7 - Shear Stress - Vocational Training Counciltycnw01.vtc.edu.hk/cbe5029/7 - Shear Stress.pdfHong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering

Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS

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SHEAR STRESS Shear stress is stress parallel to the surface on which it acts.

τ = force/area = V/A where τ = the shear stress acting on the surface

V = is the shear force acting parallel to the surface A = the area on which the shear stress acts

It is assumed that the shear stress is uniformly distributed over the surface.


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Example Two pieces of plastic are jointed by gluing overlapping areas of 50 by 70 mm as shown in the following figure. If a tensile force of 780 N applied parallel to the glued surfaces causes the glue to fail, at what shear stress did the glue fail.

Solution The shear stress acting on the glued surface:

τ = force/area = V/A =780/(50*70*10-6) = 222860 Pa = 222.86 kPa


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Example A boiler plate steel has an ultimate shear strength of 290 MPa. Compute the force required to punch a 25 mm in diameter hole if the steel plate thickness is 13 mm. Assume the shear stress is uniformly distributed.

Solution The resisting shear area is the circumference of the punch multiplied by the thickness of the plate:

A = πdt = π*25*13 = 1021 mm2

We solve for P, the required applied force that will induce an ultimate shear stress of 290 MPa, by substituting the ultimate shear stress:

P = Aτult

= 1021 x10-6 * 290 x 106

= 296090 N = 296 kN


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SHEAR STRESS IN BEAM Consider a simply supported beam with a concentrated load at mid-span. If we cut the beam at any transverse cross-section, a shear force V exists at the section to maintain equilibrium. The shear force V is distributed in the form of vertical shear stresses acting over the face of the section.

An important feature of the vertical shear stresses is that they give rise to complementary horizontal shear stresses. At any point in a beam, the horizontal and vertical shear stresses are always numerically equal in magnitude.


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Shear Stress Formula for Beams The general formula for calculating the shear stress in a beam section which is subject to a shear force if given by

Where τ = Shear stress

S = Shear force acting at the section Ap = The cross-sectional area above the ‘imaginary cut’ (the

shaded area fghi in the diagram) y = Distance from neutral axis to centroid of the shaded area A I = Moment of inertia of the entire cross-section t = Width of the section (at the ‘imaginary cut’) Q = Statical moment (Ap* y )

Notes: 1. The vertical shear stress at a point is equal to the horizontal shear stress. 2. Using the area below the ‘imaginary cut’ for Ap should give the same

result.

tIQ S

tI

y pA S==τ


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Shear Stress Distribution across a Beam Section The vertical shear stresses are not uniform in magnitude over the face of the section. It can be shown that the shear stresses: • are highest at the neutral axis of the beam, • are zero at the free surface (i.e. the top and bottom surfaces of the beam),

and • varies with the distance from the neutral axis. The maximum shear stress in a section calculated by the use of the above formula occurs at the level of the neutral axis. In rectangular, T-shaped and I-shaped beams and other commonly occurring sections the shear stress varies parabolically throughout the depth of the section, with abrupt changes of stress where the geometry of the section changes suddenly, such as where the web and flanges of an I section meet.


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Example A rectangular beam 400 mm by 800 mm supports a shear force of 5 kN. Draw the shear stress distribution across the depth of the beam section.


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Solution The moment of inertia of the section with respect to the neutral axis is given by the formula.

I = bh3/12 = 400(800)3/12 = 17.07 x 109 mm4

S = 5 x 103 N t = 400 mm Level Ap

(106 mm2) y (mm)

S/I t (10-9 N/mm5)

τ (10-3 N/mm2)

1 0.16 200 0.7323 23.43 2 0.12 250 0.7323 21.97 3 0.08 300 0.7323 17.58 4 0.04 350 0.7323 10.25 5 0 - 0.7323 0


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Methods for Calculating the Statical Moment Q

For a rectangular beam section,

1. Locate the neutral axis for the entire cross section by computing the

location of the centroid. 2. Draw in the axis where the shear stress is to be calculated. 3. Identify the partial area Ap away from the axis of interest the shade it for

emphasis. 4. Compute the magnitude of Ap. 5. Locate the centroid of the partial area. 6. Compute Q = Ap y .


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For T-shaped and I-shaped beam sections

ApyCentroidalaxis of Ap

y1y2

A

A

1

2

tN.A.

b

h


y1y2

A

A

1

2

t

b

hN.A.


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1. Locate the neutral axis for the entire cross section by computing the

location of the centroid. 2. Draw in the axis where the shear stress is to be calculated. 3. Divided Ap into component parts, which are simple areas and label them

A1, A2, and so on. Compute their values. 4. Locate the centroid of each component area. 5. Determine the distances from the neutral axis to the centroid of each

component area, calling them y1, y2, and so on. 6. Note that, by the definition of the centroid.

Ap y = A1y1 + A2y2 +…

Now, because Q = Ap y , the most convenient way to calculate Q is

Q = A1y1 + A2y2 + …


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Example Find the maximum shear stress induced in the following beam.


y 1y 2

A

A

1

2

25 mm

200 mm

25 mm150 mm

N.A.

4.5 kN 9.0 kN

AB

3m1.5m 1.5m

5.625 kN 7.875 kN

CD


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Solution: Determine the shear force diagram.

Determine the location of the centroidal axis.

y = (150*25*12.5 + 200*25*125)/(150*25 + 200*25) y = 76.8 mm

Determine the moment of inertia of the section.

I =150*253/12 + 150*25*(76.8-12.5)2 +25*2003/12 + 25*200*(125-76.8)2

I = 43.98 *106 mm4

Determine the value of Q.

Consider the shaded area,

Q = A1y1 + A2y2 = 150*25*(76.8-12.5) + 25*(76.8-25)*(76.8-25)/2 = 0.2745 * 106 mm3

A B C D

5.625 kN

1.125 kN

-7.875 kN

Shear ForcekN


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Consider the unshaded area, Q = 25*(225-76.8)*(225-76.8)/2

= 0.2745 * 106 mm3 Both values of Q are the same. Considering of the unshaded area of the section gives simpler calculation. The maximum shear stress,

τmax = 7.875*0.2745*106/(43.98 *106*25) = 1.966*10-3 kN/mm2


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Example A simply supported beam AB is subject to a uniformly distributed load of 28 kN/m as shown in the following figure. The cross section of the beam is rectangular with width 25 mm and height 100 mm. Calculate the shear stress acting at the point C (200 mm from the support B) in the beam AB.

A B

1m

28 kN/m 25 mm

25 mm

C

200 mm

C 100

mm

Solution Shear force at point C, V = -28*1/2 + 28*0.2 = -8.4 kN Moment inertia of the section, I = 25*1003/12 = 2083*103 mm4


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Q = 25*25*37.5 = 23440 mm3 Shear stress at point C, τ = 8400*23440/(2083*103*25) = 3.8 N/mm2

= 3.8 MPa Example An I-shaped beam is subject to a shear force of 100 kN. Plot a curve to show the variation of shear stress across the section of the beam and hence determine the ratio of the maximum shear stress to the mean shear stress. Solution

100 mm

12 mm

126 mm

12 mm

12 mm

44 mm


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Moment of inertia of the section, I = 100*1503/12 – 2*44*1263/12 = 13.46*106 mm4 The distribution of shear stress across the section is τ = 100*103 *A y /(13.46*106 * t) = 7.43*10-3 A y /t N/mm2 Section A

(mm2) y (mm)

t (mm)

τ (N/mm2)

0 0 - - - 1 100*6=600 72 100 3.2 2 100*12=1200 69 100 6.2 2 1200 69 12 51.3 3 1320 68 12 55.6 4 1440 66.3 12 59.1 5 1680 61.6 12 64.1 6 1956 54.5 12 66.0 It should be noted that two values of shear stresses are required at section 2 to take account of the change in breadth at this section. The values of A and Y for sections 3,4,5 and 6 are those of a I-section beam.


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The mean shear stress across the section is:

τ mean = shear force / total area = 100*103 /3.912*103

= 25.6 N/mm2

Max. shear stress/mean shear stress = 66/25.6 = 2.58

100 mm

12 mm

126 mm

12 mm

44 mm

53 mm43 mm

23 mm

01234

5

6

Shear stress distribution


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Example The following beam is simply supported at its ends and carries a point load of 500 kN at mid-span and a uniformly distributed load of 300 kN/m over the entire span of 3 m. Draw the shear stress distribution diagram for a section 1 m from the left-hand support.

A B

3 m

300 kN/m

1 m

C

500 kN

700 kN 700 kN

100 mm

100 mm

25 mm25 mm

50 mm

50 mm

Solution Shear force at C, V = 700 – 300*1 = 400 kN Distance of centroidal axis from the bottom surface of the section,

y = (50*25*25*2 + 100*50*75)/(100*50+25*50*2) = 58.4 mm Moment inertia of the section, I = 2*25*503/12 +2*25*50*(58.4-25)2 + 100*503/12 + 100*50*(75-58.4)2 = 5.72 *106 mm4


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Therefore, τ = 400*103 *A y /(5.72*106*t) = 7*10-2 *A y /t

Section A (mm2)

y (mm)

t (mm)

τ (N/mm2)

0 0 - - 0 1 1500 34.1 100 35.8 2 3000 26.6 100 55.8 3 4160 20.8 100 60.6 4 2500 33.4 100 58.4 4 2500 33.4 50 116.8 5 2000 38.4 50 107.5 6 1000 48.4 50 67.7 7 0 - - 0

50 mm

0

1

2

345

6

7

15 3041.6

50 4020

Shear stress distribution


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PROBLEMS for shear stress


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Documents

7 - Shear Stress - Vocational Training Counciltycnw01.vtc.edu.hk/cbe5029/7 - Shear Stress.pdfHong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering