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.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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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
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
Page 7 -4
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
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
Page 7 -5
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.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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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 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
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
Page 7 -9
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.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
Page 7 -10
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.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
Page 7 -11
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
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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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 .
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
Page 7 -13
For T-shaped and I-shaped beam sections
ApyCentroidalaxis of Ap
y1y2
A
A
1
2
tN.A.
b
h
ApyCentroidalaxis of Ap
y1y2
A
A
1
2
t
b
hN.A.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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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 + …
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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Example Find the maximum shear stress induced in the following beam.
ApyCentroidalaxis of Ap
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
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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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
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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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
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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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
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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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
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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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.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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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
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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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
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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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
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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PROBLEMS for shear stress
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 7 – SHEAR STRESS
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