| Introduction | ||
 Support Bracket for Hanging Load | A new steel bracket was designed to support various loads being hung from the end. The designer would like to know what is the largest load that can be hung on the bracket. What is known:
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| Question | ||
| What is the maximum load that can be placed on the bracket using the maximum distortion energy theory? | ||
| Approach | ||
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| Failure Thoeries | |||||
 Uniaxial Stress-Strain Curve with Yield Stress and Ultimate Stress | If all structures where loaded in only one direction, it would be easy to predict failure. All that would be needed was a single uniaxial test to find the yield stress and ultimate stress levels. If it is a brittle material, then the ultimate stress will determine failure. For ductile material, failure is assumed to be when the material starts to yield and permanently deform. However, when a structure has multiple stresses at a given local (σx, σy and τxy for 2D as discussed in Stresses at a Point section), then the interaction between those stresses may effect the final failure. This section presents three basic failure theories that can be used for different types of materials to help predict failure when multiple stresses are applied. For simplification, all theories are based on principal stresses (σ1, σ2) which can be determined from any (σx, σy and τxy) stress state. This removes the shear stress terms since the shear stress is zero at the principal directions. Using principal stresses does not change the results from the failure theories. | ||||
| Maximum Normal Stress Theory | |||||
 Maximum Normal Stress Criterion (Blue Regoin is Safe) | The simpliest theory ignores any interaction between the normal principal stresses, and assumes that failure occurs when either of the normal stresses exceed the ultimate stress. This is written as
This failure criteria is really good for brittle materials and should not used for ductile material like steel, aluminum, and plastics. | ||||
| Maximum Shear Stress Theory (Tresca's Yield Criterion) | |||||
 Maximum Shear Stress Criterion or Tresca's Yield Criterion (Blue Regoin is Safe) | The maximum shear stress assumes failure occurs when the maximum shear stress exceeds the shear stress in a simple uniaxial test. In a unixial test, the principal stresses are σ1 = σx (axial direction) and σ2 = 0 (transverse to axial direction). Using the stress transformation equations, the maximum shear stress for this stress state is τmax = (σx - σy)/2 = σx/2 = σyld/2 Thus, for any combination of loading for σ1 and σ2, the shear stress cannot exceed = σyld/2. This condition gives three seperate possible situations that need to checked,
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| Maximum Distortion Energy Theory (von Mises' Yield Criterion) | |||||
 Maximum Distortion Energy Criterion or von Mises' Yield Criterion (Blue Regoin is Safe) | The third theory looks at the total energy at failure and compares that with the total energy in a unixial test at failure. Any elastic member under load acts like a spring and stores energy. This is commonly called distortational energy and can be calculated as The G is the shear modulus. The distortion energy for a general stress state can be compared to distortion energy for a uniaxial test that fails at σx = σ1 = σyld. This gives,  Thus the general criterion for failure would be
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 Support Bracket for Hanging Load P | The load that can be safely carried by a bracket needs to be determined. The bracket is made from standard steel with a yield stress of 36 ksi. Since there are both bending stress and torsional shear stress in the circular bar section, the stress state will not be uniaxial. This will require the use of the stress transformation equations to find the principal stresses. With the principal stresses, the maximum distortion energy failure criteria can be applied, and the total allowable load, P, can be determined. For safety, a factor of safety of 2 is required. Thus, the bracket stress should not exceed 36/2 = 18 ksi. It is assumed that the maximum torsional shear stresses and bending stresses will occur in the circular section at the wall. | |
| Torsional Shear Stress | ||
 Equivalent Torque in the Circular Section due to the Force P in the Rectangular Section | The load P will cause a bending moment at the joint between the circular and rectangular sections. The equivalent moment or torque, T, at the joint will be T = (6 in) P This torque will cause a constant twist in the circular section which will produce a torsional shear stress,  | |
| Bending Stress | ||
 Bending Moment M and Twisting Torque T at Wall | The load P will cause the bracket to bend in two directions. The first will be about the circular section, but there is no torsional stress in this section. Second, there will be a moment reaction at the wall that will bend the circular cross section. This moment, M, is shown in the diagram at the left and is equals to M = (6 in) P The bending stress will be,  | |
| Combined Stress State | ||
 Stress State at top of Circular Sections near wall  Initial and Principal Stress States | Both the torsional moment, T and the bending moment M, cause shear and bending normal stresses, respectively. If an element at the top of the circular section is analyzed as shown in the diagram, the stress state would be σx = 144.87 P τxy = 72.43 P The principal stresses will be  σ1 = 174.88 P and σ2 = -30.00 P | |
| Maximum Distortion Energy Criteria | ||
 Stress Location on Failure Envelop with Maximum Load P of 93.9 lb | Now that the principal stresses are known in terms of the load P, the failure criteria can be applied. For the maximum distortion energy criteria, the following relationship must be meet,  Substituting the principal stresses and yield stress (do not forget the factor of safety), gives (174.88P)2 + (-30.00P)2 - (174.88P)(-30.00P) = (36,000/2)2 Solving for P, gives P = 93.91 lb The final principal stresses are σ1 = 174.88 P = 16.42 ksi σ2 = -30.00 P = -2.82 ksi These principal normal stresses can be normalized with the design stress, 18 ksi, to give σ1/σdesign = 16.42/18 = 0.9122 σ2/σdesign = -2.82/18 = -0.1567 This point are plotted on the failure envelop diagram at the left | |
| Example | ||
 Gas Tank under Pressure | During testing, a thin-walled tank was pressurized with gas and the stresses in the x and y direction were found to be σx = 150 MPa, σy = 75 MPa, and τxy = 0. However, in actual use, the tank must also withstand a torque on the cap that will introduce a possible shear stress in the tank walls. This shear stress is in addition to the stresses due to the tank pressure. What is the maximum torque that can be applied if the vessel material can only withstand a shear stress of 100 MPa? The wall thickness is 15 mm and the outside diameter is 35 cm. Also, the maximum allowable shear stress for the vessel material is 100 MPa. | |
| Solution | ||
 Mohr's Circle | Mohr's circle can be used to understand the solve for this stress state by first finding the largest radius R for which the shearing stress does not exceed 100 MPa and then determine the allowable torque. To draw the Mohr's circle, first the center should be determined. Since, (σx + σy)/2 = (150 + 75)/2 = 112.5, this will be at (112.5 , 0). Assuming a radius R, the Mohr's circle is plotted on the left. From the diagram principal stresses are, σ1 = (σx + σy)/2 - R = 112.5 - R σ2 = (σx + σy)/2 + R = 112.5 + R where R = | (σ1 - σ2)/2 | | |
The goal is to determine the largest radius R for which the shear stress does not exceed 100 MPa. This can be found using "Maximum Shear Stress Theory" that states that for any combination of loading for σ1 and σ2, the shear stress cannot exceed σyld/2 = τmax. This condition gives three separate possible situations that need to be checked, In this problem, the maximum allowable shear stress is given, and not the yield stress. However, the yield stress is simply two times the yield stress for a uniaxial test. The three conditions become, 1st: | σ1 | = σyld = 2τmax = 2(100) | 112.5 - R | = 200 R = 312.5 MPa 2nd: | σ2 | = σyld = 2τmax = 2(100) | 112.5 + R | = 200 MPa R = 87.5 MPa 3rd: | σ1 -σ2 | = σyld = 2τmax = 2(100) R = | (σ1 - σ2)/2 | = 100 MPa | ||
 Mohr's Circle for τ | Comparing these three situations, the maximum value of R is the minimum of above three. R = 87.5 MPa From the Mohr's Circle diagram corresponding torsional shear stress is found as, τ = ( R2 - (112.5 - 75) 2 )1/2 = 79.06 MPa Hence, the torque that will cause this torsional shearing stress, T = τJ / r = [(79.06) (π/2) (ro4 - ri4)] / (0.35/2) = 124.2 [(0.35/2)4 - (0.35/2 - 0.015)4] / 0.175 = 0.2005 MN-m = 200.5 kN-m | |
