2005 pearson education south asia pte ltd 7. transverse shear 1 chapter objectives develop a method...

2005 Pearson Education South Asia Pte Ltd

7. Transverse Shear

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CHAPTER OBJECTIVES

• Develop a method for finding the shear stress in a beam having a prismatic cross-section and made from homogeneous material that behaves in a linear-elastic manner

• This method of analysis is limited to special cases of cross-sectional geometry


7. Transverse Shear

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CHAPTER OUTLINE

1. Shear in Straight Members

2. The Shear Formula

3. Shear Stresses in Beams


7. Transverse Shear

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7.1 SHEAR IN STRAIGHT MEMBERS

• Shear V is the result of a transverse shear-stress distribution that acts over the beam’s cross-section.

• Due to complementary property of shear, associated longitudinal shear stresses also act along longitudinal planes of beam


7. Transverse Shear

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• As shown below, if top and bottom surfaces of each board are smooth and not bonded together, then application of load P will cause the boards to slide relative to one another.

• However, if boards are bonded together, longitudinal shear stresses will develop and distort cross-section in a complex manner

Boards not bonded together Boards bonded together


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• As shown, when shear V is applied, the non-uniform shear-strain distribution over cross-section will cause it to warp, i.e., not remain in plane.

Before deformationBefore deformation

After deformation


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7.2 THE SHEAR FORMULA• By first principles, flexure formula and V = dM/dx, we obtain

=VQ

ItEq. 7-3

= shear stress in member at the point located a distance y’ from the neutral axis. Assumed to be constant and therefore averaged across the width t of member

V = internal resultant shear force, determined from method of sections and equations of equilibrium


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7.2 THE SHEAR FORMULA

I = moment of inertia of entire cross-sectional area computed about the neutral axis

t = width of the member’s cross-sectional area, measured at the point where is to be determined

Q = ∫A’ y dA’ = y’A’, where A’ is the top (or bottom) portion of member’s cross-sectional area, defined from section where t is measured, and y’ is distance of centroid of A’, measured from neutral axis

=VQ

ItEq. 7-3


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7.2 THE SHEAR FORMULA

• The equation derived is called the shear formula

• Since Eqn 7-3 is derived indirectly from the flexure formula, the material must behave in a linear-elastic manner and have a modulus of elasticity that is the same in tension and in compression

• Shear stress in composite members can also be obtained using the shear formula

• To do so, compute Q and I from the transformed section of the

member as discussed in section 6.6. Thickness t in formula

remains the actual width t of cross-section at the pt where is to be calculated


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7.3 SHEAR STRESSES IN BEAMS

Rectangular cross-section

• Consider beam to have rectangular cross-section of width b and height h as shown.

• Distribution of shear stress throughout cross-section can be determined by computing shear stress at arbitrary height “y” from neutral axis, and plotting the function. Hence,

Q = y’A’ = y2 b1

2

h2

4( )


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Rectangular cross-section

= y2( ) 6V

bh3

h2

4Eq. 7-4

• Eqn 7-4 indicates that shear-stress distribution over cross-section is parabolic.

• After deriving Q and applying the shear formula, we have

Shear stress distribution


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max = 1.5V

AEq. 7-5

• By comparison, max is 50% greater

than the average shear stress

determined from avg = V/A.

• At y = 0, we have

= y2( ) 6V

bh3

h2

4Eq. 7-4


7. Transverse Shear

EXAMPLE-1

The beam shown in the figure is made of wood and is subjected to a resultant vertical shear force of V = 3 kN.

Determine:

a)The shear stress in the beam at point P.

b)The maximumshear stress in the beam.


7. Transverse Shear

EXAMPLE-1

Section Properties:Moment of inertia about the neutral axis

I = = = 16.28x106 mm4bh3

12

(100)(125)3

12

A horizontal section line is drawn through point P and the partial area is shown shaded. Hence, the first moment with respect to the neutral axis is

'A'yQ [12.5+(1/2)(50)][(50)(100)] mm3

= 18.75x104 mm4

Part a)


7. Transverse Shear

EXAMPLE-1

Shear stress at point P:

P = = = 0.346 N/mm2VQ

I t

(3x103)(18.75x104)

(16.28x106)(100)

= 0.346 MPa


7. Transverse Shear

EXAMPLE-1

Maximum shear stress occurs at the neutral axis. Hence, the first moment with respect to the neutral axis is largest, thus

'A'yQ [62.5/2](100)(62.5)= 19.53x104 mm4

Part b)

Shear stress maximum:

max = = = 0.36 N/mm2VQ

I t

(3x103)(19.53x104)

(16.28x106)(100)

= 0.36 MPa

Note that this equivalent to max = 1.5 = 1.5 = 0.36 N/mm2V

A3x103

(100)(125)


7. Transverse Shear

EXAMPLE-2

The steel rod has radius of 30 mm. If it is subjected to a shear force of V = 25 kN, determine the maximum shear stress.

Section Property:

Moment of inertia about the neutral axis

I = r4 = (304) = 0.636x106 mm44

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7. Transverse Shear

Maximum shear stress occurs at the neutral axis. Hence, the height of the centroid is

'y r = (30) = 12.73 mm4

34

3

The first moment with respect to the neutral axis is largest, thus

'A'yQ [12.73][(302)/2]= 18x103 mm4

Shear stress maximum:

max = = = 11.8 N/mm2VQ

I t

(25x103)(18x103)

(0.636x106)(60)

= 11.8 MPa

EXAMPLE-2

y’_

Neutral axis

V

2005 pearson education south asia pte ltd 7. transverse shear 1 chapter objectives develop a method...

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