Section 7.1 Internal Loadings

Engineering Mechanics: Statics Fifteenth Edition Chapter 7 Internal Forces Section 7.1 Internal Loadings Developed in Structural Members Copyright © 2022 Pearson Education, Inc. All Rights Reserved Internal Forces Today’s Objective: 1. Use the method of sections for determining internal forces in 2-D load cases. Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Applications (1 of 3) Beams are structural members designed to support loads applied perpendicularly to their axes. Beams can be used to support the span of bridges. They are often thicker at the supports than at the center of the span. Internal forces are important in making such a design decision. In this lesson, you will learn about these forces and how to determine them. Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Applications (2 of 3) A fixed column supports these rectangular billboards. Usually such columns are wider/thicker at the bottom than at the top. Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Applications (3 of 3) 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. Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Internal Forces (1 of 2) The design of any structural member requires finding the forces acting within the member to make sure the material can resist those loads. For example, we want to determine the internal forces acting on the cross section at B. But, first, we first need to determine the support reactions. Then we need to cut the beam at B and draw 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 unknowns using the E-of-E. Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Internal Forces (2 of 2) 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 reconnected, the net loads are zero at the section. Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Steps for Determining Internal Forces 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 FB D 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 FB D (drawn in step 3) and solve for the unknown internal loads. Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Example (1 of 3) 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 (from the cut at C to point B) since the geometry is simpler and there are no external loads. Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Example (2 of 3) 2. We need to determine B y. Use a F B D of the entire frame and solve the E-of-E for B y. Applying the E-of-E to this F B D, we get → +  FX = BX = 0; Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Example (3 of 3) 3. Now draw a F B D of the right section. Assume directions for V C, N C and M C. 4. Applying the E-of-E to this F B D, we get → +  FX = NC = 0;  +  Fy = − VC − 6 = 0; NC = 0 VC = −6 kip Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Group Problem Solving I (1 of 3) 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 left section (point A to the cut at C) since the geometry is simpler. Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Group Problem Solving I (2 of 3) 2. First, we need to determine A x and A y using a F B D of the entire frame. Applying the E-of-E to this F B D, we get → +  FX = A X + 400 = 0; A X = −400 N Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Group Problem Solving I (3 of 3) 3. Now draw a F B D of the left section. Assume directions for V C, N C and M C as shown. Applying the E-of-E to this F B D, we get → +  FX = NC − 400 = 0;  +  Fy = − VC − 96 = 0; N C = 400 N VC = −96 N Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Group Problem Solving II (1 of 3) 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 left section (point A to the cut at C) since the geometry is simpler Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Group Problem Solving II (2 of 3) 2. Determine the reactions at A, using a F B D and the E-ofE for the entire frame. → +  FX = A X − 4320 = 0;  +  Fy = A y − 1800 = 0; A X = 4320 lb A y = 1800 lb Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Group Problem Solving II (3 of 3) 3. A F B D of section A C is shown below. 4. Applying the E-of-E to the F B D, we get → +  FX = NC + 4320 = 0; NC = −4320 lb  +  Fy = 1800 − 450 − VC = 0; VC = 1350 lb Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Attention Quiz (1 of 2) 1. Determine the magnitude of the internal loads (normal, shear, and bending moment) at point C. A. (100 N, 80 N, 80 N m) B. (100 N, 80 N, 40 N m) C. (80 N, 100 N, 40 N m) D. (80 N, 100 N, 0 N m) Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Attention Quiz (1 of 2) 2. A column is loaded with a horizontal 100 N force. At which section are the internal loads the lowest? A. P B. Q C. R D. S Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Reading Quiz (1 of 2) 1. In a multiforce member, the member is generally subjected to an internal ________. A. Normal force B. Shear force C. Bending moment D. All of the above. Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved Reading Quiz (2 of 2) 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 Copyright © 2016, 2013, 2010 Pearson Education, Inc. All Rights Reserved