ME 437 – 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 (4/10)=0.4 of the torque and the right support takes 0.6 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 Po=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:

[This number is off by a factor of 4 compared to the posted answer – Error is in using radii as diameters]

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 #S20 - Challenge

The main factors influencing the speed are the yield strength and the density of the material. The higher this ratio, the higher the speed can be. Some typical materials:

Material / Yield strength / Density / Ratio
Al 7075-T6 / 73 ksi / 0.098 lbs/in3 / 745
Steel 4140 / 250 / 0.282 / 886
Titanium Ti-3A / 175 / 0.163 / 1080
Steel 6150 / 315 / 0.282 / 1120

Since 4140 is most available and least expensive, that is chosen. A reasonable inner radius is 1.25 inches. With these parameters, the speed can be as high as 120000 rpm leading to a tip velocity of more than 1000 mph. Using the 6150 steel with an inner radius of 0.3 inches leads to a speed of 145000 rpm or a tip velocity of 1240 mph.