Shear and moment diagram

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.

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]

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, 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.

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.

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:

${\displaystyle \sum F=-10-V_{1}=0}$

and

${\displaystyle \sum M_{A}=-V_{1}x+M_{1}=0\,.}$

Therefore,

${\displaystyle V_{1}=-10\quad {\text{and}}\quad M_{1}=-10x\,.}$

Free-body diagram of segment 2

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,

${\displaystyle V_{2}=R_{a}-x\quad {\text{and}}\quad M_{2}=-50+R_{a}(x-10)-{\frac {x^{2}}{2}}\,.}$

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,

${\displaystyle M_{2}=-50+R_{a}(x-10)-{\frac {x^{2}}{2}}\,.}$

Free-body diagram of segment 3

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.

Free-body diagram of segment 4

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

${\displaystyle M_{4}=M_{c}=-937.5+40R_{a}+25R_{b}\,.}$

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

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

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

${\displaystyle {\begin{aligned}w_{1}&={\frac {5}{3EI}}\,x^{3}+C_{1}+C_{2}\,x\\w_{2}&={\frac {1}{24EI}}\,x^{2}\,\left[x^{2}+600-4R_{a}(x-30)\right]+C_{3}+C_{4}\,x\\w_{3}&={\frac {1}{100EI}}\left[{\frac {x^{3}}{3}}(-625+30R_{a}-2M_{c})-50x^{2}(-675+30R_{a}-M_{c})\right]+C_{5}+C_{6}\,x\\w_{4}&={\frac {1}{100EI}}\left[{\frac {x^{3}}{3}}(-625+30R_{a}-2M_{c})-50x^{2}(-625+30R_{a}-M_{c})\right]+C_{7}+C_{8}\,x\end{aligned}}}$

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

${\displaystyle C_{7}=-{\frac {1250}{3EI}}(-625+M_{c}+30R_{a})\quad {\text{and}}\quad C_{8}={\frac {125}{EI}}(-125+6R_{a})\,.}$

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

${\displaystyle C_{5}=-{\frac {625}{12EI}}(-5675+8M_{c}+240R_{a})\quad {\text{and}}\quad C_{6}={\frac {250}{EI}}\left(3R_{a}-70\right)\,.}$

Substitution of these constants into the expression for w₃ gives us

${\displaystyle {\begin{aligned}w_{3}={\frac {1}{300EI}}{\Bigl [}&30R_{a}(-50+x)^{3}-2M_{c}(-50+x)^{2}(25+x)-\\&625(-141875+x(8400+(-162+x)x)){\Bigr ]}\,.\end{aligned}}}$

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

${\displaystyle C_{3}=-{\frac {3125}{24EI}}(-1645+4M_{c}+64R_{a})\quad {\text{and}}\quad C_{4}={\frac {25}{12EI}}\left(-40325+6M_{c}+120R_{a}\right)\,.}$

Therefore,

${\displaystyle {\begin{aligned}w_{2}={\frac {1}{24EI}}{\Bigl [}&-3125(-1645+4M_{c}+64R_{a})+\\&50(-4025+6M_{c}+120R_{a})x+120(5+R_{a})x^{2}-4R_{a}x^{3}+x^{4}{\Bigr ]}\,.\end{aligned}}}$

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

${\displaystyle C_{1}=-{\frac {125}{24EI}}(-40145+100M_{c}+1632R_{a})\quad {\text{and}}\quad C_{2}={\frac {25}{4EI}}(-1315+2M_{c}+48R_{a})\,.}$

Therefore,

${\displaystyle w_{1}={\frac {5}{24EI}}\left[1026125-39450x+8x^{3}+20M_{c}(-125+3x)+480R_{a}(-85+3x)\right]\,.}$

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

${\displaystyle R_{a}=25.278\quad \implies \quad M_{c}=-14.585\,.}$

Free-body diagram

Shear force diagram

Bending moment diagram

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.

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.