moment of inertia of a composite section about y-axis.

Problem:
Find the moment of inertia about y-axis.


Solution:

Solved Example - Cable and Determinate Frame - How to find Forces at the Internal Hinges and supports

Hi,
Consider the frame in the diagram. If the tension in the cable that runs between H and I is 50 lbs, determine the loads acting on the frame at A and at C, and on member EH.


Solution:


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If you have any doubt, please leave a comment. 

Solved Example - Beam with Hinge - Shear Force and Bending Moment Diagram

Given the determinate continuous beam in the following, not that load 24k is applied at the mid-point of BC(you may define it as D.)

Plot the shear and moment diagram for the beam A-B-C.

Draw the Shear force and Bending moment diagram for the determinate beam

A cable applied with concentrated loads, find the sag - Solved Example.

Example: Given cable with three concentrated load.
If P1= 5.9 lbs, P2 = 5.9 lbs, and P3 = 10.9 lbs, Determine the reaction at E and the distances dB and dD.




If you have any doubt, please ask in the comment box.

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How to Solve a simple Truss problem using Method of Joints - solved example.

Here is the problem of a determinate truss.


And here is its joint by joint solution.

Redundant of curved(Circular) beam using Castigliano's theorem.

Hi,
Problem: The given beam consist of straight beam AC where A is fixed,  and curved(circular) beam CE. Load P is horizontal. (a) Draw Free Body diagram (b) Write moment equations for AC and CE part. (c) Write the expression to find the redundant at E, using the Castigliano's theorem for deflection.

curved/Circular beam redundant using Castigliano's theorem

Check the relevant books to buy.

   

Characteristic Equation for a 4*4 Matrix (Eigenvalues)

Whenever a 4*4 matrix is given, and it is to find the characteristic equation. Always expand along the column or row, which has the maximum number of zeroes. Look for the solved example given below.
Given Matrix A is a 4*4 matrix, with 9, 7, 5 and 9 as the diagonal elements. First column contains the maximum zeroes, so expand along it. The final equation is the characteristic equation. Soling it gives 9, 5 and 7 as the eigenvalues.

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Solved - Virtual Work Method - Deflection of Beam - Propped Cantilever with Internal hinge

Hi

Consider the compound beam shown in fig. EI is constant. Use the principle of Virtual Work. Determine the displacement at point D.
Please refer the image below for the solution.

Virtual Work Method - Deflection of beam - Propped cantilever with internal hinge

Virtual Work Method - Deflection of Beam - Propped Cantilever with Internal Hinge

Solved - Moment Distribution Method -Frame with pinned and fixed ends)

hi,
The given frame has two roller supports and one fixed, support. One must know the stiffness and distribution factors to distribute the unbalanced moments. Take the following example. Solution is handwritten below.

Solved -Shear Flow equation- Wooden Beam and Nail Spacing using

Hi,
Here is an example of solved problem, when you are asked to find the shear capacity of nailed wooden beams, or asked about the spacing of nails. Leave your suggestions or doubts in the comment box below.

Problem: The two cross sections (a) and (b) of a wooden beam are shown below. Both are subjected to a vertical shear force of V. Each nail can support a shear force of 20 kN, and are spaced at 125 mm along the length of the beam. Section details:
a= 250 mm, b= 35 mm, c= 265 mm, d = 180 mm, and e= 300mm. Assume the sections are uniform along the length of the beam. Find:
(A) Maximum applicable shear force on member (a)
(B) Maximum applicable shear force on member (b)
(C) Adjust the spacing of the weaker member for the maximum of the forces found in part (A) and part (B).

Solution:


Thanks!

Solved- ILD(Influence Line Diagram) - Shear Force and Bending Moment Overhang Beam

Hi,

The overhang beam shown is pinned at A and roller supported at B. Using equilibrium approach, derive and draw the influence line for
(a) horizontal and vertical support reactions at A.
(b) the vertical support reaction at B.
(c)The shear at C.
(d)The moment at C.

Force method for Indeterminate continuous Beam - Matrix Approach

Hi,
In force method of structural analysis,

• the first step is to find the Redundancy/indeterminacy of the structure. For example for the given beam in the following example there are total five number of support reactions, therefore the redundancy is 2.
• Second step is to convert the indeterminate structure into a basic determinate by replacing the the two of the unknown support reactions with redundants.  
• Third step is to find out the displacements of the basic determinate structure along the redundants due to the given loading conditions. 
• fourth step is to find out the flexibility matrix by applying the unit loads along the redundants and finding out the corresponding displacements at these two places.
• Next, form the equations as per the compatibility conditions and form the matrices. 
• Solve the matrix problem to find out the unknown reaction forces.

Example: Let us consider a continuous beam with a total span of 10m, EI constant, udl acting on the first half and a couple acting at 7.5m from the left. There are hinge and roller support at B and C, while beam is fixed at the left end A. Find out the support reactions at B and C using the force method (matrix approach.)

Solution: 

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Tension member connected to Gusset plate(fail in Rupture/Fracture or yield?)

Hi,
A tension member - a rectangular plate of dimension 1/2"*8" connected with two rows of bolts, determine the bolt size so that the tension member fails in rupture but not yield. Steel of grade A36 grade is used.
It must be kept in mind that the capacity factors are different for the rupture and yield, its lower for finding the rupture capacity, the section where bolts are inserted into the member are considered critical. Solution is written in the image shown below.

Solved - Moment of Inertia- Spoked Wheel with inner and outer Ring.

Hi,
I was stuck at this problem of finding the mass moment of inertia of a spoked wheel/ring, which is necessary to find out the angular acceleration or the kinetic energy of the wheel.

It's easy to determine the moment of inertia of a spoked wheel with the given weights of the outer rim/ring, inner rim/ring and weight of spokes. The moment of inertia about the centroid and an axis parallel to the centroidal axis can also be easily found. Please refer the image below for knowing the process.



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Flexure Stress and Normal Stress

Hello there,

Its a long time since I have posted in my Structural Engineering blog, so today I thought to discuss something in here. If I ask you the type of stresses, you may come up with the following types:

1. Normal Stress
2. Shear Stress
3. Flexure Stress

   

   Flexure Stress Vs Normal Stress

These are the three types of stresses which we deal all along our structural engineering, arising out of so many different types of loading on the structure. I asked this question to one of my student, if there is any similarity between the first one and the third one. 

Normal stress is calculated simply as Force divided by the cross section area under action. Flexure stress is calculated with the help of the flexure formula. Now, to find out the similarility between the two, you have to consider the nature of them.

We know normal stress are either tensile or compressive, similarly, you know that the flexure stresses are also either tensile or compressive.

The main difference between the two, is that the normal stress remains constant throughout the section, but the flexure stress varies along the cross section. This shows that the similarity between the two is that the flexure stresses are also the normal stresses but with a constant variation along the cross section on which applied. 

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List of Civil Engineering Softwares used these days

Hi, I was going through the online website of the Autodesk and found a list of the softwares, which caught my interest because many of them are new to me. I have worked with the AutoCad and am little aware of Revit, but there is this huge list of the softwares, titled as given below:

1. Building Design suite
2. AutoCAD
3. Revit
4. Simulation Mechanical
5. AutoCAD 360
6. Buzzsaw
7. Robot Structural Analysis Professional
8. Vault
9. Advance Steel
10. Point Layout
11. Simulation 360
12. AutoCAD LT
13. Advance Concrete

    

    License page for downloading the trial versions of Autodesk 

Now, if I was in the software industry related with the Civil Engineering projects, hopefully I would have been associated with the other 11 softwares too, but I have ignored them till now. Now that I am thinking of pursuing my M.Tech. in Structural Engineering, I think I should have the knowledge of these softwares. 

I think to learn about all these softwares, you need a lot of free time. At the same time, interest matters a lot. If you are interested, you can go the website and try them for free, and if they fulfill your thirst, you can buy them and yes they are very costly.

What do you think, is it important to have the knowledge of every new software product in the market?

Welding | Design of Steel Structure

Hi,

Welding is a method in which metals are joined together by melting them. This post discusses fillet weld and slot weld, in brief.

Fillet Welds

• Size of fillet weld. It should not be less than the minimum allowable value given in the table  in the image above. Size of the fillet weld used along tee of an angle or rounded edge of a flange should not exceed three-forth the nominal thickness of an angle of flange leg.

• Throat of fillet weld. It is length of perpendicular from right angle corner to the hypotenuse.

Effective throat thickness = k*fillet size

Value of k depends upon angle between fusion faces. Value of k decreases with increase in angle between fusion face.

In most cases, a right angled fillet weld is used, for which k = 0.7.

• Effective length of fillet weld. It is equal to its overall length minus twice the weld size. Effective length of a fillet weld designed to transmit the loading should not be less than four times weld size. Equal return should be made equal to twice the size of the weld.

• Overlap. Overlap in lap joint should not be less than five times the thickness of the thinner plate as shown in the figure to right.
• Side Fillet. In a lap joint made by a side or longitudinal fillet weld, length of each fillet weld should not be less than perpendicular distance between them; the perpendicular distance between side fillets should not exceed sixteen times thickness of the thinner part connected. 

• Intermittent fillet weld. Any section of an intermittent fillet weld should have an effective length of not less than four time the weld size or 40 mm, whichever is greater. Clear spacing between ends or effective lengths of intermittent weld carrying stresses should not exceed 12.t for compression and 16.t for tension and in no case should be more than 20cm, 't; is the thickness of thinner part joined.

• Permissible stress and strength of fillet weld. Permissible stress in fillet weld is 108 MPa or 1100 kgf/cm2.

Permissible stresses in shear and tension are reduced to 80% for the fillet welds made during erection. Permissible stresses are increased by 25% if wind or earthquake load are taken into account. However, size of the weld should not be less than the size required when the wind or earthquake load is considered or neglected.

Slot or Plug Weld

Following specifications are used for the design of slot or Plug welds:

1. Width or diameter of slot should not be less than three times the thickness of the part in which slot is formed or 25 mm, whichever is greater.
2. Corners at the enclosed ends should be rounded to a radius not less than 1.5 times the thickness of upper plate or 12 mm, whichever is greater.
3. Distance between edges of the plates and slot or between edges of adjacent slots should not be less than twice the thickness of the upper plate.
4. Permissible stress is taken as 108 MPa or 1100 kgf/cm2.

Thanks!

Two Way Slabs - IS Codes Specification

Hi,

Indian Standard codes give following specification for two way slabs for their analysis.

• Case 1: Simply supported slabs which do not have adequate provision to resist torsion at corners, and to prevent the corners from lifting. 

Bending moment for short span,  Mx= Ax.w.Lx^2
Bending moment for long span, My = Ay.w.Ly^2
Where, Ax and Ay are moment co-efficients depending on the ratio Lx/Ly.

• Case 2. Simply supported on the four edges and corners of the slabs are held down.

If Mx and My are maximum bending moments per unit width in the middle strip of the slab in short and long spans respectively, then
                                                   Mx = Ax.w.Lx^2
                                                  My = Ay.w.Ly^2
Where Ax and Ay are co-efficients depending on the ratio Ly/Lx.
At the corners, top and bottom reinforcement should be provided for torsion.

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Structure Engineering notes for GATE and PSUs - part 13

Hello there,
How have you been!?

 Here is our next part for your preparation for the GATE and PSUs examinations.

All information is learned through books and practical exercises.

1. The symmetrical two-hinged parabolic arch when subjected to a uniformly distributed load on the whole span, is subject to normal thrust only.
2. In a two-hinged arch, an increase in temperature induces a maximum bending moment at the crown.
3. The normal thrust at any section of the arch is the component of interacting forces on the section along the tangent to the centerline of the arch.
4. The radial shear at any section of the arch is the component of the interacting forces on the section along the normal to the centerline of the arch.
5. The intercept between a given arch and the linear arch at a section is proportional to the bending moment at the section.
6. A fixed beam AB is subjected to a triangular load varying from zero at end A to 'w' per unit length at end B. The ratio of fixed end moment at end B to that at end A is 3/2.
7. The horizontal thrust due to the rise in temperature in a semi-circular two-hinged arch of radius R is proportional to 1/R2.

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Structure Engineering notes for GATE and PSUs - part 12

Hello there,
How have you been? Here is our next part for your preparation of GATE and PSUs examinations.

All information is learned through books and practical exercises.

• A T-section is used as a simply supported beam with uniform loading. The maximum bending stress for a given load will occur at the bottom of the section.
• Mohr's circle becomes zero when the both the axial stresses are equal in magnitude and of same nature and both planes being free from shear.
• For beams, I-sections are more economical than rectangular sections because most of the material is concentrated away from the centroid.
• The shear stress distribution over a rectangular cross section of a beam follows a parabolic path.
• Shear stress distribution diagram varies for different cross sectional shapes.
• Bending stress distribution diagram is similar for different sections.
• Polar moment of inertia is sum of MI about the XX and YY axes.
• A cantilever beam of uniform EI has span equal to 'L'. An upward force W acts at the mid point of the beam and a downward force P acts at the free end. In order that the deflection at the free end is zero, the relation between P and W should be W=16P/5.
• If K is defined as the ratio of Young's modulus of elasticity and the permissible stress in compression of a material used in the construction of a column, then the Rankine's constant used in finding the load carrying capacity of columns is proportional to 1/K.

Reference:

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