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

Engineering Mechanics: StaticsEngineering Mechanics: Statics

Chapter 7: Internal Forces

Chapter ObjectivesChapter 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 OutlineChapter 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

MembersThe 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

Members

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

Members

MembersSince 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

Members

MembersForce component N, acting normal to the

beam at the cut session and V, acting tangent to the session are known as normal or axial force and the shear force

Couple moment M is referred as the bending moment

Members

MembersFor 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

Members

MembersFor 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

Members

MembersFree 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

Members

MembersFree 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 dismemberthe frame and determine reactions at each connection

Members

MembersFree 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

Members

MembersProcedure for AnalysisSupport 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

Members

MembersProcedure for AnalysisFree-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

Members

MembersProcedure for AnalysisFree-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

Members

MembersProcedure for AnalysisEquations 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

Members

MembersThe 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

Members

MembersExample 7.1The bar is fixed at its end and is loaded. Determine the internal

normal force at points B and C.

Members

MembersSolutionSupport 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

Members

MembersSolutionFBD of the sectioned barNo 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

Members

MembersSolutionSegment AB

+↑∑ Fy = 0; 8kN – NB = 0

NB = 8kN

Segment DC+↑∑ Fy = 0; NC – 4kN= 0

NC = 4kN

Members

MembersExample 7.2The circular shaft is subjected to three concentrated torques. Determine the

internal torques at points B and C.

Members

MembersSolutionSupport Reactions Shaft subjected to only collinear torques

∑ Mx = 0;

-10N.m + 15N.m + 20N.m –TD = 0

TD = 25N.m

Members

MembersSolutionFBD of shaft segments AB and CD

Members

∑ Mx = 0; -10N.m + 15N.m – TB = 0

TB = 5N.m

Segment CD∑ Mx = 0; TC – 25N.m= 0

TC = 25N.m

Members

MembersExample 7.3The 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.

Members

MembersSolutionSupport Reactions9kN.m is a free vector and can be place

anywhere in the FBD+↑∑ Fy = 0; 9kN.m + (6kN)(6m) - Ay(9m) = 0

Ay = 5kN

Members

MembersSolutionFBD of the segments AB and AC9kN.couple moment must be kept in

original position until after the section is made and appropriate body isolated

Members

+→∑ Fx = 0; NB = 0+↑∑ Fy = 0; 5kN – VB = 0

VB = 5kN∑ MB = 0; -(5kN)(3m) + MB = 0

MB = 15kN.mSegment AC

+→∑ Fx = 0; NC = 0+↑∑ Fy = 0; 5kN - 6kN + VC = 0

VC = 1kN∑ MC = 0; -(5kN)(3m) + MC = 0

MC = 15kN.m

Members

MembersExample 7.4Determine the internal force, shear force

and the bending moment acting at point B of the two-member frame.

Members

MembersSolutionSupport ReactionsFBD of each member

Members

MembersSolutionMember 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

Members

MembersSolutionFBD of segments AB and BC Important to keep distributed

loading exactly as it is after the section is made

Members

MembersSolutionMember 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

Members

MembersExample 7.5Determine the normal

force, shear force and the

bending moment acting at point

E of the frame loaded.

Members

MembersSolutionSupport Reactions Members AC and CD are two force

members+↑∑ Fy = 0;

Rsin45° – 600N = 0R = 848.5N

Members

MembersSolutionFBD of segment CE

Members

MembersSolution

+→∑ Fx = 0; 848.5cos45°N - VE = 0VE = 600 N

+↑∑ Fy = 0; -848.5sin45°N + NE = 0NE = 600 N

∑ ME = 0; 848.5cos45°N(0.5m) - ME = 0ME = 300 N.m

Results indicate a poor design Member AC should be straight to eliminate

bending within the member

Members

MembersExample 7.6The 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

Members

MembersSolution 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 forceFw = 900N/m2(6m)(2.5m) = 13.5kN

View Free Body Diagram

Members

MembersSolutionFBD of the loadings

Members

MembersSolution

mkNkjiM

.}5.409.701.19{

376.605.13

25.530

}38.65.13{

03475.65.13

Members

MembersSolutionFAz = {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

22yx MMM

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

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