Question

The uniform beam is supported by two rods AB and CD that have cross-sectional areas of 10 mm 2 and 15 mm, 2 respectively. Determine the intensity w of the distributed load so that the average normal stress in each rod does not exceed 300 kPa.

Answer 1

Answer:

2 N / m

Explanation:

It is necessary to determine the maximum w so that the normal stress in AB and CD rods does not exceed the allowable stress.

The cross-section of respective rods are:

A_ab = 10 mm^2

A_cd = 15 mm^2

Step 1 :The maximum load is determined from the condition that the normal stress (б)

is not higher than (б_allow).

б_ab = F_ab / A_ab  =< б_allow

б_cd = F_cd / A_cd  =< б_allow

Step 2 : Compute the static size

We apply the Equilibrium conditions:

1) Sum of Moments (A) = 0

-w * 6 * 0.5 * 6 * 0.75 + 6*F_cd = 0

F_cd = (13.5 / 6) * w = 2.25*w KN

2) Sum of Forces (y-direction) = 0

F_cd + F_ab - 0.5*6*w = 0

F_ab = 3*w - 2.25*w = 0.75*w KN

Step 3 : Compute Load w

1) Sector AB :

F_ab / A_ab  =< б_allow

0.75*w / 10 * 10^-6 =< 300KPa

0.75*w =< 3

w =< 4 N / m

2) Sector CD :

F_cd / A_cd  =< б_allow

2.25*w / 15 * 10^-6 =< 300KPa

2.25*w =< 4.5

w =< 2 N / m

Answer: w = 2 N / m

Answer 2

The intensity of the distributed load so that the average normal stress in each rod does not exceed 300 kPa is 3 N/m in rod AB and 2.25 N/m in rod CD.

Area of rod AB = 10 mm² = 10 × 10⁻⁶ m

Area of rod CD = 15 mm² = 15 × 10⁻⁶ m

Normal stress (σ) = 300 kPa = 300 * 10³ Pa

Summation of moment about point A is zero:

$T_{CD}=2w\\\\For\ rod\ CD:\\\sigma=\frac{P}{A} \\\\300*10^3=\frac{2w}{15*10^{-6}} \\\\w=2.25\ N/m$

Summation of vertical forces is zero:

$T_{AB}=w\\\\For\ rod\ AB:\\\sigma=\frac{P}{A} \\\\300*10^3=\frac{w}{10*10^{-6}} \\\\w=3\ N/m$

The intensity of the distributed load so that the average normal stress in each rod does not exceed 300 kPa is 3 N/m in rod AB and 2.25 N/m in rod CD.

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