Chapter 8 Tension, shear and bending moment

Engineers often want to know how the internal forces vary from point to point in a structure. How do you find these internal forces? You could draw a variety of free-body diagrams with cuts at the points of interest. Another approach, which we present now, is to leave the position, say $x$, of the free-body-diagram cut a variable, and then calculate the internal forces in terms of $x$.

Recall that in the first example of sec. 5.4 we found $T$ without ever using information about the location of the free-body diagram cut. So the location does not affect the tension. For a two-force body the tension is a constant along the length.

The uniform $1{\mathrm{cm}}^2$ steel square rod with density $\rho =7.7\text{gm}/{\mathrm{cm}}^3$ and length $\mathrm{\ell }=100\mathrm{m}$ has total weight $W=mg=\rho \mathrm{\ell }Ag$ (see fig. 8.1). What is the tension a distance $x_{\mathrm{D}}$ from the top? Using the free-body diagram with cut at $x_{\mathrm{D}}$ we get:

So, at the bottom end at $x_{\mathrm{D}}=100\mathrm{m}$ we get $T=0$ and at the top end where $x_{\mathrm{D}}=0\mathrm{m}$ we get $T=750\text{N}$ and in the middle at $x_{\mathrm{D}}=50\mathrm{m}$ we get $T=375\text{N}$.

Because the free-body diagram cut location is variable, we can plot the internal forces as a function of position. This is most useful in civil engineering where an engineer wants to know the internal forces in a horizontal beam carrying vertical loads. Common examples include bridge platforms and floor joists.

A cantilever beam is mounted firmly at one end and has various loads orthogonal to its length, in this case a downwards load $F$ at the end (fig. 8.2a). By drawing a free-body diagram with a cut at the arbitrary point C (fig. 8.2b) we can find the internal forces as functions of the position of C.

That the axial tension is zero in these problems is so well known that the tension is often not drawn on the free-body diagram, and isn’t calculated. We can now plot $V(x)$ and $M(x)$ as in figs. 8.2c and 8.2d. In this case the shear force is a constant and the bending moment varies from its maximum magnitude at the wall ($M=-F\mathrm{\ell }$) to 0 at the end. It is the big value of $|M|$ at the fixed support, the maximum over the length, that makes cantilever beams typically break at the ends.

Often one is interested in distributed loads from gravity on the structure itself or from a distribution (say of people on a floor). The method is the same.

A cantilever beam has a downwards uniformly distributed load of $w$ per unit length, e.g., the weight of the beam, (fig. 8.3a). Using the free-body diagram shown (fig. 8.3b) we can find:

The integrals were used because of their general applicability for distributed loads. For this problem we could have avoided the integrals by using an equivalent downwards force $w\cdot (\mathrm{\ell }-x)$ applied a distance $(\mathrm{\ell }-x)/2$ to the right of the cut. Shear and bending moment diagrams are shown in figs. 8.3a and 8.3b.

As for all problems based on the equilibrium equations and a given geometry, the principle of superposition applies.

Consider a cantilever beam that simultaneously has both of the loads from the previous two examples. By the principle of superposition:

The shear force at every point is the sum of the shear forces from the previous examples. The bending moment at every point is the sum of the bending moments.

If there are concentrated loads in the middle of the region of interest the calculation gets more elaborate; the concentrated force may or may not show up on the free-body diagram of the cut bar, depending on the location of the cut.

A simply supported beam is mounted with pivots at both ends (fig. 8.4a). First we draw a free-body diagram of the whole beam (fig. 8.4a) and then two more, one with a cut to the left of the applied force and one with a cut to the right of the applied force (figs. 8.4c and 8.4d). With the free-body diagram 8.4c we can find $V(x)$ and $M(x)$ for and with the free-body diagram 8.4d we can find $V(x)$ and $M(x)$ for $x>\mathrm{\ell }/2$.

These relations can be plotted as in figs. 8.4e and 8.4f. Some observations: For this beam the biggest bending moment is in the middle, the place where simply supported beams often break. Instead of the free-body diagrams shown in (c) and (d) we could have drawn a free-body diagram of the bar to the right of the cut and would have gotten the same $V(x)$ and $M(x)$. We avoided drawing a free-body diagram cut at the applied load where $V(x)$ has a discontinuity.

Here are some guidelines for finding internal forces and drawing shear and bending moment diagrams.

• •

  Draw a free-body diagram of the whole bar.

• •

  Using the free-body diagram above, find the reaction forces.

• •

  Draw a free-body diagram(s) of the cut bar.

  • –

    For each region between concentrated loads, draw one free-body diagram.

  • –

    Show the piece from the cut to one or the other end (So that all but the internal forces are known).

  • –

    Don’t make cuts at intermediate points of connection or load application.

• •

  Use the equilibrium equations to find $T$, $V$, or $M$ (Moment balance about a point at the cut is a good way to find $M$.)

• •

  Use the results above to plot $V(x)$ and $M(x).$ $T(x)$ is rarely plotted.

  • –

    Use the same $x$ scale for this plot as for the free-body diagram of the whole bar.

  • –

    Put the plots directly under the free-body diagram of the bar (so you can most easily relate features of the loads to features of the $V$ and $M$ diagrams).

For a given load, if you replace one bar in tension with two bars side-by-side, you would imagine the tension in each bar would go down by a factor of 2. Thus the pair of bars should be twice as strong as a single bar. If you glued these side-by-side bars together you would again have one bar, but it would be twice as strong as the original bar. Why? Because it has twice the cross sectional area.

What makes a solid break is the force per unit area carried by the material. For an applied tension load $T$, the force per unit area on an interior free-body diagram cut is $T/A$. The force per unit area, that is normal to an internal free-body diagram cut, is called tension stress and denoted $\sigma$ (lower case ‘sigma’, the Greek letter s).

Look at the hanging bar in the example on page 8.1. We can find the tension stress in this bar as a function of position along the bar as:

Note that the stress for this bar doesn’t depend on the cross sectional area. The bigger the area the bigger the volume and hence the load. But also, the bigger the area on which to carry it.

For reasons that are beyond this book, the tension stress tends to be uniform in homogeneous (all one material) bars, no matter what their cross sectional shape, so that the average tension stress $\frac{T}{A}$ is actually the tension stress all across the cross section.

We can similarly define the average shear stress ${\tau }_{\mathrm{ave}}$ (‘tau’) on a free body diagram cut as the average force per unit area tangent to the cut,

For reasons you may learn in a strength of materials class, shear stress is not so uniformly distributed across the cross section. But the average shear stress ${\tau }_{\mathrm{ave}}$ does give an indication of the actual shear stress in the bar (e.g., for a rectangular elastic bar, the peak shear stress is 50% larger than ${\tau }_{\mathrm{ave}}$).

The biggest stresses typically come from the bending moment. Motivating formulas for these stresses here is too big a digression. The formulas for the stresses due to bending moment are a key part of elementary strength of materials. But just knowing that these stresses tend to be larger, gives you the important notion that bending moment is a common cause of structural failure.

‘Internal forces’ are the scalars that describe the force and moment on potential internal free-body diagram cuts. They are found by applying the equilibrium equations to free-body diagrams that have cuts at the points of interest. The internal forces are intimately associated with the internal stresses (force per unit area) and thus are important for determining the strength of structures.

Support reactions on a simply supported beam: A uniform beam of length 3 $\mathrm{m}$ is simply supported at A and B as shown in the figure. A uniformly distributed vertical load $q=100\mathrm{N}/\mathrm{m}$ acts over the entire length of the beam. In addition, a concentrated load $P=150\text{N}$ acts at a distance $d=1\mathrm{m}$ from the left end. Find the support reactions.

The force equilibrium, ${\displaystyle \sum \stackrel{\rightharpoonup }{𝑭}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, gives

Support reactions on a cantilever beam: A 2  kN horizontal force acts at the tip of an ’L’-shaped cantilever beam as shown in the figure. Find the support reactions at A.

The force equilibrium, ${\displaystyle \sum \stackrel{\rightharpoonup }{𝑭}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, gives

Net force of a uniformly distributed system: A uniformly distributed vertical load of intensity 100  N/macts on a cantilever beam of length $\mathrm{\ell }=2\mathrm{m}$ as shown in the figure.

1. 1.

   Find the net force acting on the beam.

2. 2.

   Find an equivalent force-couple system at the mid-point of the beam.

3. 3.

   Find an equivalent force-couple system at the right end of the beam.

1. 1.

   The net force:

   Filename:sfig8-3-1a

   Because the load is uniformly distributed along the length, we can find the total (or net) load by calculating the load on an infinitesimal segment of length $dx$ of the beam and then integrating over the entire length of the beam. Let the load intensity (load per unit length) be $q$ ($q=100\mathrm{N}/\mathrm{m}$, as given). Then the vertical load on segment $dx$ is (see Fig. 8.11),

   	$d\stackrel{\rightharpoonup }{𝑭}$	$=$	$qdx(-\hat{\mathit{ȷ}}).$
   	Therefore, the net force is,
   	${\stackrel{\rightharpoonup }{𝑭}}_{\mathrm{net}}$	$=$	${\displaystyle {\int }_0^{\mathrm{\ell }}}q𝑑x(-\hat{\mathit{ȷ}})=q\mathrm{\ell }\hat{\mathit{ȷ}}=-100\mathrm{N}/\mathrm{m}\cdot 2\mathrm{m}\hat{\mathit{ȷ}}=-200\text{N}\hat{\mathit{ȷ}}.$

   Answer: ${\stackrel{\rightharpoonup }{F}}_{\mathrm{net}}=-200\text{N}\hat{ȷ}$

2. 2.

   The equivalent system at the mid-point:

   Filename:sfig8-5-wiper

   We have already calculated the net force that can replace the uniformly distributed load. Now we need to calculate the couple at the mid-point of the beam to get the equivalent force-couple system. Again, consider a small segment of the beam of length $dx$ located at distance $x$ from the mid-point C (see Fig. 8.12). The moment about point C due to the load on $dx$ is $(qdx)x(-\widehat{𝒌})$. But, we can find a similar segment on the other side of C with exactly the same length $dx$, at exactly the same distance $x$, that produces a moment of $(qdx)x(+\widehat{𝒌})$. The two contributions cancel each other and we have a net zero moment about C. Now, you can imagine the whole beam made up of these pairs that contribute equal and opposite moment about C and thus the net moment about the mid-point is zero.

   You can also find the same result by straight integration:

   $${\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{C}}={\int }_{-\mathrm{\ell }/2}^{+\mathrm{\ell }/2}qx𝑑x(-\widehat{𝒌})={\frac{q{x}^2}{2}|}_{-\mathrm{\ell }/2}^{+\mathrm{\ell }/2}(-\widehat{𝒌})=\stackrel{\rightharpoonup }{\mathrm{𝟎}}.$$

   Answer: ${\stackrel{\rightharpoonup }{F}}_{\mathrm{net}}=-200\text{N}\hat{ȷ}$, and ${\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{C}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$

3. 3.

   The equivalent system at the end: The net force remains the same as above. We compute the net moment about the end point B, referring to Fig. 8.13, as follows.

   Filename:sfig8-5-wiper-a

   	${\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{B}}$	$=$	${\displaystyle {\int }_0^{\mathrm{\ell }}}(-x\hat{\mathit{ı}})\times (-qdx\hat{\mathit{ȷ}})=-q{\displaystyle {\int }_0^{\mathrm{\ell }}}x𝑑x\widehat{𝒌}$
   		$=$	$-{\displaystyle \frac{q{\mathrm{\ell }}^2}{2}}\widehat{𝒌}=-{\displaystyle \frac{100\mathrm{N}/\mathrm{m}\cdot 4{\mathrm{m}}^2}{2}}\widehat{𝒌}=-200\mathrm{N}\mathrm{\cdot }\mathrm{m}\widehat{𝒌}.$

   Answer: ${\stackrel{\rightharpoonup }{F}}_{\mathrm{net}}=-200\text{N}\hat{ȷ}$ and ${\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{B}}=-200\mathrm{N}\mathrm{\cdot }\mathrm{m}\widehat{𝒌}$

For the uniformly loaded, simply supported beam shown in the figure, find the shear force and the bending moment at the mid-section c-c of the beam.

To determine the shear force $V$ and the bending moment $M$ at the mid-section c-c, we cut the beam at c-c and draw its free-body diagram as shown in fig. 8.15. For writing force- and moment-balance equations we use the second figure where we have replaced the distributed load with an equivalent single load $F=(q\mathrm{\ell })/2$ acting vertically downward at distance $\mathrm{\ell }/4$ from end A.

The force balance, ${\displaystyle \sum \stackrel{\rightharpoonup }{𝑭}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, implies that

Dotting with ˆ ı and ˆ ȷ , respectively, we get

From the moment equilibrium about point A, ${\displaystyle \sum {\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{A}}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, we get

Thus, to find $V$ and $M$ we need to know the support reaction $\stackrel{\rightharpoonup }{𝑨}$. From the free-body diagram of the beam in fig. 8.16 and the moment equilibrium equation about point B, ${\displaystyle \sum {\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{B}}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, we get

Thus $\stackrel{\rightharpoonup }{𝑨}=500\text{N}\hat{\mathit{ȷ}}$. Substituting $\stackrel{\rightharpoonup }{𝑨}$ in eqns. (8.2) and (8.3), we get

The cantilever beam AD is loaded as shown in the figure where $W=200\mathrm{lbf}$. Find the shear force and bending moment on a section just left of point B and another section just right of point B.

The uniformly distributed load $2W$ of the block sitting on the beam can be replaced by an equivalent concentrated load $2W$ acting at point E, at distance $a/2$ from the end D of the beam.

Let us denote the shear force by $V^{+}$ and the bending moment by $M^{+}$ at the section of our interest. Now, from the force equilibrium of the part-beam BD we get

The moment equilibrium about point B, ${\displaystyle \sum {\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{B}}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, gives

Now, we determine the internal forces at a section just to the left of point B. Let the shear and bending moment at this section be $V^{-}$ and $M^{-}$, respectively, as shown in the free-body diagram (fig. 8.19). Note that load $W$ acting at B is now included in the free-body diagram since the beam is now cut just a teeny bit left of this load.

From the force equilibrium of the part-beam, we have

Note that the bending moment remains the same on either side of point B but the shear force jumps by $V^{+}-V^{-}=200\mathrm{lbf}=W$ as we go from right to the left. This jump is expected because a concentrated load $W$ acts at B, in between the two sections we consider. Concentrated external forces cause a jump in shear, and concentrated external moments cause a jump in the bending moment.

Tension in a bar: A T-shaped bar is fixed in a wall at one end and is acted on by three forces as shown in the figure. Find the tension in the rod at

1. 1.

   section $a$-$a$, and

2. 2.

   section $b$-$b$.

1. 1.

   Filename:pfigure-s95f3a

   Let us cut the bar at section $a$-$a$ and consider the part of the bar to the right of the cut-section. The free-body diagram of this part of the bar is shown in fig. 8.21. The scalar force balance in the horizontal direction gives

   	$-T-F+2F$	$=$	$0$
   	$\Rightarrow T$	$=$	$F$
   		$=$	$2\mathrm{kN}.$

   Answer: $\text{ At section }a\text{-}a\text{:}T=2\mathrm{kN}$

   
   

2. 2.

   Now, we cut the bar at section $b$-$b$ and again consider the section of the bar to the right of the cut-section. The free-body diagram of this part of the bar is shown in fig. 8.22. Again, the force balance in the horizontal direction gives

   	$-T+2F$	$=$	$0$
   	$\Rightarrow T$	$=$	$2F$
   		$=$	$4\mathrm{kN}.$

   Answer: $\text{ At section }b\text{-}b\text{:}T=4\mathrm{kN}$

   Filename:Danef94s1q2

Tension in a tapered bar due to self weight: A tapered bar of height $1\mathrm{m}$, top width $10\mathrm{cm}$, bottom width $4\mathrm{cm}$ and uniform thickness $4\mathrm{cm}$ hangs upside down from a ceiling. If the density of the material is $7500\mathrm{kg}/{\mathrm{m}}^3$, find the tension in the rod halfway from the top. Use $g\approx 10\mathrm{m}/{\mathrm{s}}^2$.

The width of the bar at the cut section is $c=(a+b)/2$ where $a=4\mathrm{cm}$ and $b=10\mathrm{cm}$. The frontal area of the bar-part is $A=(a+c)/2\cdot (h/2)$ where $h=1\mathrm{m}$. Thus,

Thus, $T=82.5\text{N}$.

A simple frame: A 2$\mathrm{m}$ high and 1.5$\mathrm{m}$ wide rectangular frame ABCD is loaded with a 1.5  kN horizontal force at B and a 2  kN vertical force at C. Find the internal forces and moments at the mid-section e-e of the vertical leg AB.

The moment balance about point A, ${\displaystyle \sum {\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{A}}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, gives

From force equilibrium, ${\displaystyle \sum \stackrel{\rightharpoonup }{𝑭}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, we have

Now we draw the free-body diagram of AE to find the shear force $V$, axial (tensile) force $T$, and the bending moment $M$ at section e-e.

From the force equilibrium of part AE, we get

From the moment equilibrium about point A, ${\displaystyle \sum {\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{A}}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, we have

Shear force and bending moment diagrams: A simply supported beam of length $\mathrm{\ell }=2\mathrm{m}$ carries a concentrated vertical load $F=100\text{N}$ at a distance $a$ from its left end. Find and plot the shear force and the bending moment along the length of the beam for $a=\mathrm{\ell }/4$.

Now, from the moment equilibrium about point B, ${\displaystyle \sum {\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{B}}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, we get

and from the force equilibrium in the vertical direction, $({\displaystyle \sum \stackrel{\rightharpoonup }{𝑭}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}})\cdot \hat{\mathit{ȷ}}$, we get

Now we make a cut at an arbitrary (variable) distance $x$ from A where (see fig. 8.30). Carrying out the force balance and the moment balance about point A, we get, for ,

Thus $V$ is constant for all but $M$ varies linearly with $x$.

Now we make a cut at an arbitrary $x$ to the right of load $F$, i.e., . Again, from the force balance in the vertical direction, we get

and from the moment balance about point A,

Although eqn. (8.5) is strictly valid for and eqn. (8.7) is strictly valid for $x>a$, substituting $x=a$ in these two equations gives the same value for $M(=Fa(1-a/\mathrm{\ell }))$ as it must because there is no reason to have a jump in the bending moment at any point along the length of the beam. The shear force $V$, however, does jump because of the concentrated load $F$ at $x=a$.

Now, we plug in $a=\mathrm{\ell }/4=0.5\mathrm{m}$, and $F=100\text{N}$, in eqns. (8.4)–(8.7) and plot $V$ and $M$ along the length of the beam by varying $x$. The plots of $V(x)$ and $M(x)$ are shown in fig. 8.31.

Shear force and bending moment diagrams by superposition: For the cantilever beam and the loading shown in the figure, draw the shear force and the bending moment diagrams by

1. 1.

   considering all the loads together, and

2. 2.

   considering each load (of one type) at a time and using superposition.

1. 1.

   $V(x)$ and $M(x)$ with all forces considered together: The horizontal forces acting at the end of the cantilever are equal and opposite and, therefore, produce a couple. So, we first replace these forces by an equivalent couple $M_{\mathrm{applied}}=100\text{N}\cdot 1\mathrm{m}=100\mathrm{N}\mathrm{\cdot }\mathrm{m}$. Since we have a cantilever beam, we can consider the right hand side of the beam after making a cut anywhere for finding $V$ and $M$ without first finding the support reactions.

   Filename:sfig8-7-2

   Let us cut the beam at an arbitrary distance $x$ from the right hand side. The free-body diagram of the right segment of the beam is shown in fig. 8.33. From the force balance, ${\displaystyle \sum \stackrel{\rightharpoonup }{𝑭}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, we find that

   	$-V\hat{\mathit{ȷ}}+qx\hat{\mathit{ȷ}}$	$=$	$\stackrel{\rightharpoonup }{\mathrm{𝟎}}$
   	$\Rightarrow V$	$=$	$qx$
   		$=$	$(50\mathrm{N}/\mathrm{m})x.$

   Thus the shear force varies linearly along the length of the beam with

   	$V(x=0)$	$=$	$0,$
   	$\text{and}V(x=3\mathrm{m})$	$=$	$150\text{N}.$

   The moment balance about point C, ${\displaystyle \sum {\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{C}}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, gives

   $$-M\widehat{𝒌}-qx\cdot \frac{x}{2}\widehat{𝒌}+M_{\mathrm{applied}}\widehat{𝒌}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$$

   where the moment due to the distributed load is most easily computed by considering an equivalent concentrated load $qx$ acting at $x/2$ from the end B. Thus,

   	$\Rightarrow M$	$=$	$M_{\mathrm{applied}}-q{\displaystyle \frac{x^2}{2}}$
   		$=$	$100\mathrm{N}\mathrm{\cdot }\mathrm{m}-50\mathrm{N}/\mathrm{m}\cdot {\displaystyle \frac{x^2}{2}}.$

   Filename:sfig8-7-2a

   Thus, the bending moment varies quadratically with $x$ along the length of the beam. In particular, the values at the ends are

   	$M(x=0)$	$=$	$100\mathrm{N}\mathrm{\cdot }\mathrm{m}$
   	$\text{and}M(x=3\mathrm{m})$	$=$	$-125\mathrm{N}\mathrm{\cdot }\mathrm{m}.$

   The shear force and the bending moment diagrams obtained from eqns. (1) and (1) are shown in fig. 8.34. Note that $M=0$ at $x=2\mathrm{m}$ as given by eqn. (1).

   Filename:sfig8-7-2again

2. 2.

   $V(x)$ and $M(x)$ by superposition: Now we consider the cantilever beam with only one type of load at a time. That is, we first consider the beam only with the uniformly distributed load and then only with the end couple. We draw the shear force and the bending moment diagrams for each case separately and then just add them up. That is superposition.

   So, first let us consider the beam with the uniformly distributed load. The free-body diagram of a segment CB, obtained by cutting the beam at a distance $x$ from the end B, is shown in fig. 8.35. Once again, from force balance, we get

   Filename:sfig8-7-2disks

   $$V=qx\text{for }0\le x\le \mathrm{\ell }$$

   (8.11)

   and from the moment balance about point C, ${\displaystyle \sum {\stackrel{\rightharpoonup }{𝑴}}_{\mathrm{C}}}=\stackrel{\rightharpoonup }{\mathrm{𝟎}}$, we get

   $$M=-qx\cdot \frac{x}{2}=-q\frac{x^2}{2}\text{for }0\le x\le \mathrm{\ell }.$$

   (8.12)

   Figure 8.36 shows the plots of $V$ and $M$ obtained from eqns. (8.11) and (8.12), respectively, with the values computed from $x=0$ to $x=3\mathrm{m}$ with $q=50\mathrm{N}/\mathrm{m}$ as given.

   Filename:sfig8-4-4

   Now we take the beam with only the end couple and repeat our analysis. A cut section of the beam is shown in fig. 8.37. In this case, it should be obvious that from force balance and moment balance about any point, we get

   	$V$	$=$	$0$
   	$\text{and}M$	$=$	$M_{\mathrm{applied}}.$

   Thus, both the shear force and the bending moment are constant along the length of the beam as shown in fig. 8.37.

   Now superimposing (adding) the shear force diagrams from Figs. 8.36 and 8.37, and similarly, the bending moment diagrams from Figs. 8.36 and 8.37, we get the same diagrams as in fig. 8.38.

   Filename:sfig8-4-4a

In general if you know the downwards load per unit length $w(x)$ on a beam you can integrate 4 times to find

These apply for any distributed loading $w(x)$.

But what kind of function $w(x)$ do you use for a concentrated load $P$; or a reaction force $F$; or for a load that is constant in some regions and zero in others; or when there is an applied couple? What are the functions $w(x)$ for such loads and how do you integrate them?

The answers are: singularity functions.

Without singularity functions you have to find separate expressions for $V,M,$etc. for the regions to the left and to the right of such discontinuities. And then, as described in most books about beams, you have to pick integration constants in the two regions so that there is an appropriate jump, or not, of the $V$, $M$, $u^{\prime }$ or $u$. Using singularity functions you can skip all this matching. You just follow the integration rules and the solution is properly expressed on both sides of the discontinuity. That is, singularity functions allow you to do ordinary calculus even with functions that are discontinuous. Sometimes this is called operational calculus.

What function is it?

In the classical mathematics sense the delta function $\delta (x-a)$ isn’t a function. Mathematicians were upset about this for a while. There are lots of ways to think about this function (See fig. 8.56). One way to think of $\delta (x-a)$ is as a function of $x$ that is zero except for very close to $x=a$. Near $x=a$ it is as tall as it is narrow (i.e., it’s very tall and very narrow). So the area underneath is one. Actually $\delta (x-a)$ is the limit of infinitely tall and infinitely narrow, but the area underneath is still equal to one.

But if the concentrated load is very high in value and very narrow in spatial extent, it’s just like the delta function. Thus for a concentrated load $P$ at $a$ the associated distributed load $w$ is

1. 1.

   The step function is the integral of the delta function, informally:

   $$\int \delta (x-a)𝑑x=H(x-a)$$

   or more formally

   $${\int }_{-\mathrm{\infty }}^x\delta (x^{\prime }-a)𝑑x^{\prime }=H(x-a).$$

   That is, the Heaviside step function is the cumulative area under the delta function curve.

2. 2.

   The Heaviside step function is that function of $x$ that is zero to the left of $a$ and one to the right of $a$,

This is pictured in the first of fig. 8.57.

So, integrating,

If there is a concentrated load $P$ at $a$ then the shear force function $V(x)$ has a step down of size $P$ at $a$.

The brackets mean that you should think of the whole expression as being zero if . For $x>a$ the brackets are like ordinary parentheses (at least for $n\ge 0$).

It’s easy to see that