440:221 Intro to Engineering Mechanics: Statics Instructor: Alberto Cuitino TA: Pam Carabetta and Jingru Zhang Lt 21 d 22Lecture an 22 Fall 2009 1 Based on Textbook Material: Engineering Mechanics Statics, 12 th Edition, R.C. Hibbeler, Pearson 2010. 440:221 Lectures Internal Forces School of Engineering, Fall 2009 2 440:221 Lectures Topics to be covered to be covered • What is a beam? • What are internal forces? • What are internal forces important in structural analysis? School of Engineering, Fall 2009 3 440:221 Lectures Reading Quiz 1. In a multiforce member, the member is generally subjected to an internal multiforce member the to an internal _________. A) Normal force B) Shear force C) Bending moment D) All of the above. D) 2. In mechanics, the force component V acting tangent to, or along the face of, the section is called the _________ . A) Axial force B) Shear force C) Normal force D) Bending moment School of Engineering, Fall 2009 force B) 440:221 Lectures Applications Beams are structural members designed to support loads applied perpendicular to their axes. Beams are often used to support the span of bridges. They can be thicker at the supports than at the center of center the span. Why are the beams tapered? Internal forces are important in making such a design decision. In this lesson, you will learn about these forces and how to determine them School of Engineering, Fall 2009 . 440:221 Lectures Applications (cont ) A fixed column supports this rectangular billboard. (cont. Usually such columns are wider/thicker at the bottom than at the top Why? top. School of Engineering, Fall 2009 440:221 Lectures Applications (cont ) (cont. The shop crane is used to move heavy machine tools around the shop. The picture shows that an additional frame around the joint is added. Why might have this been done? School of Engineering, Fall 2009 440:221 Lectures Internal Forces The design of any structural member requires finding the forces acting within the member to forces acting the make sure the material can resist those loads. B For example, we want to determine the internal forces acting on the cross section at B. But, first, we first need to determine the Then we need to cut the beam at B and draw support reactions. B the a FBD of one of the halves of the beam. This FBD will include the internal forces acting at B. Finally, we need to solve for these kithEf E School of Engineering, Fall 2009 unknowns using the -o - . 440:221 Lectures Internal Forces In two-dimensional cases, typical internal loads are: • normal or axial forces (N, acting perpendicular to the section), • shear forces (V, acting along the surface), and • the bending moment (M) . The loads on the left and right sides of the section at B are equal in magnitude but opposite in direction This is because when the two sides are School of Engineering, Fall 2009 . This is when two reconnected, the net loads are zero at the section. 440:221 Lectures Sign convention (positive normal shear bending) (positive normal, shear, School of Engineering, Fall 2009 10 440:221 Lectures Steps for solving beam problems solving 1. Take an imaginary cut at the place where you need to determine the internal forces. Then, decide which resulting section or piece will be easier to analyze. 2. If necessary, determine any support reactions or joint forces you need by drawing a FBD of the entire structure and solving for the unknown reactions. 3. Draw a FBD of the piece of the structure you’ve decided to analyze. Remember to show the N V and M loads at the “cut” surface , , . 4 Apply the E-of-E to the FBD (drawn in step 3) and solve for the School of Engineering, Fall 2009 . Apply the to the (drawn in step 3) for unknown internal loads. 440:221 Lectures Solving steps School of Engineering, Fall 2009 440:221 Lectures Solving steps School of Engineering, Fall 2009 440:221 Lectures Solving steps School of Engineering, Fall 2009 440:221 Lectures Solving steps School of Engineering, Fall 2009 440:221 Lectures Solving steps Internal Forces School of Engineering, Fall 2009 440:221 Lectures Solving steps Pick substructure – Left or Right School of Engineering, Fall 2009 440:221 Lectures Solving steps School of Engineering, Fall 2009 440:221 Lectures Solving steps School of Engineering, Fall 2009 440:221 Lectures Example Given: The loading on the beam. Find: The internal forces at point C. Plan: Follow the procedure!! Solution 1. Plan on taking the imaginary cut at C. It will be easier to work with the right section (the cut at C to point B) since the geometry is simpler and there are no external loads. School of Engineering, Fall 2009 440:221 Lectures Example (cont ) 2. We need to determine B y . Use a FBD of the entire frame and solve the E- f Ef B . o -E for y . FBD of the entire beam: B x 3 ft 9 ft 18 kip 3 ft of the A y B y Applying the E-of-E to this FBD, we get + F x = B x = 0; School of Engineering, Fall 2009 + M A = − B y ( 9 ) + 18 ( 3 ) = 0 ; B y = 6 kip 440:221 Lectures Example (cont ) 3. Now draw a FBD of the right section. Assume directions for V C , N C and M . C . N C 4.5 ft 6 kip V C M C C B 4. Applying the E-of-E to this FBD, we get + F =N =0; N =0 x = N C = 0; C = 0 + F y = – V C – 6 = 0; V C = – 6 kip School of Engineering, Fall 2009 + M C = – 6 (4.5) – M C = 0 ; M C = – 27 kip ft 440:221 Lectures Concept Quiz Quiz 1. A column is loaded with a vertical 100 N force. At which sections are the internal loads the same? A) P, Q, and R B) P and Q • P Q 100 N C) Q and R D) None of the above. R C) 2. A column is loaded with a horizontal 100 N force. At which section are the internal loads largest? A) P B)Q P 100 N ) Q C) R D) S Q R S School of Engineering, Fall 2009 D) 440:221 Lectures Representative Problem Given: The loading on the beam. Find: The internal forces at point F. Plan: Follow the procedure!! the Solution 1. Make an imaginary cut at F. Why there? Which section will you pick to analyze via the FBD? School of Engineering, Fall 2009 Why will it be easier to work with segment FB? 440:221 Lectures Representative Problem (cont ) 2. We need to determine the cable tension, T, using a FBD and the E- (cont. ,, of-E for the entire frame. T T A x 3 m 3 m A 45 y 1800 N + F x = A x = 0 + M A = T ( 6 ) + T sin 45 ( 6 ) − 1800 (3) = 0 ; School of Engineering, Fall 2009 T = 665 N 440:221 Lectures Representative Problem (cont ) 3. A FBD of section FB is shown below. (cont. 665 450 N M N F 0.75 m F B N 0.75 m 4 Applying the E of E to the FBD we get V F F FBD of Section FB . Applying - - to the , we + F x = N F = 0 + F y = – 450 + 665 – V F = 0 ; V F = 215 N School of Engineering, Fall 2009 + M C = 665 (1.5) – 450 (0.75) – M F = 0 ; M F = 660 N m 440:221 Lectures Example 1 School of Engineering, Fall 2009 27 440:221 Lectures Example 1 (cont) (cont) School of Engineering, Fall 2009 28 440:221 Lectures Example 2 School of Engineering, Fall 2009 29 440:221 Lectures Example 2 (cont) (cont) School of Engineering, Fall 2009 30 440:221 Lectures Attention Quiz 1. Determine the magnitude of the internal loads 100 N (normal, shear, and bending moment) at point C. A) (100 N, 80 N, 80 N m) • 0.5m 1 m 80 N B) (100 N, 80 N, 40 N m) C) (80 N, 100 N, 40 N m) D) (80 N, 100 N, 0 N m ) C B) 2. A column is loaded with a horizontal 100 N force. At which section are the internal P 100N section the loads the lowest? A) P B) Q Q R S A) School of Engineering, Fall 2009 B) C) R D) S 440:221 Lectures Take home message- message • Internal forces allow us to assess material failure • In order to compute the value of the internal forces, we need to produce a virtual cut at the desired location in order to expose their action/reaction • Beams are the most utilized structure in many engineering applications – Learn this material well! School of Engineering, Fall 2009 32 amc Microsoft PowerPoint - Lecture 21-22