Engineering Mechanics: Statics Chapter 7: Internal Forces Chapter 7: Internal Forces.
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Transcript of Engineering Mechanics: Statics Chapter 7: Internal Forces Chapter 7: Internal Forces.
Engineering Mechanics: StaticsEngineering Mechanics: Statics
Chapter 7: Internal Forces
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
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
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
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
MembersExample 7.1The 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
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
MembersSolutionSegment 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
7.1 Internal Forces Developed in Structural
MembersExample 7.2The 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
7.1 Internal Forces Developed in Structural
MembersSolutionSupport 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
7.1 Internal Forces Developed in Structural
MembersSolutionFBD of shaft segments AB and CD
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
MembersSolutionSegment 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
7.1 Internal Forces Developed in Structural
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.
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
MembersSolutionSegment AB
+→∑ 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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
MembersExample 7.4Determine 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
7.1 Internal Forces Developed in Structural
MembersSolutionSupport ReactionsFBD of each member
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
MembersSolutionFBD 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
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
MembersExample 7.5Determine 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
7.1 Internal Forces Developed in Structural
MembersSolutionSupport Reactions Members AC and CD are two force
members+↑∑ Fy = 0;
Rsin45° – 600N = 0R = 848.5N
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
MembersSolutionFBD of segment CE
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
A
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
MembersSolutionFBD of the loadings
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
MembersSolution
mkNkjiM
kji
M
WFXrM
M
kNkiF
kiF
F
A
A
wA
A
A
A
.}5.409.701.19{
0
376.605.13
25.530
0)(
;0
}38.65.13{
03475.65.13
;0
7.1 Internal Forces Developed in Structural
Members
7.1 Internal Forces Developed in Structural
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