lecture 3 - shear stresses

7/29/2019 LECTURE 3 - Shear Stresses

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Beams -Horizontal Shear Stress

In addition to the bending (axial) stress which develops in a loaded

beam, there is also a shear stress which develops, including both aVertical Shear Stress, and a Horizontal (longitudinal) Shear

Stress. It can be shown that at any given point in the beam, the

values of vertical shear stress and the horizontal shear stress must beequal, at that point, for static equilibrium. As a result it is usual to

discuss and calculate the horizontal shear stress in a beam (and simplyremember that the vertical shearing stress is equal in value to the

horizontal shear stress at any given point).We will take a moment to

derive the formula for the Horizontal Shear Stress. In Diagram 1, wehave shown a simply supported loaded beam.

In Diagram 2a, we have cut a section dx long out of the left end of the

beam, and have shown the internal horizontal forces acting on the

section.


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In Diagram 2b, we have shown a side view of section dx. Notice thatthe bending moment is larger on the right hand face of the section by

an amount dM. (This is clear if we make the bending moment diagramfor the beam, in which we see the bending moment increases from a

value of zero at the left end to a maximum at the center of the beam.)

In Diagram 2c, we have shown a top slice of section dx. Since theforces are different between the top of the section and the bottom of

the section (less at the bottom) there is a differential (shearing) force

which tries to shear the section, shown in Diagram 2c, horizontally.This means there is a shear stress on the section, and in terms of the

shear stress, the differential shearing force, F, can be written as F= times the longitudinal area of the section (b dx). A second way of

expressing the shear force is by expressing the forces in terms of the

bending stress, that is F1 = (My/I) dA, and F2 = (M+dM)y/I dA,

then the differential force is (dM y/I)dA. If we now combine the two

F = expressions, we have:

F = * b dx = (dM y/I)dA, and then rewriting to solve for the

shear stress:

= [(dM/dx)/Ib] y dA, however dM/dx is equal to the shear force

V (as discussed in the previous topic), and y dA is the first moment

of the area of the section, and may be written as A y', where A is thearea of the section and y' is the distance from the centroid of the areaA to the neutral axis of the beam cross section. Rewriting in a final

form we have:


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Horizontal Shear Stress: = VAy'/Ib, whereV = Shear force at location along the beam where we wish to

find from the horizontal shear stressA = cross sectional area, from point where we wish to find the

shear stress at, to an outer edge of the beam cross section (top

or bottom)y' = distance from neutral axis to the centroid of the area A.

I = moment of inertia for the beam cross section.b = width of the beam at the point we wish to determine the

shear stress.

(In some texts, the product Ay' is given the symbol Q and used in theshear stress equation)

If we consider our shear relationship a little, we observe that the

Horizontal Shear Stress is zero at the outer edge of the beam -

since the area A is zero there. The Horizontal Shear Stress is(normally) a maximum at the neutral axis of the beam. This is

the opposite of the behavior of the Bending Stress which is maximumat the other edge of the beam, and zero at the neutral axis.

To help clarify the Horizontal Shear Stress equation we will now look atat several example of calculating the Horizontal Shear Stress.

Example 1


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In Diagram 1, we have shown a simply supported 20 ft. beam with aload of 10,000 lb. acting downward at the center of the beam. The

beam used is a rectangular 2" by 4" steel beam. We would like todetermine the maximum Horizontal Shear Stress which develops in the

beam due to the loading. We will also determine the Horizontal Shear

Stress 3 inches above the bottom of the beam at the position in thebeam where the shear force is a maximum.

Step 1: Out first step in solving this problem is to apply staticequilibrium conditions to determine the external support reactions. In

this particular example, because of the symmetry of the problem, wewill not go through the statics in detail, but point out that the two

support forces will support the load at the center equally with forces of5000 lb. each as shown in Diagram 2.

Step 2: The second step is to draw the shear force and bendingmoment diagrams for the beam.

Step 3. We will now apply the Horizontal Shear Stress formula:

Shear Stress = Vay'/Ib


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We wish to find the maximum shear stress, which occurs at theneutral axis of the beam:

V = maximum shear force = 5,000 ft-lb. (from the shear force

diagram)

I = moment of inertia of cross section; for rectangleI = (1/12) bd3 = 1/12 (2" * 4"3) = 10.67 in4.

b = width of beam section where we wish to find shear stressat; b= 2 in.

a = area from point we wish to find shear stress at (neutral

axis) to an outer edge of beama= (2" x 2")= 4 in2.

y' = distance from neutral axis to the centroid of the area "a"which we used; y'= 1 in.

(See Diagram 4)

Placing the values into the equation, we find:MaximumHorizontal Shear Stress = Vay'/Ib = (5000 lb)*(4

in2)*(1 in)/(10.67 in4)(2 in)= 937 lb/in2

Part II We now would also like to determine the Horizontal ShearStress 3 inches above the bottom of the beam at the position in the

beam where the shear force is a maximum (which is actually through

out the beam, since the value of shear force is either +5000 lb., or -5000 lb. through out the beam.)We again apply the Horizontal Shear Stress formula: Horizontal

Shear Stress = Vay'/Ib

We wish to find the shear stress 3 inches above the bottom of thebeam cross section. (See Diagram 5)

V = shear force = 5,000 ft-lb. (from the shear force diagram)


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I = moment of inertia of cross section; for rectangle I = (1/12)

bd3 = 1/12 (2" * 4"3) = 10.67 in4.b = width of beam section where we wish to find shear stress

at; b = 2 in.

a = area from point we wish to find shear stress at (3" above

bottom of the beam) to an outer edge of beam. We will go tothe top edge of the beam, then a = (2" x 1")= 2 in2.y' = distance from neutral axis to the centroid of the area "a"

which we used; y'= 1.5 in. (See Diagram 5)

Then the horizontal shear stress 3 inches above the bottom of thebeam is:

Horizontal Shear Stress = Vay'/Ib = (5000 lb)*(2 in2)*(1.5in)/ (10.67 in4)(2 in)= 703 lb/in2

Notice, as we expect, the horizontal shear stress value becomessmaller as we move toward the outer edge of the beam cross section.

Horizontal Shear Stress - Example 2

A loaded, simply supported beam is shown in Diagram 1. For two

different beam cross sections (a WT 8 x 25 T-beam, and a W 10 x 45beam) we will determine the maximum Horizontal Shear Stress which

would develop in the beam due to the loading. We will also determinethe Horizontal Shear Stress 3 inches above the bottom of the beam at

the position in the beam where the shear force is a maximum.


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STEP 1: Apply Static Equilibrium Principles and determine the external

support reactions:1.) FBD of structure (See Diagram 2)

2.) Resolve all forces into x/y components

3.) Apply equilibrium conditions:Sum Fx = 0 none

Sum Fy = By + Cy - 1,000 lbs/ft (4 ft) - 1,500 lbs/ft (4 ft) = 0

Sum TB = 1,000 lbs/ft (4 ft) (2 ft) - 1,500 lbs/ft (4 ft) (8 ft) +Cy(6 ft) = 0

Solving: By = 3,330 lb.; Cy = 6,670 lb.

Step 2: The second step is to draw the shear force and bendingmoment diagrams for the beam


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Beam Table for WT 8 x 25

Designation Area of T Width thick thick -x-x

axis

x-x

axis

x-x

axis

x-x

axis

- A d bf tf tw d/tw I S r y

- in2 in in in in - in4 in3 in in

WT 8x25 7.36 8.13 7.073 0.628 0.380 21.40 42.20 6.770 2.400 1.890

Step 3. For the WT 8 x 25 T-beam (table above) we will now apply

the Horizontal Shear Stress formula:

Shear Stress = Vay'/Ib, to find the maximum shear stress, which

occurs at the neutral axis of the beam:

V = maximum shear force = 6,000 lb. (from the shear force

diagram)I = moment of inertia of cross section, from beam table; I =42.20 in4.

b = width of beam where we wish to find shear stress (neutral

axis for maximum) from table; b = .38 in.a = area from point we wish to find shear stress at (neutral

axis) to an outer edge of beam. In this case we will go tobottom of beam. Then a = (.38" * 6.24" )= 2.37 in2.


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y' = distance from neutral axis to the centroid of the area "a" ;y' = 3.12 in. (See Diagram 5) Then placing values into our

expression we find:Maximum Horizontal Shear Stress= Vay'/Ib = (6000 lb)*(2.37

in2)*(3.12 in)/ (42.20 in4)(.38 in) = 2770 lb/in2

We now would also like to determine the Horizontal Shear Stress 3inches above the bottom of the beam at the position in the beam

where the shear force is a maximumWe again apply the Horizontal Shear Stress formula: Shear Stress =

Vay'/Ib

We wish to find the shear stress 3 inches above the bottom of the

beam cross section, where the shear force is a maximum. (SeeDiagram 6)\

V = maximum shear force = 6,000 lb. (from the shear force

diagram)

I = moment of inertia of cross section, from beam table; I =42.20 in4.

b = width of beam where we wish to find shear stress (3"above bottom of beam) from table; b = .38 in.

a = area from point we wish to find shear stress at (neutral

axis) to an outer edge of beam. In this case we will go tobottom of beam. Then a = (.38" * 3" )= 1.14 in2. (See Diagram

6)y' = distance from neutral axis to the centroid of the area "a" ;


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y' = 4.74 in. (See Diagram 6)Then the horizontal shear stress 3 inches above the bottom of the

beam is:Horizontal Shear Stress = Vay'/Ib = (6000 lb)*(1.14 in2)*(4.74

in)/(42.2 in4)(.38 in)= 2020 lb/in2

Notice, as we expect, the horizontal shear stress value becomes

smaller as we move toward an outer edge of the beam cross section.

Part 2: W 10 x 45 beam: We would like to again determine the

maximum horizontal shear stress, and the shear stress 3 inches abovethe bottom of the beam (at the point where the shear force is a

maximum), but now find these values for a W 10 x 45 I-beam.

Beam Table for W 10 x 45 I-Beam

- - - Flange Flange Web Cross Section Info. Cross Section Info.

Designation Area Depth Width thick thickx-x

axisx-x axis

x-x

axis

y-y

axisy-y axis

y-y

axis

- A d bf tf tw I S r I S r

- in2 in in in in in4 in3 in in4 in3 in

W 10 x 45 13.20 10.12 8.022 0.618 0.350 249.0 49.1 4.33 53.20 13.30 2.00

We have already done the statics, and the shear force and bendingmoment diagrams are shown in the first part of this example above, so

we continue at the point where we apply the horizontal shear stress

formula to find the values we desire.For the WT 8 x 25 T-beam we apply the Horizontal Shear Stress

formula: Shear Stress = Vay'/Ib, however since we are looking forthe maximum shear stress in the I-Beam, we can use the


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approximate formula for I-beam, Maximum Horizontal ShearStress = Vmax/Aweb. This says the approximate maximum shear stress

in an I - Beam is equal to the maximum shear force divided by thearea of the web of the I-Beam. Applying this we have:

Vmax = maximum shear force = 6,000 lb. (from the shear force

diagram)Amax = area of web: A = (.35" * 8.88" )= 3.11 in

2. (See Diagram 7)

Then Maximum Horizontal Shear Stress = (6000 lb)/(3.11 in2)= 1930 lb/in2

As long as this approximate value is reasonably below the allowableshear stress for the beam material there is no need to use the exact

formula for the maximum shear stress. Please remember, however,the approximate formula is only for the maximum horizontal shear

stress (which occurs are the neutral axis) in an I-Beam. If we need to

know the shear stress at any other location, we must use the standardformula as we will do in the next part.


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We now wish to find the shear stress 3 inches above the bottom of thebeam cross section, where the shear force is a maximum. (See

Diagram 8). To do so, we apply the standard horizontal shear stressformula: Shear Stress = Vay'/Ib

V= maximum shear force =6,000 lb. (from the shear force diagram)

I= moment of inertia of cross section, from beam table; I = 249.0

in4.b = width of beam where we wish to find shear stress (3" above

bottom of beam) from table; b = .35 in.

a = area from point we wish to find shear stress at (3" from the

bottom) to an outer edge of beam. In this case we will go to bottom ofbeam. Notice that the area is composed of the area of the flange (A1)and part of the area of the web (A2). (See Diagrams 8 and 9.) Thena

= (A1 + A2)= (.618" x 8.022") + (2.383 in2 x .35in2) = 4.96 in2+ .834 in2= 5.794 in2 (See Diagram 9)

y' = distance from neutral axis to the centroid of the area "a" Notice

that in this case, for an I-Beam, this is not a entirely simple matter.The area we wish to find the centroid of is not a simple rectangle, but

rather two rectangles. To find the centroid of this compound area we

use: y' = (A1 y1 + A2 y2)/(A1 +A2); where A1 and A2 are the twoareas, and y1 and y2 are the distances from the neutral axis of the


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beam to the centroid of each of the respective areas. (Which is simplythe distance from the neutral axis to the center of each of the

respective areas, since for a rectangle the centroid is at the center.)Using the values shown in Diagram 9, we have:

y' = (A1 y1 + A2 y2)/(A1 +A2) = (4.96 in2 x 4.75 in + .834 in2 x 2.94in)/(4.96 in2 + .834 in2) = 4.49 in.Then the horizontal shear stress 3 inches above the bottom of thebeam is:Horizontal Shear Stress = Vay'/Ib = (6000 lb)*(5.794 in2)*(4.49 in)/(249.0 in4)(.35 in)= 1790 lb/in2

Notice, as we expect, the horizontal shear stress value becomessmaller as we move toward an outer edge of the beam cross section.

lecture 3 - shear stresses

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