; [ If you are unsure how the third term in the equation was obtained, please see example 2.}

When we substitute in the value for

**V3 =**

**[-2,000x + 30,000] lb**. we obtain the equation:

**Sum of Torque= -4000 lb.* 4 ft. -3000 lb.* 8 ft. – 2000 lb/ft *(x-8′)*[(x-8)/2 + 8′] –**

**[-2,000x + 30,000] lb**

*** x + 200,000 ft-lb + M2 = 0**And solving for

**M3 = [-1,000x**

^{2}+30,000x – 224,000] ft-lb.By Integration: , and then

**M3 =-1,000x**

^{2}+ 30,000x + C3Since the right hand end of the beam is a free end, the bending moment must go to zero as x goes approaches 14 ft. So our boundary condition to find the integration constant, C3, is:

**at x = 14 ft M = 0 ft-lb**

Applying the boundary condition:

**0 = -1,000(14)**, and solving

^{2}+ 30,000(14) + C3**C3 = -224,000 ft-lbs**

So

**M3 = [-1,000x**for 8 < x < 14 ft.

^{2}+30,000x – 224,000] ft-lbs**Part C:** Shear Force and Bending Moment Diagrams: Now using the expressions found in Part B above, we can draw the shear force and bending moment diagrams for our loaded beam.**V1 = 21,000 lb., V2 = 17,000 lb., V3 = [-2,000x+30,000] lb.****M1 =[21,000×-200,000] ft-lb., M2 = [17,000×-184,000] ft-lb., M3 = [-1,000x ^{2}+30,000x – 224,000] ft-lb. **