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MAE243 Section 1 Fall 2006 Class 2

MAE243 Section 1 Fall 2006 Class 2. darran.cairns@mail.wvu.edu. Snowstorm collapses roof hours after hockey game. No one hurt. Designers required to retake Mechanics of Materials!!.

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MAE243 Section 1 Fall 2006 Class 2

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  1. MAE243 Section 1 Fall 2006Class 2 darran.cairns@mail.wvu.edu

  2. Snowstorm collapses roof hours after hockey game. No one hurt. Designers required to retake Mechanics of Materials!! On January 18, 1978 the Hartford Arena experienced the largest snowstorm of its five-year life. At 4:15 A.M. with a loud crack the center of the arena's roof plummeted the 83-feet to the floor of the arena throwing the corners into the air. Just hours earlier the arena had been packed for a hockey game. Luckily it was empty by the time of the collapse, and no one was hurt Three major design errors coupled with the underestimation of the dead load by 20% (estimated frame weight = 18 psf, actual frame weight = 23 psf) allowed the weight of the accumulated snow to collapse the roof. The load on the day of collapse was 66-73 psf, while the arena should have had a design capacity of at least 140 psf. The top layer's exterior compression members on the east and the west faces were overloaded by 852%. The top layer's exterior compression members on the north and the south faces were overloaded by 213%. The top layer's interior compression members in the east-west direction were overloaded by 72%.

  3. For those who can’t get enough of MAE 243

  4. For the truly ambitious

  5. FIG. 1-1 Structural members subjected to axial loads

  6. Resultant forces at O

  7. To determine the load at any point in a material we consider it to be made up of many tiny cubes (finite elements). In general we assume that the material is continuous and cohesive. For each element we have three normal forces Fx, Fyand Fz. As we decrease the area the forces and area approach zero. However, the force/area normal to the area approaches a finite value. These are normal stresses (σ). Similarly, the force per unit area parallel to the surface are shear stresses (τ).

  8. General state of stress around the chosen point. Units SI N/m2or Pa very often kPa, MPa & GPa are used. US psi, kpsi

  9. Localized distortions at the ends. Will not consider. If Bar is long compared to ends then don’t need to worry too much. Assumptions Bar is straight Load is along centroid Material is homogenous Material is isotropic (or anisotropy is aligned with loading axis)

  10. If bar is subject to constant uniform deformation then deformation is due to a constant normal stress

  11. Normal Strain Tensile +ve Compression -ve

  12. FIG. 1-2Prismatic bar in tension:(a) free-body diagram of a segment of the bar, (b) segment of the bar before loading, (c) segment of the bar after loading, and (d) normal stresses in the bar

  13. A short post constructed from a hollow circular tube of aluminum supports a compressive load of 26 kips. The inner and outer diameter of the tube are 4.0 in and 4.5 in and its length is 16in. The shortening of the post due to the load is 0.0012 in. Determine the compressive stress and strain in the post. (You may neglect the weight of the post and assume that buckling does not occur.)

  14. The 80kg lamp is supported by two rods AB and BC as shown. If AB has a diameter of 10mm and BC has a diameter of 8mm, determine the average normal stress in each rod. Get in groups and calculate the average normal stresses.

  15. Internal Loading X direction FBC(4/5)-FBAcos(60°)=0 Y direction FBC(3/5)+FBAsin(60°)-80(9.81)=0 FBC=395.2 N FBA=632.4 N Average Normal Stress

  16. A circular steel rod of length L and diameter d hangs in a mine shaft and holds an ore bucket of weight W at its lower end. • Obtain a formula for the maximum stress in the rod, taking into account the weight of the rod itself. • Calculate the maximum stress if L=40 m, d=8mm and W=1.5 kN.

  17. The slender rod is subjected to an increase of temperature along its axis, which creates a normal strain in the rod of Where z is given in meters. Determine (a) the displacement of the end of the rod B due to the temperature increase and (b) the average normal strain in the rod.

  18. Deformed length of small segment dz’ Integrate from 0 to 0.2 m to find total deformed length z’ Increase in length = 2.39mm Average strain = 2.39/200=0.0119

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