Question

A hydraulic cylinder pushes a heavy tool during the outward stroke, placing a compressive load of 400lb in the piston rod. During the return stroke, the rod pulls on the tool with a force of 1500lb. Calculate the resulting design factor for the 0.6 in-diameter rod when subjected to this pattern of forces for many cycles. Discuss your result and justify if the design factor is appropriate, too low or too high. The material is SAE 4130 WQT 1300 steel. Assume wrought steel, machined, and 99% reliability. Piston Rod Extension stroke Pump flow Piston seal Return flow Piston Rod Return stroke Return flow Piston seal Pump flow

Answer 1

Given Information :

1. Compressive load acting : 400 lb
2. Tensile load acting : 1500 lb
3. Diamter of the rod : 0.6 in
4. Material used : SAE 4130 wrought steel, machined and 99 % reliability.

Assumptions :

For the ease of calculations the given information is converted to metric format, which yields the same value of design factor and in no way affects the final answer.

The properties of SAE 4130 Steel :

Yield Stress :
Syt = 453 M Pa

Ultimate Stress :
Sut = 670 MPa

Solution :

Step 1: Conversion of the quantitites from imperial to metric format :

Compressive load acting : 400 lb == 181.437 kg

Tensile load acting : 1500 lb == 680.3886 kg

Diameter of rod : 0.6 in == 15.24 mm

Step 2: Calculation of the stresses acting on the rod during the operation :

The cross section area of the rod is given by the formula :

$A= \frac{\pi}{4}*D^2$

Where

D : Diameter of rod

Substituting the values in the above equation :

$A= \frac{\pi}{4}*D^2$

A = *15.245

A = 182.414 mm

The stresses induced in the rod due to the load is given by the formula :

$\sigma=\frac{P}{A}$
Where
$\sigma:$   Stress induced
Load acting
Cross sectional area of the rod

Substitute the values of the compressive and tensile loads in the following formula

$\sigma=\frac{P}{A}$

$\sigma_c=\frac{181.437*9.81}{182.414}$

$\sigma_c=-9.75\ MPa$

$\sigma_t=\frac{680.3886*9.81}{182.414}$

$\sigma_t=36.59\ MPa$

Hence the values of stresses are :

Tensile stress : $\sigma_t=36.59\ MPa$

Compressive stress : $\sigma_c=-9.75\ MPa$

Step 3: Calculation of Endurance stress.

Endurance stress of the given rod is calculated by the following formula :

$\sigma_e={S_e}^**K_a*K_b*K_c*K_e$

Where
${S_e}^*:$ Endurance limit of Material
$K_a:$ Machining Factor
$K_b:$ Size Factor
$K_c:$ Load Factor
$K_e:$ Reliability Factor

Since the given material is steel the endurance limit of material is given by :

Since there is no mention of machining factor we can assume it to be unity with little loss in accuracy.

$K_a=1$

The size factor is given by :

d=diameter is in inches

$K_b=0.928$

The load factor is given by :

$K_c=0.85$ for axial loading

The relability factor is given by :

$K_e=0.814$ for 99 % reliability

Combining all the factors and calculating the endurance stress:

$\sigma_e={S_e}^**K_a*K_b*K_c*K_e$

$\sigma_e=335*1*0.928*0.85*0.814$

$\sigma_e=215.09\ MPa$

Step 4 : Calculating Stress Amplitude, Mean Stress :

From the tensile and the compressive stresses we can find the following :

Stress amplitude is given by :

$\sigma_a=\frac{\sigma_t-\sigma_c}{2}$

Mean Stress is given by :

$\sigma_m=\frac{\sigma_t+\sigma_c}{2}$

Substitute the values in the above equations :

$\sigma_a=23.17\ MPa$

$\sigma_m=13.42\ MPa$

Step 5 : Calculating the design factor from Soderberg Line equation :

The Soderberg Line equation is taken for calculation since Steel is a ductile material .

The equation is as follows :

$\frac{\sigma_a}{\sigma_e}+\frac{\sigma_m}{S_{yt}}=\frac{1}{N}$
Where
Design Factor

Substitute the values in the above equation

$\frac{\sigma_a}{\sigma_e}+\frac{\sigma_m}{S_{yt}}=\frac{1}{N}$

$\frac{23.17}{215.09 }+\frac{13.42}{453}=\frac{1}{N}$

$\frac{1}{N}=0.1373$

$N=7.28$

Therefore the design factor is  $N=7.28$

The value of the design factor obtained from Soderberg Line equation is very high. The reason is that during the calculation,it has been considered that the rod is free from notches and hence free from stress concentration effects.