STATICS: CE201. Chapter 7 Internal Forces. Notes are prepared based on: Engineering Mechanics, Statics by R. C. Hibbeler, 12E Pearson

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1 STATICS: CE201 Chapter 7 Internal Forces Notes are prepared based on: Engineering Mechanics, Statics by R. C. Hibbeler, 12E Pearson Dr M. Touahmia & Dr M. Boukendakdji Civil Engineering Department, University of Hail (2012/2013) 7. Internal Forces Chapter Objective: Use the method of sections to determine internal forces in 2-D load cases. Contents: 7.1 Internal Forces Development in Structural Members 7.2 Shear and Moment Equations and Diagram 7.3 Cables 1

2 7.1 Internal Forces Development in Structural Members The design of any structural member requires to know both the external loads acting on the member and the internal forces acting within the member in order to be sure the material can resist these loading. The concrete supporting a bridge has fractured: What might have caused it to do this? Is it because of the internal forces? If so, what are they and how can we design these structures to make them safer? 7.1 Internal Forces Development in Structural Members If a coplanar force system acts on a member, then in general a resultant internal normal force N (acting perpendicular to the section), shear force V (acting along the surface), and bending moment M will act at any cross section along the member. 2

3 Steps for Determining Internal Loadings: Internal Loadings can be determined by using the method of section. The following example explains the steps that we should follow to determine the internal forces acting on the cross section at point C. 1. Before the member is sectioned, it is first necessary to determine its support reactions by drawing a FBD of the entire structure and solving for the unknown reactions. Steps for Determining Internal Loadings: 3. Pass an imaginary section a-a perpendicular to the axis of the beam through point C and separate the beam into 2 segments. The internal loadings acting at C will be exposed and become external on the FBD of each segment: Then, decide which resulting section or piece will be easier to analyze. 3

4 Steps for Determining Internal Loadings: 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 Equations of Equilibrium to the FBD (drawn in step 3) and solve for the unknown internal loads. x y C F 0 F 0 M 0 Sign Convention: 4

5 Example 1 Determine the normal force, shear force and bending moment acting just to the left, point B, and just to the right, point C, of the 6 kn force on the beam. Solution 1 1. Support Reactions: Can be determined from the freebody diagram of the beam. M D 0 6 kn6 m A 9 m 0 9 kn.m A y 5 kn y 5

6 Solution 1 2. Free-Body Diagrams: The free-body diagrams of the left segments AB and AC of the beam are: 3. Equations of Equilibrium: Segment AB: x F 0 0 N B y F 0 5 kn V 0 B B M 0 5 kn3m M 0 B V B 5 kn M B 15 kn.m Solution 1 Segment AC: Fx 0 N C 0 Fy 0 5 kn 6 kn V 0 C V C 1kN M c 0 5 kn3 m M 0 C kn.m 15 M C 6

7 Example 2 Determine the normal force, shear force and bending moment at point C of the beam. Solution 2 Note: It is not necessary to find the support reactions at A since segment BC of the beam can be used to determine the internal loadings at C. w C 1.5 m 3 m 1200 N/m 600 N/m Free-Body Diagram: The distributed load acting on segment BC can be replaced by its resultant force: 7

8 Solution 2 Equations of Equilibrium: x F 0 0 N C y F 0 V 450 N 0 C M 0 M 450 N0.5 m 0 C C V C 450 N M C 225 N.m Note: The negative sign indicates that M C acts in the opposite sense to that shown on the free body diagram. 7.2 Shear and Moment Diagrams Beams are designed to support loads perpendicular to their axes. The design of a beam requires a detailed knowledge of the variation of the internal shear force V and bending moment M acting at each point along the axis of the beam. 8

9 7.2 Shear and Moment Diagrams To construct the shear and moment diagrams, it is necessary to section the member at an arbitrary point, located at distance x from the left end. The variations of V and M as functions of the position x along the beam s axis can be obtained using the method of sections. The graphical variations of V and M as functions of x are termed the shear diagram and bending moment diagram respectively. 7.2 Shear and Moment Diagrams shear diagram and bending moment diagram of the beam: 9

10 Example 3 Draw the shear and moment diagrams for the shaft shown in the figure below. The support at A is a thrust bearing and the support at C is a journal bearing. Solution 3 Support Reactions: C y M 0 A 4 m 5 kn 2 m 0 y F 0 5 kn 2.5 kn C 0 Shear and Moment Functions: The shaft is sectioned at an arbitrary distance x from point A, extending within the segment AB. The FBD of the left segment is: Applying the equations of equilibrium: y A y 2.5 kn C y 2.5 kn y F 0 V 2.5 kn (1) M knx M 0 M 2.5x kn.m (2) 10

11 Solution 3 A free body diagram for the left segment of the shaft extending a distance x within the region BC is shown: Applying the equilibrium equations yields: Fy 0 M kn 5 kn V 0 V 2.5 kn (3) x 2m 2.5 knx 0 M 5 kn M x kn.m (4) Solution 3 Shear and Moment Diagrams: When equations (1) to (4) are plotted within the regions in which they are valid, the shear and moment diagrams are obtained. The shear diagram indicates that the internal shear force is always 2.5 kn (positive) within segment AB. Just to the right of point B, the shear force changes sign and remains at a constant value of -2.5 kn for segment BC. The moment diagram starts at zero, increases linearly to point B at x = 2 m, where M max = 2.5 kn (2 m) = 5 kn.m, and therefore decreases back to zero. 11

12 Example 4 Draw the shear and moment diagrams for the beam shown in the figure below. Solution 4 Support Reactions: The support reactions are shown on the beam s free-body diagram. Shear and Moment Functions: 1 2 F 0 9 x V x V 9 kn 3 y (1) 1 2 x M 0 M x 9x x M 9x kn.m 9 (2) 12

13 Solution 3 Shear and Moment Diagrams: The shear and moment diagrams are obtained by plotting Eqs. (1) (2). The point of zero shear can be found using Eq. (1): 2 x V 9 3 0, x 5.20 m Maximum moment can be found using the value of x = 5.2 m in Eq. (2): M max kn.m QUIZ Draw the shear and moment diagrams for the beam shown in the figure below. 13

14 QUIZ Draw the shear and moment diagrams for the beam shown in the figure below. QUIZ Draw complete shear force and bending moment diagrams for the beam below. 14

15 QUIZ Draw complete shear force and bending moment diagrams for the beam shown in the figure. 7.3 Cables Flexible cables and chains combine strength with lightness and often are used in structures for support and to transmit loads from one member to another. Cables are used to support suspension bridges and trolley wheels. 15

16 7.3 Cables Two assumptions are considered in the force analysis of these systems: 1. The weight of the cable is negligible. 2. The cable is perfectly flexible and inextensible. Cable Subjected to Concentrated loads: When a cable of negligible weight supports several concentrated loads, the cable takes the form of several straight line segments, each of which is subjected to a constant tensile. 7.3 Cables The equilibrium analysis is performed by writing down a sufficient number of equilibrium equations and equations describing the geometry of the cable to solve for all the unknowns leading to a description of the tension in each segment of the cable. Example: Consider the cable shown in the figure, where h, L 1, L 2, L 3, P 1 and P 2 are known: The problem here is to determine the nine unknowns: Tension in each of the three segments, the four components of reaction at A and B and the two sags y C and y D at point C and D. 16

17 Example 5 Determine the tension in each segment of the cable shown in the figure below: Solution 5 Unknowns: There are 4 unknown external reactions (A x, A y, E x, E y ), 4 unknown cable tensions (one in each cable segment) and 2 unknown sags (y B and y D ). Consider the free-body diagram for the entire cable: F x 0, A E x x 0A x E x M E A y 0 18 m 4 kn15 m15 kn10 m 0 3 kn 2 m A y 12 kn F y 0, 12kN 4kN 15kN 3kN E 0 E y 10kN y 17

18 Solution 5 Consider the leftmost segment, which cuts cable BC: M C 0 A x 12 m12 kn8 m 4 kn5 m 0 x A x E x 6.33 kn F 0 T cos 6.33 kn 0 y BC BC F 0 12 kn 4 kn T sin 0 BC BC Thus: BC 51.6 T BC 10.2 kn Solution 5 Proceeding now to analyze the equilibrium of point A, C, and E in sequence, we have: Point A: x y F 0 T cos 6.33 kn 0 AB AB F 0 T sin 12 kn 0 AB AB Point C: AB 62.2 T AB 13.6 kn Fx Fy 0 0 TCD TCD cos sin CD CD 12.2cos 51.6 kn sin 51.6 kn 15 kn 0 CD 47.9 kn 9.44 T CD 18

19 Solution 5 Point E: Fx Fy kn T cos 0 ED ED 10 kn T sin 0 ED ED ED 57.7 T ED 11.8 kn Sags y B and y D : y B y D 3 tan m 2tan m 19