Question

Link BD consists of a single bar 36 mm wide and 18 mm thick. Knowing that each pin has a 12-mm diameter, determine the maximum value of the average normal stress in link BD if (a) θ 5 0, (b) θ 5 90°.

Answer 1

Answer:

hello the diagram attached to your question is missing attached below is the missing diagram

answer :

a) 48.11 MPa

b) - 55.55 MPa

Explanation:

First we consider the equilibrium moments about point A

∑ Ma = 0

( Fbd * 300cos30° ) + ( 24sin∅ * 450cos30° ) - ( 24cos∅ * 450sin30° ) = 0

therefore ; Fbd = 36 ( cos ∅tan30° - sin∅ ) kN  ----- ( 1 )

A ) when ∅ = 0

Fbd = 20.7846 kN

link BD will be under tension when ∅ = 0, hence we will calculate the loading area using this equation

A = ( b - d ) t

b = 12 mm

d = 36 mm

t = 18

therefore loading area ( A ) = 432 mm^2

determine the maximum value of average normal stress in link BD  using the relation below

бbd = $\frac{Fbd}{A}$  = 20.7846 kN / 432 mm^2  =  48.11 MPa

b) when ∅ = 90°

Fbd = -36 kN

the negativity indicate that the loading direction is in contrast to the assumed direction of loading

There is compression in link BD

next we have to calculate the loading area using this equation ;

A = b * t

b = 36mm

t = 18mm

hence loading area = 36 * 18 = 648 mm^2

determine the maximum value of average normal stress in link BD  using the relation below

бbd = $\frac{Fbd}{A}$ = -36 kN / 648mm^2 = -55.55 MPa

Answer 2

The maximum value of the average normal stress in link BD at the given angles are;

At θ = 0°; 64.15 MPa

At θ = 90°; 66.66 MPa

Average Normal Stress

The image of the link and the single bar is missing and so i have attached it.

From the image of the link and single bar attached, i have drawn a free body diagram of link ABC that will help us to solve this question.

Taking Moments about point A and summing to zero, we can solve for F_bd at the given angles as;

A) At θ = 0°;

From the diagram, AC = 450 mm = 0.45 m and force acting at point C is 24 kN or 24000 N. Thus;

(0.45 * sin 30)(24000) - F_bd(0.3 * cos 30) = 0

Thus;

(0.45 * sin 30)(24000) = F_bd(0.3 * cos 30)

⇒ 5400 = 0.2598F_bd

F_bd = 5400/0.2598

F_bd = 20785.22 N

Area at tension Loading is;

A = (0.03 - 0.012)0.018

A = 324 × 10⁻⁶ m²

Thus;

Average Normal stress is;

σ = 20785.22/(324 × 10⁻⁶)

σ = 64.15 × 10⁶ Pa = 64.15 MPa

B)  At θ = 90°;

(0.45 * cos 30)(24000) + F_bd(0.3 * cos 30) = 0

Thus;

-(0.45 * cos 30)(24000) = F_bd(0.3 * cos 30)

F_bd = -36000 N

Area at compression Loading is;

A = 0.03 * 0.018

A = 540 × 10⁻⁶ m²

Thus;

Average Normal stress is;

σ =  -36000/(540 × 10⁻⁶)

σ = 66.66 × 10⁶ Pa = 66.66 MPa

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