Question

Each of the two vertical links CF connecting the two horizontal members AD and EG has a uniform rectangular cross section with dimensions 10 mm times 40 mm, and is made of a steel with an ultimate strength in tension of 400 MPa. Each of the pins at C and F has a 20 mm diameter and is made of a steel with an ultimate strength in shear of 150 MPa. Ignoring bearing stress, determine the factor of safety based on the links CF as well as the pins connecting them to the horizontal members. (Final Answer: 2.41)

Answer 1

2.4166

Step-by-step explanation:

Given:

- Section CF: A_cf = 10 x 40 mm^2 , UT = 400 MPa

- Bolt at F and E: A_p = pi*(0.02)^2 / 4 , ST = 150 MPa

- Force P = 24 KN

Find:

Factor of safety (FS) based on the links CF as well as the pins connecting them to the horizontal members.

Solution:

Step 1: We will make a Free body diagram (FBD) of the member EG and evaluate the force F_cf in member at point F.

- Taking sum of moments about point E equal to zero:

- F_cf * 400 + 24*650 = 0

F_cf = 39 KN

Step 2: Find the tensile stress in member CF and shear stress at pin F:

- Tensile Stress T_cf in member CF :

T_cf = F_cf / 2*A_cf

T_cf = 19.5 KN / 2*(0.01*0.04)

T_cf = 24.375 MPa

- Shear Stress S_p in pin at F:

T_cf = F_cf / 2*A_p

T_cf = 19.5KN*4 / 2*pi*0.02^2

T_cf = 62.070 MPa

Step 3: Determine the factor of safety for both tensile stress at member CF and shear stress in pin at F

- Factor of safety for CF (F.S_cf):

F.S_cf = UT / T_cf

F.S_cf = 400 / 24.375

F.S_cf = 16.41

- Factor of safety for pin F (F.S_p):

F.S_p = UT / T_cf

F.S_p = 150 / 62.070

F.S_p = 2.4166

Step 4: Failure Analysis

- The factor of safety of the pin is 2.4166 while that of member is 16.41. So the entire integrity of the structure is governed by the failure of pin at F as compared to member CF. Hence, overall Factor of safety is min (16.41 , 2.4166) = 2.4166