mae243 section 1 fall 2006 class 2 [email protected]
TRANSCRIPT
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%.
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
Snowstorm collapses roof hours after hockey game. No one hurt. Designers required to retake Mechanics of Materials!!
For those who can’t get enough of MAE 243
For the truly ambitious
FIG. 1-1 Structural members subjected to
axial loads
Resultant forces at O
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, Fy and 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).
AFz
Az
0lim
AFx
Azx
0lim
A
FyAzy
0lim
General state of stress around the chosen point.
Units
SI N/m2 or Pa very often kPa, MPa & GPa are used.
US psi, kpsi
Assumptions
Bar is straight
Load is along centroid
Material is homogenous
Material is isotropic (or anisotropy is aligned with loading axis)
Localized distortions at the ends. Will not consider. If Bar is long compared to ends then don’t need to worry too much.
If bar is subject to constant uniform deformation then deformation is due to a constant normal stress
APAP
dAdfA
Normal Strain
sss
avg
sss
AB
lim
Tensile +ve
Compression -ve
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
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.)
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.
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
MPam
N
A
F
MPam
N
A
F
BA
BABA
BC
BCBC
05.8)005.0(
4.632
86.7)004.0(
2.395
2
2
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.
(a) Obtain a formula for the maximum stress in the rod, taking into account the weight of the rod itself.
(b) Calculate the maximum stress if L=40 m, d=8mm and W=1.5 kN.
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.
2/13)10(40 zz
2/13)10(40 zz
dzzzd 2/13)10(401
Deformed length of small segment dz’
Integrate from 0 to 0.2 m to find total deformed length z’
mz
zzz
m
m
dzzz
20239.0
)3
2)(10(40
2.0
0
2/13
2.0
0
2/33
)10(401
Increase in length = 2.39mm
Average strain = 2.39/200=0.0119