created at April 12, 2021

Lesson 6 - Solving problems of statics

Internal and external loads

Regarding to the specific mechanical system, all loads can be divided into two groups: internal and external. As mentioned in lesson 3, solving of mechanical problem starts from mentally isolating mechanical system from the outer world, and replacing impacts from the surrounding bodies on our system with forces and torques. By "mechanical system" are meant a single body or a group of bodies, which are somehow interacting with the outer world.

Those loads, which results from interaction with external bodies (which are not a part of our system) are external loads. In opposite, loads, which occurs from interaction between the target system parts are internal loads. To make it clear let's take a look on the train:

fig. 1.6.1 External and internal loads

The third Newton's law says: "The force of action is equal to the force of reaction". If first body impacts the second body with force F, then the second body also impacts the first one with force F', which is of the same magnitude, but opposite directed. The Locomotive is pulling the car forward with force F, so the car is pulling the locomotive backward with the same but opposite force F', making it harder for locomotive to accelerate. Those two forces F and F' are obviously compensating each other.

Now if we solve a case for the locomotive only, we mentally discard the car and replace car's impact with force F'. Thus, relative to the locomotive force F' will be external load. On the other hand, if the target mechanical system is "Locomotive + car", then forces F and F' will be considered as internal loads.

The point is: internal loads can be omitted since they compensate each other. When solving the case we should be concerned on external loads only - they contribute equilibrium equations.

Solving Problems

Now we can proceed with solving a problems of statics. The subject of statics is equilibrium state of mechanical system. The goal is to find such a combination of forces and torques, which makes the system to stay in equilibrium state. Usually, external loads are given and reactions should be found. However, the opposite task can also be considered.

The equilibrium equations are used in solving problems of statics. In general they looks as follows:

kRk=0(1)\sum_{k}^{}R_{k}=0\quad (1)
kMk=0(2)\sum_{k}^{}M_{k}=0\quad (2)

In three-dimensional space both these equations can be divided into projections for three axis, thus giving:

kRxk=0;kRyk=0;kRzk=0;\sum_{k}^{}R_{xk}=0;\quad \sum_{k}^{}R_{yk}=0;\quad \sum_{k}^{}R_{zk}=0;
kMxk=0;kMyk=0;kMzk=0;\sum_{k}^{}M_{xk}=0;\quad \sum_{k}^{}M_{yk}=0;\quad \sum_{k}^{}M_{zk}=0;

So, we have six equations, which should be solved as a system.

Many practical cases can be considered as in-plane problems (2D problems). This is the case when all forces and torques are located in one common plane. This is a particular case of the three-dimensional problem and it lets us simplify solution. The third axis (orthogonal to the common plane) can be omitted and we obtain only two equations for the forces, and one equation for torques (total: 3 equations).

Let's start with solving 2D problems as they are simpler.

Example 1

Consider equilibrium state of the bridge with the man standing on it. The bridge is attached to the pillar with cylindrical joint on the left side, so the bridge can rotate in view plane. On the right side bridge lies on the sloped surface. The gravity forces of both the bridge and the man are vertical (green vectors on fig. 1.6.2). The support reactions are thus located in vertical plane as well. So, we have 2D problem:

fig. 1.6.2 Equilibrium of the bridge

Two active forces are impacting the bridge: both bridge and man's gravity forces. They are considered as given values. Also reaction forces of both supports are impacting the bridge. They are unknown and should be found.

Before we write down equations, let's choose the direction of coordinate system axis. We can choose it as we want. In our case it's convenient to direct x axis along the bridge, and y axis will be vertical.

On the left side we have unknown support's reaction NB. Both direction and magnitude are unknown. Let's project NB on the axis, thus obtaining two unknown reaction forces XB and YB. Dividing one force into projections is handy while solving equations.

On the right side we have support's reaction NA. However, since the support's surface is considered to be flat (no friction), we know the direction of NA - it's orthogonal to the surface. Thus, projections XA and YA are related to NA in following way:

XA=NAsin(α);YA=NAcos(α);X_{A}=N_{A}\cdot sin(\alpha );\quad Y_{A}=N_{A}\cdot cos(\alpha );

i.e. they are expressed through one unknown value of NA.

Let's write equation for x axis. In order to do that we put together all force components, which are acting in x axis direction, and make them all equal to zero (since we have equilibrium state and resulting force must be zero).

XBNAsin(α)=0X_{B}-N_{A}\cdot sin(\alpha)=0

In the same manner we compose equation for y axis:

YBFG+NAcos(α)=0Y_{B}-F-G+N_{A}\cdot cos(\alpha)=0

As you can see, force contributes equation with "plus" sign when force direction coincides with axis direction. Sign is "minus" otherwise.

Two equations for translation are ready, one equation for torques (rotation) still remaining. Before we compose it, some target point should be selected. We can choose it in either way, however, best practice is to do it so that the number of unknown values in equation will be reduced. We will choose point B, since both forces XB and YB go through this point and, thus, the torque contribution will be zero (zero arm). Now we gather all the forces and multiply them by according distance between force impact line and target point:

NAcos(α)LGL2Fl=0N_{A}\cdot cos(\alpha)\cdot L - G\cdot \frac{L}{2}-F\cdot l=0

Take attention to the signs of the equation terms. Usually torque is considered to be positive, if the force tries to rotate the body counter-clockwise.

And finally we obtained the following system:

{X:XBNAsin(α)=0Y:YBFG+NAcos(α)=0MB:NAcos(α)LGL2Fl=0\begin{cases} X:\quad X_{B}-N_{A}\cdot sin(\alpha)=0 \\Y:\quad Y_{B}-F-G+N_{A}\cdot cos(\alpha)=0 \\ M_{B}:\quad N_{A}\cdot cos(\alpha)\cdot L - G\cdot \frac{L}{2} - F\cdot l = 0 \end{cases}

Now we can find NA from the third equation:

NA=GL2+Flcos(α)LN_{A}=\frac{G\cdot \frac{L}{2} + F\cdot l}{cos(\alpha)\cdot L}

Then knowing NA let's find out XB from the first equation:

XB=NAsin(α)X_{B} = N_{A}\cdot sin(\alpha)

And finally we obtain YB from the second equation:

YB=F+GNAcos(α)Y_{B} = F + G - N_{A}\cdot cos(\alpha)

So, support reactions are found and the problem is solved.

Example 2

Consider equilibrium state of the pipe, clamped into the wall. The bucket filled with water is hanging on the end. Also there is a valve, which can rotate in vertical plane. Consider someone is currently rotating the valve thus applying some torque to the pipe.

fig. 1.6.3 Equilibrium of the pipe

Since the problem is planar, clamp reaction (on the left side) has three unknown components: two forces XA and YA, and torque MA. You wonder why do we have MA? Well, clamp is not a joint, it disables pipe's rotation relative to point A. The gravity of the bucket produces a torque, which tries to rotate pipe relative to A. However, the pipe remains static because of reaction MA, which should be found.

Let's direct x axis along the pipe, and y axis to the top. There is no forces acting in horizontal direction, so there is no point to derive equations for x axis. This also means that XA reaction is zero.

Now we compose equation for y axis, and here yields YA:

YAG1G2=0;YA=G1+G2Y_{A} - G1 - G2 = 0;\quad \rightarrow \quad Y_{A} = G1 + G2

Equation for torques relative to point A:

MA+MG1L2G2L=0M_{A} + M - G1\cdot \frac{L}{2} - G2\cdot L = 0

The torque produced by pipe's gravity force Gl we multiply by arm L/2, since center of gravity of the uniform cross-section pipe is in the middle. Also the torque M applied to the pipe contributes equation as is. Now it's easy to derive reaction MA:

MA=G1L2+G2LMM_{A} = G1\cdot \frac{L}{2} + G2\cdot L - M

Here we dealt with two equations and two unknown values.

important note

We usually don't know the actual direction of reactions beforehand. What we do is choosing any direction we want as positive and start solving the problem. If the solved reaction value turned out to be positive, then initially assumed direction is correct. Otherwise the actual reaction force acts in opposite direction. It also applies to torques.

statically indefinable tasks

As you might noticed, there is a number of equations that can be derived for a specific problem, and there is a set of unknown reactions to be found. Obviously, when number of unknown reactions exceeds number of possible equations, it becomes impossible to solve the problem in that way. Such problems are called statically indefinable. This doesn't means that one can't solve them in general, simply another yet more difficult approach is being used for that. We will not dig deeper in it for now, let's just bring example of such a task:

fig. 1.6.4 A bridge with "redundant" supports

Here only three equations can be composed, whereas we have five unknown support reactions. Hence we deal with a problem with degree of static indefinability of two (5 - 3 = 2). Static indefinability can be seen as a result of applying a "redundant" connections. In order to ensure a static structure (fully fixed), there is enough with only one left support, and any one from remaining supports (say, support A). All other supports are to be considered as "redundant".

What is important to understand is that these additional supports can be considered as "redundant" only from a perspective of solving problem of static. They are yet required from the perspective of structure functionality. More supports are needed to ensure required strength of the bridge.