| Introduction | ||
 Shelf System | A new shelf system is planned to store large crates. The shelves are relatively short but carry a large distributed load. Each "T" beam is firmly fixed to the wall and acts as a cantilever beam. The design team would like to know what the largest shear stress in the "T" at the joint between the top flange and bottom web (point A). What is known:
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| Question | ||
| What is the maximum shearing stress at point A anywhere along the beam length? What is the maximum normal stress? | ||
| Approach | ||
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| | Maximum and Minimum Normal Stress | |||||||
 Rotating Stresses from x-y Coordinate System to new x'-y' Coordinate System | Rotating the stress state of a stress element can give stresses for any angle. But usually, the maximum normal or shear stresses are the most important. Thus, this section will find the angle which will give the maximum (or minimum) normal stress. Start with the basic stress transformation equation for the x or y direction.  To maximize (or minimize) the stress, the derivative of σx′ with respective to the rotation angle θ is equated to zero. This gives, dσx′ / dθ = 0 - (σx - σx) sin2θp + 2τxy cos2θp = 0 where subscript p represents the principal angle that produces the maximum or minimum. Rearranging gives,
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 Principal Stresses, σ1 and σ2, at Principal Angle, θp | The angle θp can be substituted back into the rotation stress equation to give the actual maximum and minimum stress values. These stresses are commonly referred to as σ1 (maximum) and σ2 (minimum),
For convenience, the principal stresses, σ1 and σ2, are generally written as,
It is interesting to note that the shear stress, τx′y′ will go to zero when the stress element is rotated θp. | |||||||
| Maximum Shear Stress | ||||||||
 Maximum Shear Stresses, τmax, at Angle, θτ-max | Like the normal stress, the shear stress will also have a maximum at a given angle, θτ-max. This angle can be determined by taking a derivative of the shear stress rotation equation with respect to the angle and set equate to zero. 
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| Plotting Stresses vs Angle | ||||||||
 Stresses as a Function of Angle | The relationships between principal normal stresses and maximum shear stress can be better understood by examining a plot of the stresses as a function of the rotation angle. Notice that there are multiple θp and θτ-max angles because of the periodical nature of the equations. However, they will give the same absolute values. At the principal stress angle, θp, the shear stress will always be zero, as shown in the diagram. And the maximum shear stress will occur when the two principal normal stresses, σ1 and σ2, are equal | |||||||
 Beam Loading and Cross Section | Before the stress state at point A can be analyzed for maximum shear and normal stresses, the maximum bending moment and shear load in the beam needs to be determined. From the maximum bending moment, the maximum bending stress can be found. Also, from the maximum shear load, the maximum shear stress can be determined. Both of these stresses are in the beam oriented coordinate system. After bending and shear stresses are found, the stress element can be rotated to given the absolute maximum shear stress. This stress state will most likely be something other than the natural beam coordinate system. There are a number of preliminary steps required to find both the beam shear stress and bending stress. These include determining the moment-shear diagram, the neutral axis, moment of inertia, I, and the first moment of the area, Q. | |
| Moment -Shear Diagrams | ||
 Shear-Moment Diagrams | To find the maximum bending moment and shear load, the moment and shear equations can be used. Cutting the structure and summing the forces and moments, gives, ΣFy = 0 = V - 50(0.2 - x) V = (10 - 50x) kN and ΣM = 0 = M - 50 (0.2 - x) (0.2 - x)/2 M = (1 - 10 x + 25 x2) kN-m The maximum shear and moment will be at the wall, Vmax = 10 kN and Mmax = 1 kN-m | |
| Neutral Axis and Moment of Inertia | ||
 Neutral Axis Location | The neutral axis is critical in finding the cross section moment of inertia and the first moment of the area.  The moment of inertia for full cross section is, I = I1 + y12 A1 + I2 + y22 A2 = 40(10)3/12 + (50 - 38.33 + 10/2)2(40)(10) + 10(50)3/12 + (38.33 - 50/2)2(50)(10) = 307,500 mm4 = 3.075 × 10-7 m4 | |
| First Moment of the Area, Q | ||
 First Moment of Area 1 | The first moment of the area, Q, is needed to determine the shear stress at point A. The area below or above point A can be used to calculate Q. If the area above is used, then Q is Q = y1A1 = (50 - 38.33 + 10/2)(40)(10) = 16.67 (400) = 6,668 mm3 = 6.668 × 10-6 m3 | |
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| Normal and Shear Stress at the Wall (at Point A) | ||
 Distance from NA to Point A | The maximum moment and shear occur at the wall and thus the maximum stresses will also occur at the wall. The normal stress at point A is σb = My/I = (1 kN-m)(0.01167 m)/(3.075 × 10-7 m4) = 37.95 MPa The shear stress at A, is  = 21.68 MPa | |
| Maximum Shear Stress | ||
 Stress State at Point A Near Wall (all arrows pointing in positive directions)  Rotated Stress State at Point A for Maximum Shear Stress | The bending stress and shear stress at point A is shown on a stress element at the left. The bending stress is considered to be acting in the x direction. There is no normal vertical stress, so σy is zero. The shear stress is acting down on the right edge of the stress element. Thus, the stress is negative and the shear stress on the right edge is drawn in the up direction. The maximum shear stress is  = ± 28.81 MPa This occurs at an angle of  θτ-max = 20.60o The rotated normal stresses are equal when the shear stress is a maximum, giving σx′ = σy′ = (σx + σy)/2 = 37.95/2 = 18.98 MPa All rotated stresses are labeled on the stress element at the left. Notice, the shear stress is actually negative when the shear stress rotation equation is used. On the other hand, the maximum shear stress equation above can be either positive or negative due to the square root. | |
| Maximum Normal Stresses | ||
 Rotated Stress State at Point A for Maximum Normal Stress  Rotated Stresses as a Function of Angle | The maximum normal stress, or principal stresses σ1 and σ2, are  = 18.98 ± 28.81 MPa = 47.79, -9.83 MPa This occurs at an angle of  θp = -24.41o It is interesting to visualize the stresses for any angle by plotting the stresses as a function of θ. This is shown in the diagram at the left. Note, the stresses have a period of π, or 180o. This is due to the symmetric nature of the stresses | |
 Stress Element | Example | |
| What is the maximum shear stress in the stress element shown? | ||
| Solution | ||
| The stress state shows the normal stresses, σx and σy, to be positive (tension) but the shear stress arrows are in the negative direction. Thus, the stress state is σx = 10.1 ksi σy = 6.2 ksi τxy = -3.7 ksi | ||
 Rotated Stress Element | The maximum shear stress is,  This could be either positive or negative due to the square root. Angle that the max shear stress acts at is,  2θτ-max = 27.79o (or -152.21 ) θp = 13.90o τmax = 4.182 ksi @ 13.90o It is interesting to note, the new normal stresses are, σx' = σy' = (σx+ σy)/2 = 8.15 ksi | |
