ENGR-1100 Introduction to Engineering Analysis Lecture 15 3-D FREE-BODY DIAGRAMS, EQUILIBRIUM EQUATIONS, CONSTRAINTS AND STATICAL DETERMINACY Todays Objective: Students will be able to: a) Identify support reactions in 3-D and draw a free-body diagram, and, b) Apply the equations of equilibrium. In-Class Activities: Check Homework, if any

Reading Quiz Applications Support Reactions in 3-D Equations of Equilibrium Concept Quiz Group Problem Solving Attention quiz APPLICATIONS Ball-and-socket joints and journal bearings are often used in mechanical systems. To design the joints or bearings, the support reactions at these joints and the loads must be determined.

APPLICATIONS (continued) The tie rod from point A is used to support the overhang at the entrance of a building. It is pin connected to the wall at A and to the center of the overhang B. If A is moved to a lower position D, will the force in the rod change or remain the same? By making such a change without understanding if there is a change in forces, failure might occur. APPLICATIONS (continued) The crane, which weighs 350 lb, is supporting a oil drum.

How do you determine the largest oil drum weight that the crane can support without overturning ? SUPPORT REACTIONS IN 3-D (Table 5-2) A few examples are shown above. Other support reactions are given in your text book (Table 5-2). As a general rule, if a support prevents translation of a body in a given direction, then a reaction force acting in the opposite direction is developed on the body. Similarly, if rotation is prevented, a couple moment is exerted on the body by the support. IMPORTANT NOTE

A single bearing or hinge can prevent rotation by providing a resistive couple moment. However, it is usually preferred to use two or more properly aligned bearings or hinges. Thus, in these cases, only force reactions are generated and there are no moment reactions created. EQUATIONS OF EQUILIBRIUM (Section 5.6) As stated earlier, when a body is in equilibrium, the net force and the net moment equal zero, i.e., F = 0 and MO = 0 . These two vector equations can be written as six scalar equations of equilibrium (E-of-E). These are FX =

FY MX = MY = = FZ = 0 MZ = 0 The moment equations can be determined about any point. Usually, choosing the point where the maximum number of unknown forces are present simplifies the solution. Any forces

occurring at the point where moments are taken do not appear in the moment equation since they pass through the point. CONSTRAINTS AND STATICAL DETERMINACY (Section 5.7) Redundant Constraints: When a body has more supports than necessary to hold it in equilibrium, it becomes statically indeterminate. A problem that is statically indeterminate has more unknowns than equations of equilibrium. Are statically indeterminate structures used in practice? Why or why not? IMPROPER CONSTRAINTS

Here, while we have 6 unknowns, there is nothing restricting rotation about the AB axis! In some cases, there may be as many unknown reactions as there are equations of equilibrium. M A 0 However, if the supports are not properly constrained, the body may

become unstable for some loading cases. EXAMPLE I Given:The rod, supported by thrust bearing at A and cable BC, is subjected to an 80 lb force. Find: Reactions at the thrust bearing A and cable BC. Plan: a) b) c) d)

Use the established x, y and z axes. Draw a FBD of the rod. Write the forces using scalar equations. Apply scalar equations of equilibrium to solve for the unknown forces. EXAMPLE I (continued) FBD of the rod: Applying scalar equations of equilibrium in appropriate order, we get F X = AX = 0; AX = 0 F Z = AZ + FBC 80 = 0;

M Y = 80 ( 1.5 ) + FBC ( 3.0 ) = 0; Solving the last two equations: FBC = 40 lb, AZ = 40 lb EXAMPLE I (continued) FBD of the rod = 40 lb Now write scalar moment equations about what point? M X = ( MA) X + 40 (6) 80 (6) = 0 ; M Z = ( MA) Z = 0 ; (MA ) Z= 0

Point A! (MA ) X= 240 lb ft CCW EXAMPLE II Given:The uniform plate has a weight of 500 lb, supported by three cables. Find: The tension in each of the supporting cables. Plan: a) b)

c) d) Use established x, y and z axes. Draw a FBD of the plate. Write the forces using scalar equations. Apply scalar equations of equilibrium to solve for the unknown forces. EXAMPLE II (continued) FBD of the plate: TA 500 lb TC

TB 200 lb 1.5 ft Applying scalar equations of equilibrium : Fz = TA + TB + TC 200 500 = 0 (1) Mx = TA (3) + TC (3) 500 (1.5) 200 (3) = 0 (2)

My = -TB (4) TC (4) + 500 (2) + 200 (2) = 0 (3) EXAMPLE II (continued) Fz = TA + TB + TC 200 500 = 0 (1) Mx = TA (3) + TC (3) 500 (1.5) 200 (3) = 0 (2) My = -TB (4) TC (4) + 500 (2) + 200 (2) = 0

(3) Using Eqs. (2) and (3), express TA and TB in terms of TC: Eq. (2) TA = 450 TC Eq. (3) TB = 350 TC Substituting the results into Eq. (1) & solving for TC: Eq. (1) (450 TC ) + (350 TC) + TC 200 500 = 0 TC = 100 lb TA = 350 lb and TA = 250 lb

READING QUIZ 1. If a support prevents rotation of a body about an axis, then the support exerts a ________ on the body about that axis. A) Couple moment B) Force C) Both A and B. D) None of the above. 2. When doing a 3-D problem analysis, you have ________ scalar equations of equilibrium. A) 3 B) 4

C) 5 D) 6 CONCEPT QUIZ The rod AB is supported using two cables at B and a ball-and-socket joint at A. How many unknown support reactions exist in this problem? A) 5 force and 1 moment reaction B) 5 force reactions C) 3 force and 3 moment reactions

D) 4 force and 2 moment reactions ATTENTION QUIZ A plate is supported by a ball-andsocket joint at A, a roller joint at B, and a cable at C. How many unknown support reactions are there in this problem? A) 4 forces and 2 moments B) 6 forces C) 5 forces D) 4 forces and 1 moment GROUP PROBLEM SOLVING

Given: A bent rod is supported by the roller at B, and the smooth collar at A. Find: The reactions at all the supports for the loading shown. Plan: a) Draw a FBD of the rod. b) Apply scalar equations of equilibrium to solve for the unknowns. GROUP PROBLEM SOLVING (continued) A FBD of the rod Applying scalar equations of equilibrium in appropriate order, we get

Fy = Ay = 0 ; Ay = 0 N Mx = NB (0.8+0.8) 800 (0.8) 600 (0.8+0.8) = 0 ; NB = 1000 N Mz = (MA)z= 0 ; (MA)z = 0 Nm GROUP PROBLEM SOLVING (continued) A FBD of the rod: 1000 N Applying scalar equations of equilibrium in appropriate order, we get Fz = Az + 1000 800 600 = 0 ;

Az = 400 N My = (MA)y 600 (0.4) + 1000 (0.8) = 0 ; (MA)y = -560 Nm or 560 Nm CW