Question

Renang Time! 1 hour, 10 minutes, 56 seconds. Question Completion Status: 1 2. 3 QUESTION 1 A 0.5 m long rod is supported at both of its ends. A fluctuating load of a min. 20,000 N and a max. of 50,000 N acted on the middle of the rod. If the rod has a diameter of 75 mm, and a concentration factor, kt-1.0, find the safety factor of the rod. The material properties of bar are as follows: ultimate strength of 650 MPa, yield strength of 500 MPa and estimated actual endurance strength of 267.75 MPa. For circular rod, moment of inertia, 1 - r*14 Centroid, cd/2 Choose the closest answer from the below choices: O 2.1 O 2.6 O 32 O 3.7 O 3.9 O45 O 5.7 Click Save and Submit to save and submit. Click Save All Answers to save all answers. LO2 (Part 2 LO 3 Part 1 10 J....pdf LO 3 Part 19 Ju...pdf @ L02 (Part 2 Pra...pdf

Answer 1

Given:

• length:0.5 m
• Fmin=20,00oN
• Fmax=50,000N
• Kt=1.0
• ultimate tensile strength 650MPa
• yield strength 500 MPa
• endurance stength 267.75 MPa

minimum bending moment $M_{min}=\frac{F_{min}\times l}{4}=\frac{20000\times 0.5}{4}=2500N-m$

maximum bending moment

Fmar x 1 Мутах 50000 х 0,5 4 6250N – т 4

$\therefore$ mean bending moment $M_m=\frac{M_{max}+M_{min}}{2}=\frac{6250+2500}{2}=4375N-m$

and variable bending moment $M_v=\frac{M_{max}-M_{min}}{2}=\frac{6250-2500}{2}=1875N-m$

Now, the section modulus of the circular bar $Z=I/c=\frac{\pi r^3}{4}=\frac{\pi d^3}{32}=\frac{\pi 0.075^3}{32}=4.14\times 10^{-5}m^3$

mean bending stress $\sigma_m=\frac{M_m}{Z}=105.67 MPa$

variable bending stress $\sigma_v=\frac{M_v}{Z}=45.28 MPa$

• According to Goodman's formula:

$\frac{1}{FS}=\frac{\sigma_m}{\sigma_u}+\frac{\sigma_v}{\sigma_e}(\because K_t=1)$

$\\\therefore \frac{1}{FS}=\frac{105.67}{650}+\frac{45.28}{267.75}\\\\ \therefore FS=3.014$

• According to Soderberg's formula:

$\frac{1}{FS}=\frac{\sigma_m}{\sigma_y}+\frac{\sigma_v}{\sigma_e}(\because K_t=1)$

$\\\therefore \frac{1}{FS}=\frac{105.67}{500}+\frac{45.28}{267.75}\\\\ \therefore FS=2.63$