6161103 7.1 internal forces developed in structural members
TRANSCRIPT
7.1 Internal Forces Developed in Structural Members
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 memberthe 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 twothe 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 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 momentcomponents 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 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 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 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 componentsthe 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 Members
� For most applications, these resultant loadings will act at the geometric center or centroid (C) of the section’s cross sectional areacross 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 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 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 Members
Free Body Diagrams
� FBD of the sectioned frame
� At each sectioned member, there is an unknown normal force, shear force and bending momentnormal 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
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 unknownsequations 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 Members
Procedure for AnalysisSupport Reactions
� Before the member is cut or sectioned, determine the member’s support reactionsdetermine 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 Members
Procedure for Analysis
Free-Body Diagrams
� Keep all distributed loadings, couple moments and forces acting on the member 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 Members
Procedure for Analysis
Free-Body Diagrams
� Indicate the z, y, z components of the force and couple moments and the resultant couple 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 Members
Procedure for AnalysisEquations of Equilibrium
� Moments should be summed at the section about the axes passing through the centroid or 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 Members
� The link on the backhoe is a two force member
� It is subjected to both bending and axial load at its 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 Members
Example 7.1
The bar is fixed at its end and is
loaded. Determine the internal normal 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 Members
Solution
Support Reactions
� FBD of the entire bar
By inspection, only normal force A� 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 Members
Solution
� FBD of the sectioned bar
� No shear or moment act on the sections since they are 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 Members
Solution
Segment AB
+↑∑ Fy = 0; 8kN – NB = 0
N = 8kNNB = 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 Members
Example 7.2
The circular shaft is subjected to three
concentrated torques. Determine the internal 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 Members
Solution
Support Reactions
� Shaft subjected to only collinear torques
∑ Mx = 0;∑ 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 Members
Solution
� FBD of shaft segments AB and CD
7.1 Internal Forces Developed in Structural Members
7.1 Internal Forces Developed in Structural Members
Solution
Segment AB
∑ Mx = 0; -10N.m + 15N.m – TB = 00
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 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 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 Members
Solution
Support Reactions
� 9kN.m is a free vector and can be place � 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
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 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 Members
SolutionSegment AB
+→∑ Fx = 0; NB = 0+↑∑ Fy = 0; 5kN – VB = 0
VB = 5kNVB = 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 Members
Example 7.4
Determine the internal force, shear force and
the bending moment acting at point B of the
two-member frame.two-member frame.
7.1 Internal Forces Developed in Structural Members
7.1 Internal Forces Developed in Structural Members
Solution
Support Reactions
� FBD of each member
7.1 Internal Forces Developed in Structural Members
7.1 Internal Forces Developed in Structural Members
Solution
Member AC
∑ MA = 0; -400kN(4m) + (3/5)FDC(8m)= 0
F = 333.3kNFDC = 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 Members
Solution
� FBD of segments AB and BC
� Important to keep distributed loading � 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 Members
Solution
Member AB
+→∑ Fx = 0; NB – 266.7kN = 0
N = 266.7kNNB = 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 Members
Example 7.5
Determine the normal force,
shear force and the bending 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 Members
Solution
Support Reactions
� Members AC and CD are two force members
+↑∑ F = 0;+↑∑ Fy = 0;
Rsin45° – 600N = 0
R = 848.5N
7.1 Internal Forces Developed in Structural Members
7.1 Internal Forces Developed in Structural Members
Solution
� FBD of segment CE
7.1 Internal Forces Developed in Structural Members
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
N = 600 Ny E
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
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 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 Members
Solution� Idealized model for the sign� Consider FBD of a section above
A since it dies not involve the support reactionssupport reactions
� Sign has weight of W = 650(9.81) = 6.376kN
� Wind creates resultant forceFw = 900N/m2(6m)(2.5m)
= 13.5kN
7.1 Internal Forces Developed in Structural Members
7.1 Internal Forces Developed in Structural Members
Solution
� FBD of the loadings
7.1 Internal Forces Developed in Structural Members
7.1 Internal Forces Developed in Structural Members
Solution
kNkiF
kiF
F
A
}38.65.13{
03475.65.13
;0
rrr
rrr
r
+=
=−−
=∑
mkNkjiM
kji
M
WFXrM
M
kNkiF
A
A
wA
A
A
.}5.409.701.19{
0
376.605.13
25.530
0)(
;0
}38.65.13{
rrrr
rrr
r
rrrr
r
rrr
++−=
=
−
+
=++
=∑
+=
7.1 Internal Forces Developed in Structural Members
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
M = {40.5k}kN represents the torisonal momentMAz = {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
rrr+=