The 450 Kg Uniform I Beam

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Example - a uniform rod of length L rotating about one end. How do we evaluate the moment of inertia integral: I = ∫ r 2 dm for a uniform rod of length L rotating about an axis passing through one end of the rod, perpendicular to the rod? Align the rod with the x axis so it extends from 0 to L. Split the rod into little pieces of size dx.

  1. Example 2.2 Design a simply supported beam subjected to uniformly distributed dead load of 450 lbs/ft. And a uniformly distributed live load of 550 lbs/ft. The dead load does not include the self-weight of the beam. Calculate the factored design loads (without self-weight). WU = 1.2 wD + 1.6 wL = 1.42 kips / ft.
  2. In this tutorial we will look at how to calculate the bending stress of a beam using a bending stress formula that relates the longitudinal stress distribution in a beam to the internal bending moment acting on the beam’s cross section. We assume that the beam’s material is linear-elastic (i.e. Hooke’s Law is applicable).

For calculations total load on columns, Beam, Slab we must know about various load coming on the column.

Types of loads on the column

1- Self-weight of the column x Number of floors

The 450 Kg Uniform I Beam

2- Self-weight of beams per running meter

3- A load of walls per running meter

4- Total Load of the slab (Dead load + Live load + Self-weight)

The columns are also subjected to bending moments which have to be considered in the final design. The best way to design a good structure is to use advanced structural design software like ETABS or STAAD Pro. These tools are leagues ahead of manual methodology for structural design and highly recommended.

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In professional practice, there are some basic assumptions we use for structural loading calculations.

For Columns

Self-weight of Concrete is around 2400 kg per cubic meter, which is equivalent to 240 kN. Self-weight of Steel is around 8000 kg per cubic meter. Even if we assume a large column size of 230 mm x 600 mm with 1% steel and 3 meters standard height, the self-weight of the column is around 1000 kg per floor, which is equivalent to 10 kN. So, in my calculations, I assume self-weight of the column to be between 10 to 15 kN per floor.

For Beams

Similar calculations as above. I assume each meter of the beam has dimensions of 230 mm x 450 mm excluding slab thickness. So, self-weight can be around 2.5 kN per running meter.

For Walls

The density of bricks varies between 1500 to 2000 kg per cubic meter. For a 6″ thick wall of 3-meter height and a length of 1 meter, we can calculate the load per running meter to be equal to 0.150 x 1 x 3 x 2000 = 900 kg which is equivalent to 9 kN/meter. You can calculate load per running meter for any brick type using this technique.

The 450 Kg Uniform I Beam Conversion

For autoclaved, aerated concrete blocks like Aerocon or Siporex, the weight per cubic meter is between 550 to 700 kg per cubic meter. By using these blocks for construction, the wall loads per running meter can be as low as 4 kN/meter, which can result in a significant reduction in the cost of construction.

Penguin audio meter crack tool. For Slab

Assume the slab has a thickness of 125 mm. Now each square meter of the slab would have a self-weight of 0.125 x 1 x 2400 = 300 kg which is equivalent to 3 kN. Now, assume Finishing load to be 1 kN per meter and superimposed live load to be 2 kN per meter. So, we can calculate the slab load to be around 6 to 7 kN per square meter.

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The 450 Kg Uniform I Beam Sensor

Factor of Safety

In the end, after calculating the entire load on a column, please do not forget to add in the factor of safety. For IS 456:2000, the factor of safety is 1.5.

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1. The uniform beam shown in figure 5-15 weighs 500 N and supports a 700 N load. Find the tension in tie rope and the force of the hinge of the beam.
The answer according to the book must be 2.9 kN as the tension in tie rope and 2.0 kN as the force of the hinge of the beam.
2. The arm shown in figure 5-16 supports a 4.0 kg sphere. The mass of the hand and and forearm together is 3.0 kg and its weight acts at a point 15 cm from the elbow. Determine the force exerted by the biceps muscle.
The answer according to the book is 0.13 kM which is the force exerted by the biceps muscle.
3. The uniform, 120 N board shown in figure 5-19 is supported by two ropes as shown. A 0.40 kN weight is suspended one quarter of the way from the left end. Find FT1 (force of tension 1) and FT2 (force of tension 2) and the angle made by the left rope.
The answer according to the book is 0.19 kN for FT1, 0.37 kN for FT2 and 14 degrees for the angle made by the left rope.
I don't know how the book got these answers, so I need a full explanation and solution for these problems. As much as possible, make it step by step.
Please help me! I'm an engineering student. I already looked at some examples, compared it and did the same process but it didn't match the book's answers.
The following figures/diagrams are uploaded. Any answers will be accepted as much as it is correct and very well explained. Thank you!