6161103 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

� 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

� 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

� 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

� 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

� 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

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

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

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

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

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

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

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

� 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

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.

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

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

Solution

Segment AB

+↑∑ Fy = 0; 8kN – NB = 0

N = 8kNNB = 8kN

Segment DC

+↑∑ Fy = 0; NC – 4kN= 0

NC = 4kN

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.

Solution

Support Reactions

� Shaft subjected to only collinear torques

∑ Mx = 0;∑ Mx = 0;

-10N.m + 15N.m + 20N.m –TD = 0

TD = 25N.m

Solution

� FBD of shaft segments AB and CD

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

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.

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

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

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

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.

Solution

Support Reactions

� FBD of each member

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

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

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

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.

Solution

Support Reactions

� Members AC and CD are two force members

+↑∑ F = 0;+↑∑ Fy = 0;

Rsin45° – 600N = 0

R = 848.5N

Solution

� FBD of segment CE

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

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

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

Solution

� FBD of the loadings

Solution

}38.65.13{

03475.65.13

=−−

mkNkjiM

.}5.409.701.19{

376.605.13

25.530

}38.65.13{

++−=

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

6161103 7.1 internal forces developed in structural members

internal forces

internal shear

internal normalforce

determinethe internal

general internal force

resultant loadings

unknown loadings

unknown normaland shear

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