Internal Forces 7 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010 Pearson...

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Internal Forces77

Engineering Mechanics: Statics in SI Units, 12e

Chapter Objectives

• Method of sections for determining the internal loadings in a member

• Develop procedure by formulating equations that describe the internal shear and moment throughout a member

• Analyze the forces and study the geometry of cables supporting a load

Chapter Outline

1. Internal Forces Developed in Structural Members

2. Shear and Moment Equations and Diagrams

3. Relations between Distributed Load, Shear and Moment

4. 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

• Force component N, acting normal to the beam at the cut session

• V, acting tangent to the session are normal 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 components

• My is the torisonal or twisting moment, and Mx and Mz are the bending moment components

Procedure for AnalysisSupport Reactions• Before cut, determine the member’s support reactions• Equilibrium equations used to solve internal loadings

during sectioning

Free-Body Diagrams• Keep all distributed loadings, couple moments and

forces acting on the member in their exact locations• After session draw FBD of the segment having the

least loads

Procedure for AnalysisFree-Body Diagrams (Continue)• Indicate the z, y, z components of the force, couple

moments and resultant couple moments on FBD• Only N, V and M act at the section• Determine the sense by inspection

Equations of Equilibrium• Moments should be summed at the section • If negative result, the sense is opposite

Example 7.3

Determine the internal force, shear force and the bending moment acting at point B of the two-member frame.

Solution

Support ReactionsFBD of each memberMember 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

Solution

Support Reactions

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

7.2 Shear and Moment Equations and Diagrams

• Beams – structural members designed to support loadings perpendicular to their axes

• A simply supported beam is pinned at one end and roller supported at the other

• A cantilevered beam is fixed at one end and free at the other

Procedure for AnalysisSupport Reactions• Find all reactive forces and couple moments acting on

the beam• Resolve them into components

Shear and Moment Reactions• Specify separate coordinates x• Section the beam perpendicular to its axis• V obtained by summing the forces perpendicular to

the beam• M obtained by summing moments about the sectioned

Procedure for AnalysisShear and Moment Reactions (Continue)• Plot (V versus x) and (M versus x)• Convenient to plot the shear and the bending moment

diagrams below the FBD of the beam

Example 7.7

Draw the shear and bending moments diagrams for the shaft. The support at A is a thrust bearing and the support at C is a journal bearing.

Solution

Support Reactions

FBD of the shaft

mxkNMM

.5.2;0

xkNmxkNMM

VkNkNFy

.)5.210(

0)(5.2)2(5 ;0

055.2 ;0

Solution

Shear diagram

Internal shear force is always positive within the shaft AB.

Just to the right of B, the shear force changes sign and remains at constant value for segment BC.

Moment diagram

Starts at zero, increases linearly to B and therefore decreases to zero.

7.3 Relations between Distributed Load, Shear and Moment

Distributed Load• Consider beam AD subjected to an arbitrary load

w = w(x) and a series of concentrated forces and moments

• Distributed load assumed positive when loading acts downwards

Distributed Load• A FBD diagram for a small segment of the beam

having a length ∆x is chosen at point x along the beam which is not subjected to a concentrated force or couple moment

• Any results obtained will not apply at points of concentrated loadings

• The internal shear force and bending moments assumed positive sense

Distributed Load• Distributed loading has been replaced by a resultant

force ∆F = w(x) ∆x that acts at a fractional distance k (∆x) from the right end, where 0 < k <1

0)()(;0

xkxwxVM

MMxkxxwMxVM

VVxxwVFy

Distributed Load

)(xwdx

dVSlope of the

shear diagramNegative of distributed

load intensity

Slope of shear diagram

Shear moment diagram

VdxM BC

dxxwVBC )(

Change in shearArea under

shear diagram

Change in moment Area under shear diagram

Force and Couple Moment• FBD of a small segment of the beam• Change in shear is negative• FBD of a small segment of the beam located at the

couple moment• Change in moment is positive

Example 7.9

Draw the shear and moment diagrams for the overhang beam.

Solution

The support reactions are shown.

Shear Diagram Shear of –2 kN at end A of the beam is at x = 0.

Positive jump of 10 kN at x = 4 m due to the force.

Moment Diagram mkN 8420

7.4 Cables

• Cables and chains used to support and transmit loads from one member to another

• In force analysis, weight of cables is neglected • Assume cable is perfectly flexible and inextensible• Due to its flexibility cables has no resistance to bending• Length remains constant before and after loading

7.4 Cables

Cable Subjected to Concentrated Loads

• For a cable of negligible weight, it will subject to constant tensile force

• Known: h, L1, L2, L3 and loads P1 and P2

• Form 2 equations of equilibrium • Use Pythagorean Theorem to relate the three

segmental lengths

Example 7.11

Determine the tension in each segment of the cable.

Example 7.11

FBD for the entire cable.

EkNkNkNkN

mknmkNmkNmA

0315412

0)2(3)10(15)15(4)18(

Example 7.11

Consider leftmost section which cuts cable BC since sag yC = 12m.

mkNmkNmA

2.10,6.51

0sin412

033.6cos

0)5(4)8(12)12(

7.4 Cables

Cable Subjected to a Distributed Load• Consider weightless cable subjected to a loading

function w = w(x) measured in the x direction

7.4 Cables

Cable Subjected to a Distributed Load• For FBD of the cable having length ∆• Since the tensile force changes continuously, it is

denoted on the FBD by ∆T• Distributed load is represented by second integration,

dxdxxwF

Example 7.12

The cable of a suspension bridge supports half of the uniform road surface between the two columns at A and B. If this distributed loading wo, determine the maximum force developed in the cable and the cable’s required length. The span length L and, sag h are known.

Solution

Note w(x) = wo

Perform two integrations

Boundary Conditions at x = 0

Therefore,

Curve becomes

This is the equation of a parabola

Boundary Condition at x = L/2

dxdxwF

0/,0,0 dxdyxy

021 CC

Solution

For constant,

Tension, T = FH/cosθ

Slope at point B

Therefore

Using triangular relationship

2 4 and

LwF oH

LwFT oH

)cos( maxmax

2tantan 1

Solution

For a differential segment of cable length ds,

Determine total length by integrating,

Integrating yields,

hL 4sinh

dydydxds

7.4 Cables

Cable Subjected to its Own Weight• When weight of the cable is considered, the loading

function becomes a function of the arc length s rather than length x

• FBD of a segment of the cable is shown

7.4 Cables

Cable Subjected to its Own Weight• Apply equilibrium equations to the force system

• Replace dy/dx by ds/dx for direct integration

dsswFdx

dydxdsT

7.4 Cables

Cable Subjected to its Own Weight• Therefore

• Separating variables and integrating

1 dsswF

dsswFdx

Example 7.13

Determine the deflection curve, the length, and the maximum tension in the uniform cable. The cable weights wo = 5N/m.

Solution

For symmetry, origin located at the center of the cable.

Deflection curve expressed as y = f(x)

Integrating term in the denominator

2/11 CswF

2/122/11 dswF

Solution

Substitute

So that

Perform second integration

1/1 CswFu oH

dsFwdu Ho )/(

21sinh Cu

sinh CCswFw

Solution

Evaluate constants

dy/dx = 0 at s = 0, then C1 = 0

To obtain deflection curve, solve for s

dswFdx

H sinh

Solution

Boundary Condition y = 0 at x = 0

For deflection curve,

This equations defines a catenary curve.

3cosh CxF

1cosh x

Solution

Boundary Condition y = h at x = L/2

Since wo = 5N/m, h = 6m and L = 20m,

By trial and error,

1cosh x

50cosh

NFH 9.45

Solution

For deflection curve,

x = 10m, for half length of the cable

Maximum tension occurs when θ is maximum at

s = 12.1m

1.12109.45

/5sinh

1109.0cosh19.9

Solution

9.758.52cos

32.19.45

1.12/5tan

maxmax

max1.12

1. In a multi-force member, the member is generally subjected to an internal _________.

A) Normal force B) Shear force

C) Bending moment D) All of the above.

2. In mechanics, the force component V acting tangent to, or along the face of, the section is called the _________ . A) Axial force B) Shear force

C) Normal force D) Bending moment

3. A column is loaded with a vertical 100 N force. At which sections are the internal loads the same?

A) P, Q, and R B) P and Q

C) Q and R D) None of the above.

4. A column is loaded with a horizontal 100 N force. At which section are the internal loads largest?

A) P B) Q

C) R D) S

5. Determine the magnitude of the internal loads

(normal, shear, and bending moment) at point C.

A) (100 N, 80 N, 80 N m)

B) (100 N, 80 N, 40 N m)

C) (80 N, 100 N, 40 N m)

D) (80 N, 100 N, 0 N m )

2. A column is loaded with a horizontal 100 N force. At which section are the internal loads the lowest?

A) P B) Q

C) R D) S

Internal Forces 7 Engineering Mechanics: Statics in SI Units, 12e Copyright © 2010 Pearson...

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