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Alla [95]
3 years ago
5

A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maxi

mum shear stress is 450 N/cm2 when the maximum bending stress is 1500 N/cm2. Calculate the value of the centrally applied point load W and the span L. The overall height of the I section is 29 cm.

Engineering
1 answer:
saw5 [17]3 years ago
4 0

Answer:

W = 11,416.6879 N

L ≈ 64.417 cm

Explanation:

The maximum shear stress, \tau_{max}, is given by the following formula;

\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )

t_w = 1 cm = 0.01

h = 29 cm = 0.29 m

h_w = 25 cm = 0.25 m

b = 15 cm = 0.15 m

I_c = The centroidal moment of inertia

I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )

I_c = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴

Substituting the known values gives;

I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right )  = 1.2257083\bar 3 \times 10^{-4}

I_c = 1.2257083\bar 3 × 10⁻⁴ m⁴

From which we have;

4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )

Which gives;

W = 11,416.6879 N

\sigma _{b.max} = \dfrac{M_c}{I_c}

\sigma _{b.max} = 1500 N/cm² = 15,000,000 N/m²

M_c = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²

From Which we have;

M_{max} = \dfrac{W \cdot L}{4}

L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417

L ≈ 0.64417 m ≈ 64.417 cm.

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A train starts from rest at station A and accelerates at 0.6 m/s^2 for 60 s. Afterwards it travels with a constant velocity for
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Answer:

The distance between the station A and B will be:

x_{A-B}=55.620\: km  

Explanation:

Let's find the distance that the train traveled during 60 seconds.

x_{1}=x_{0}+v_{0}t+0.5at^{2}

We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:

x_{1}=\frac{1}{2}(0.6)(60)^{2}

x_{1}=1080\: m

Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.

v_{1}=v_{0}+at

v_{1}=(0.6)(60)=36\: m/s

Then the second distance will be:

x_{2}=v_{1}*1500

x_{2}=(36)(1500)=54000\: m        

The final distance is calculated whit the decelerate value:

v_{f}^{2}=v_{1}^{2}-2ax_{3}

The final velocity is zero because it rests at station B. The initial velocity will be v(1).

0=36^{2}-2(1.2)x_{3}

x_{3}=\frac{36^{2}}{2(1.2)}  

x_{3}=540\: m

Therefore, the distance between the station A and B will be:

x_{A-B}=x_{1}+x_{2}+x_{3}  

x_{A-B}=1080+54000+540=55.620\: km  

I hope it helps you!

 

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