Lesson 4 – Rigid Body Statics When performing static equilibrium calculations for objects, we always start by assuming the objects are rigid bodies. This assumption means that the object does not change shape when forces act on the object. This also means that the object is considered unbreakable. If you cannot change the shape of a solid object, you cannot break it. This is because the electrical bonds that hold the solid together are unchanged if you don't deform the solid. In practice, no real object is unbreakable. Strength of Material analysis is the type of analysis that takes into account the type of material in predicting whether or not the structure will fail. But, the static analysis that precedes the strength of material analysis always assumes the structure consists of rigid bodies. When you study projectile motion, or the motion of an object, you don't worry about the size of the object very much. But, when you are studying how to put an object into static equilibrium the size of the object can be a very important consideration. Taking into account finite size of rigid bodies Consider a book that weighs 1 pound. To hold this book up and put the book into static equilibrium, you have to exert a force of 1 pound in the upward direction. Those forces are vectors so both the magnitude and direction are important. That is, the force your hand exerts to hold the book up has to have the correct magnitude (1 pound) and correct direction (straight up – to oppose the force due to gravity). If the book is small enough so that your hand can only exert the force in one location, you would not care about the size of the book. But, if the book is wider than your hand, the place where you exert the force makes a big difference.
Force acting in center will keep the book in place
Placing that same force (magnitude and direction) off to the side will result in the book toppling over. To understand how to take into account the location of the applied force, we define the moment of a force. In physics textbooks, this quantity is often referred to as the torque. The moment of a force is defined by:
1 M =r ×F where r =the position vector pointing from the axis of rotation to the point of application for the force, F The magnitude of the moment of a force is given by: ∣M ∣=M =∣r∣∣F∣sin
where is the angle between r and F The direction of the moment of a force is determined by the right-hand rule when crossing r× F . Moment of a force in two-dimensions It is much easier to visualize the moment of a force in two dimensional problems. Consider a bar that is pinned on its left end. If we apply a force with a magnitude of 10 N to this bar, the resulting moment of that force depends on where the force is applied, as well as the direction of that force. Consider the following cases:
10 N
5 m In this case the angle between the moment arm, r , and F is 90º. So the magnitude of the vector is given by: M = (5 m)(10 N) sin(90º) = 50 Nm. The direction of this moment is the direction that the force would cause rotation. In this case, that direction would be clockwise (CW).
10 N 45º 5 m In this case the angle is 45º, so M = (5m)(10 N) sin(45º) = 35.36 Nm, clockwise.
10 N 5 m
In this case the angle between the moment arm and the force is zero. So, M = (5m)(10 N) sin(0º) = 0
2 Alternative forms for calculating the moment of a force Another way to calculate the magnitude of a moment is to use: M =dF where d is the perpendicular distance from the line of force to the axis of rotation This is illustrated below:
d F
r M=rFsin=rsinF=dF So, if you draw a perpendicular line from the axis of rotation to the line of force, that distance, d, multiplied by the force gives the magnitude of the moment. Another way to calculate the moment is to break the force into rectangular components.
F Fsin() Fcos() r r M =rFsin0 Fcos=rFsin Static equilibrium of a rigid body To keep a rigid body in static equilibrium, you must balance all forces acting on that rigid body and you must balance all the moments as well. In the problems that we are interested in solving, we will look at a rigid body that is constrained from moving. For a given loading, we want to be able to calculate the forces the constraints must exert to keep the rigid body in static equilibrium. There are two typical kinds of supports for structures: 1) pin joints and 2) roller supports. A pin joint prevents motion horizontally and vertically. A roller support prevents motion in one dimension only. For a two-dimensional problem, a rigid body will have one pin joint and one roller support. Any more supports than this will either not allow the system to be in static equilibrium for a general loading, or the constraint forces will be statically indeterminate. A simple way to keep this in mind is to remember that a pin joint will have two unknown reaction forces, and a roller support will have one unknown reaction force. This gives a total of three unknown forces. In two dimensions, you have three independent equations that are required for satisfying static equilibrium. That means you can only solve for three
3 unknowns. Typically we think of the three equations that must be satisfied for static equilibrium as consisting of two force equations and a moment equation. This is a good way to think of things since you need to prevent translation horizontally (by balancing forces in the horizontal direction), prevent translation vertically (by balancing forces in the vertical direction), and prevent rotation (by balancing moments). However, it is certainly possible to use three independing moment equations, instead of the two force equations and one moment equation. Problems Consider the 10 meter long beam in static equilibrium shown below. Assume the beam to have negligible weight. A 100 N force is applied 2 m from the right end. Solve for the reaction forces at A and B. There is a pin joint at A, and a roller support at B.
100 N B A
8 m 2 m
Replace the supports with the reaction forces. This would look like this:
100 N Ax
Ay By 8 m 2 m
Since A is a pin joint, it is capable of reaction forces in both the horizontal and vertical direction. Since B is a roller support, it is only capable of a vertical reaction force. We start by taking moments about the point that eliminates the most unknowns, i.e. point A.
∑ M A=−100 N 8mBy 10 m=0 the convention that we use is counterclockwise moments are positive Solving for By gives:
4 800 Nm B = =80 N y 10 m Now we can balance the vertical forces:
∑ F y= Ay−100 N B y=0
A y =100 N −B y=100 N −80 N =20 N Finally, we can balance the horizontal forces:
∑ F x=A x=0
Ax=0
Consider a 10 m long beam of negligible weight that has a 100 N force at an angle of 30º below the positive x direction at a location of 5 m from the left end of the beam. Solve for the reaction forces at A and B.
100 N 30º B A
5 m 5 m
Replace the supports with reaction forces, and break the 100 N force into x and y components.
100 N sin(30º)
B Ax 100 N cos(30º) A By Ay 5 m 5 m
Once again, start by taking moments about point A (to eliminate Ax and Ay) from the equation:
5 ∑ M A=−(100 N )(sin (30))(5m)+ B y(10m)=0 250 Nm B = =25 N ↑ y 10 m Next, we can balance vertical forces: ↑ F = A −(100 N )(sin (30))+B =0 ∑ y y y A y =50 N −20 N =30 N ↑ Finally, we can balance horizontal forces:
→∑ F x=− Ax +100 N (cos(30))=0 Ax=100 N (cos(30))=86.6 N ←
Consider the structure shown below. Find the reaction forces at A and B that will put this structure in static equilibrium. Assume the structure's weight is negligible. 100 N
2 m 5 m 5 m
Replace the supports with the proper reaction forces.
100 N By 2 m 5 m 5 m Bx
Ay
Start by taking moments about point B.
∑ M B=−A y (10 m)+(100 N )(2m)=0 200 Nm A = =20 N ↑ y 10 m Let's balance horizontal forces next:
→∑ F x=−100 N + Bx=0
B x =100 N → Finally, let's balance the vertical forces:
6 ∑ F y= AyB y=0
By =−A y=−20 N
B y=20 N Since By came out negative, it points opposite to the direction shown in our drawing. So, By points downward.
Solve for the reaction forces for the structure in static equilibrium shown below. Assume the structure's weight is negligible: 100 N
2 m
100 N
5 m 2 m 3 m
7