Engineering Mechanics: Statics Chapter 7: Internal Forces Engineering Mechanics: Statics

Chapter Objectives To show how to use the method of sections for determining the internal loadings in a member. To generalize this procedure by formulating equations that can be plotted so that they describe the internal shear and moment throughout a member. To analyze the forces and study the geometry of cables supporting a load.

Chapter Outline Internal Forces Developed in Structural Members Shear and Moment Equations and Diagrams Relations between Distributed Load, Shear and Moment Cables

7.1 Internal Forces Developed in Structural Members The design of any structural or mechanical member requires the material to be used to be able to resist the loading acting on the member These internal loadings can be determined by the method of sections

7.1 Internal Forces Developed in Structural Members Consider the “simply supported” beam To determine the internal loadings acting on the cross section at C, an imaginary section is passed through the beam, cutting it into two By doing so, the internal loadings become external on the FBD

7.1 Internal Forces Developed in Structural Members Since both segments (AC and CB) were in equilibrium before the sectioning, equilibrium of the segment is maintained by rectangular force components and a resultant couple moment Magnitude of the loadings is determined by the equilibrium equations

7.1 Internal Forces Developed in Structural Members Force component N, acting normal to the beam at the cut session and V, acting t angent to the session are known as normal or axial force and the shear force Couple moment M is referred as the bending moment

7.1 Internal Forces Developed in Structural Members For 3D, a general internal force and couple moment resultant will act at the section Ny is the normal force, and Vx and Vz are the shear components My is the torisonal or twisting moment, and Mx and Mz are the bending moment components

7.1 Internal Forces Developed in Structural Members For most applications, these resultant loadings will act at the geometric center or centroid (C) of the section’s cross sectional area Although the magnitude of each loading differs at different points along the axis of the member, the method of section can be used to determine the values

7.1 Internal Forces Developed in Structural Members Free Body Diagrams Since frames and machines are composed of multi-force members, each of these members will generally be subjected to internal shear, normal and bending loadings Consider the frame with the blue section passed through to determine the internal loadings at points H, G and F

7.1 Internal Forces Developed in Structural Members Free Body Diagrams FBD of the sectioned frame At each sectioned member, there is an unknown normal force, shear force and bending moment 3 equilibrium equations cannot be used to find 9 unknowns, thus dismember the frame and determine reactions at each connection

7.1 Internal Forces Developed in Structural Members Free Body Diagrams Once done, each member may be sectioned at its appropriate point and apply the 3 equilibrium equations to determine the unknowns Example FBD of segment DG can be used to determine the internal loadings at G provided the reactions of the pins are known

7.1 Internal Forces Developed in Structural Members Procedure for Analysis Support Reactions Before the member is cut or sectioned, determine the member’s support reactions Equilibrium equations are used to solve for internal loadings during sectioning of the members If the member is part of a frame or machine, the reactions at its connections are determined by the methods used in 6.6

7.1 Internal Forces Developed in Structural Members Procedure for Analysis Free-Body Diagrams Keep all distributed loadings, couple moments and forces acting on the member in their exact locations, then pass an imaginary section through the member, perpendicular to its axis at the point the internal loading is to be determined After the session is made, draw the FBD of the segment having the least loads

7.1 Internal Forces Developed in Structural Members Procedure for Analysis Free-Body Diagrams Indicate the z, y, z components of the force and couple moments and the resultant couple moments on the FBD If the member is subjected to a coplanar system of forces, only N, V and M act at the section Determine the sense by inspection; if not, assume the sense of the unknown loadings

7.1 Internal Forces Developed in Structural Members Procedure for Analysis Equations of Equilibrium Moments should be summed at the section about the axes passing through the centroid or geometric center of the member’s cross-sectional area in order to eliminate the unknown normal and shear forces and thereby, obtain direct solutions for the moment components If the solution yields a negative result, the sense is opposite that assume of the unknown loadings

7.1 Internal Forces Developed in Structural Members The link on the backhoe is a two force member It is subjected to both bending and axial load at its center By making the member straight, only an axial force acts within the member

7.1 Internal Forces Developed in Structural Members Example 7.1 The bar is fixed at its end and is loaded. Determine the internal normal force at points B and C.

7.1 Internal Forces Developed in Structural Members Solution Support Reactions FBD of the entire bar By inspection, only normal force Ay acts at the fixed support Ax = 0 and Az = 0 +↑∑ Fy = 0; 8kN – NB = 0 NB = 8kN

7.1 Internal Forces Developed in Structural Members Solution FBD of the sectioned bar No shear or moment act on the sections since they are not required for equilibrium Choose segment AB and DC since they contain the least number of forces

7.1 Internal Forces Developed in Structural Members Solution Segment AB +↑∑ Fy = 0; 8kN – NB = 0 NB = 8kN Segment DC +↑∑ Fy = 0; NC – 4kN= 0 NC = 4kN

7.1 Internal Forces Developed in Structural Members Example 7.2 The circular shaft is subjected to three concentrated torques. Determine the internal torques at points B and C.

7.1 Internal Forces Developed in Structural Members Solution Support Reactions Shaft subjected to only collinear torques ∑ Mx = 0; -10N.m + 15N.m + 20N.m –TD = 0 TD = 25N.m

7.1 Internal Forces Developed in Structural Members Solution FBD of shaft segments AB and CD

7.1 Internal Forces Developed in Structural Members Solution Segment AB ∑ Mx = 0; -10N.m + 15N.m – TB = 0 TB = 5N.m Segment CD ∑ Mx = 0; TC – 25N.m= 0 TC = 25N.m

7.1 Internal Forces Developed in Structural Members Example 7.3 The beam supports the loading. Determine the internal normal force, shear force and bending moment acting to the left, point B and just to the right, point C of the 6kN force.

7.1 Internal Forces Developed in Structural Members Solution Support Reactions 9kN.m is a free vector and can be place anywhere in the FBD +↑∑ Fy = 0; 9kN.m + (6kN)(6m) - Ay(9m) = 0 Ay = 5kN

7.1 Internal Forces Developed in Structural Members Solution FBD of the segments AB and AC 9kN.couple moment must be kept in original position until after the section is made and appropriate body isolated

7.1 Internal Forces Developed in Structural Members Solution Segment AB +→∑ Fx = 0; NB = 0 +↑∑ Fy = 0; 5kN – VB = 0 VB = 5kN ∑ MB = 0; -(5kN)(3m) + MB = 0 MB = 15kN.m Segment AC +→∑ Fx = 0; NC = 0 +↑∑ Fy = 0; 5kN - 6kN + VC = 0 VC = 1kN ∑ MC = 0; -(5kN)(3m) + MC = 0 MC = 15kN.m

7.1 Internal Forces Developed in Structural Members Example 7.4 Determine the internal force, shear force and the bending moment acting at point B of the two-member frame.

7.1 Internal Forces Developed in Structural Members Solution Support Reactions FBD of each member

7.1 Internal Forces Developed in Structural Members Solution Member AC ∑ MA = 0; -400kN(4m) + (3/5)FDC(8m)= 0 FDC = 333.3kN +→∑ Fx = 0; -Ax + (4/5)(333.3kN) = 0 Ax = 266.7kN +↑∑ Fy = 0; Ay – 400kN + 3/5(333.3kN) = 0 Ay = 200kN

7.1 Internal Forces Developed in Structural Members Solution FBD of segments AB and BC Important to keep distributed loading exactly as it is after the section is made

7.1 Internal Forces Developed in Structural Members Solution Member AB +→∑ Fx = 0; NB – 266.7kN = 0 NB = 266.7kN +↑∑ Fy = 0; 200kN – 200kN - VB = 0 VB = 0 ∑ MB = 0; MB – 200kN(4m) – 200kN(2m) = 0 MB = 400kN.m

7.1 Internal Forces Developed in Structural Members Example 7.5 Determine the normal force, shear force and the bending moment acting at point E of the frame loaded.

7.1 Internal Forces Developed in Structural Members Solution Support Reactions Members AC and CD are two force members +↑∑ Fy = 0; Rsin45° – 600N = 0 R = 848.5N

7.1 Internal Forces Developed in Structural Members Solution FBD of segment CE

7.1 Internal Forces Developed in Structural Members Solution +→∑ Fx = 0; 848.5cos45°N - VE = 0 VE = 600 N +↑∑ Fy = 0; -848.5sin45°N + NE = 0 NE = 600 N ∑ ME = 0; 848.5cos45°N(0.5m) - ME = 0 ME = 300 N.m Results indicate a poor design Member AC should be straight to eliminate bending within the member

7.1 Internal Forces Developed in Structural Members Example 7.6 The uniform sign has a mass of 650kg and is supported on the fixed column. Design codes indicate that the expected maximum uniform wind loading that will occur in the area where it is located is 900Pa. Determine the internal loadings at A

7.1 Internal Forces Developed in Structural Members View Free Body Diagram Solution Idealized model for the sign Consider FBD of a section above A since it dies not involve the support reactions Sign has weight of W = 650(9.81) = 6.376kN Wind creates resultant force Fw = 900N/m2(6m)(2.5m) = 13.5kN

7.1 Internal Forces Developed in Structural Members Solution FBD of the loadings

7.1 Internal Forces Developed in Structural Members Solution

7.1 Internal Forces Developed in Structural Members Solution FAz = {6.38k}kN represents the normal force N FAx= {13.5i}kN represents the shear force MAz = {40.5k}kN represents the torisonal moment Bending moment is determined from where MAx = {-19.1i}kNm and MAy = {-70.9j}kN.m