Strength of Materials Solutions
Problem #1
Principal stresses:
Substitute values from above yields:
The maximum shear stress is determined by these two principal stresses as:
Note that the other maximum shear stresses are less than this value.
Problem #2
The total strain is:
This total strain is equal to:
Substituting:
and solving for F we get:
F=-4500 lbs
The stress is 4500 psi compressive.
Problem #3
The 2000 lbs creates a bending stress at Q that is tensile and is equal to:
Where
Substituting into the bending formula, we get
The stress due to the axial load is compressive and is equal to:
The total stress is:
Problem #4
Location of centeroid:
The area moment of inertia:
The answer is:
The stress at D is:
note that c=1.5-0.667=0.833
Problem #5: Curved beam
The neutral axis radius is:
The stress at inner radius (critical point) is:
There is also an axial stress of 5000 psi acting on the cross-section making the total stress become
Problem #6: Torsional stresses
The maximum torsional shear stress is:
Form Problem # 3, the normal stress on the surface is 15600 psi. The state of stress is shown below:
The principal stresses are calculated as before using:
The maximum shear stress at point Q is:
Problem #7
The forces in the upper portion (Fu) and lower portion (FL) are:
Where
Substituting into the force expressions:
The maximum stress is (480/0.5)=960 psi
Problem #8
The torque is divided according to torsional stiffnesses. In this case the left supports picks us (6/10)=0.6 of the torque and the right support takes 0.4 of the torque.
Problem #9
The stress is
Finding the centroid is as before:
The area moment of inertia is:
Q is
and
Problem #10
Problem #11
For this thin-walled tube:
The angle of rotation is:
Problem #12
The critical point is the inner radius. The tangential stress is:
Setting r=ri and Pi=0 we get
The state of stress is simple – just this tangential stress which is also the principal stress. From theory, we know that there are no shear stresses on these surfaces when the stress element in oriented with radial edges.
The factor of safety is:
Problem #S13
The critical point is the inner radius. Using the formula:
Problem #S14
The interface pressure is:
The radial interference is 0.013 mm. Substituting all the numbers:
The answer is P=50.4 Mpa.
Problem #S15 and #S16
Problem #S15: Using the impact formula and simplifying for h>>d:
The stress is 70.7 ksi.
Problem #S16: The two bars form a pair of two springs in series. The equivalent spring is:
Problem #17
The area, moment of inertia, and radius of gyration:
The slenderness ratio is:
The limit for the use of Euler versus Johnson formula is:
Since the slenderness ratio is larger than the limit, the Euler formula applies:
The factor of Safety is:
SF=20438/5655 = 3.6
b) For this case:
The slenderness ratio is:
We have to use the Johnson Formula:
Problem #S18: Based on DET:
Based on MST:
Problem #S19
For this cast iron :
The critical point is the inner radius
Since this is a principal stress and the other principal stress is zero (radial stress is zero on the inner radius), we equate this stress to Sut.
Problem #S21
The question in this problem is the factor of safety against eventual fatigue failure. First we calculate the maximum nominal shear stress:
We would apply the fatigue stress concentration factor to the nominal stress to get the actual stress
The need to find the VonMises stress and compare it to strength
On the strength side, the estimate of the endurance limit of the rotating bending fatigue specimen is half of the tensile strength for steels:
Applying the correction factors to estimate the endurance limit of this part:
The factor of safety is: