Skip to main content

Analysis and Design of Structures - short answer type questions from AMIE exams (Winter 2018)

Answer the following (10 x 2)

State the conditions for a stable determinate structure.
  • The structures are grouped into statically determinate and statically indeterminate structures. A structural system that can be analysed by using equations of statical equilibrium only (∑Fx = 0, ∑Fy = 0 and ∑M=0) is called a statically determinate structure, e.g., beams or trusses with both ends simply supported, one end hinged and another on rollers and the cantilever type. 
  • A structure that cannot be analysed by using equations of equilibrium only is called a statically indeterminate structure, e.g., fixed beams, continuous beams, propped cantilevers. To analyse indeterminate structures, apart from using equations of equilibrium one has to determine the various deformations and make use of compatibility conditions. 
  • Indeterminate structures arc also called redundant structures.
When do you go for the method of sections?
  • In the section method, the truss is cut into two parts and equilibrium equations are written for one of the parts of the cut truss treating it as a free body. 
  • Another of the methods used in analysing a truss is the method of joints. This method entails the use of a free-body diagram of joints with the equilibrium equations ∑Fx = 0, ∑Fy = 0.
State the concepts involved in the conjugate-beam method.
The conjugate beam method is a method for determining slopes and deflections in beams. The method can also be conveniently used for continuous beams. This method is based on a mathematical correspondence that exists between moment vs. load function and deflection vs. M/EI functions in a
beam. The following figure shows these relations.


Theorems
  • The shear at any point on the conjugate beam is equal (in sign and value) to the slope at the corresponding point on the real beam.
  • The moment at any point on the conjugate beam is equal (in sign and value) to the deflection at the corresponding point on the real beam.

What are the uses of influence line diagrams?
The steps involved in determining S.F. and B.M. at different sections of a beam as the rolling loads move from one end to the other are rather cumbersome. Influence lines are interesting and are a very useful tool in dealing with rolling loads.

Definition
An influence line is defined as a function whose value at a point represents some structural quantity as a unit load is placed at that point.

When do you expect sway in frames?
in frames, the translation of some joints is common due to forces acting in the lateral direction as in Fig. (a) or due to asymmetrical forces as in Fig.(b) or due to asymmetry in the make-up of the frame even though the load is symmetrical as in Figs. (c and d).

(a)

(b)

(c)


What is the type of connections used in steel structures?
The various types of connections used for connecting the structural members are given below :
  • Riveted connections
  • Bolted connections
  • Pin connections
  • Welded connections.
These connections are named after the type of fastening (viz., rivets, bolts and nuts, pins and welds) used for connecting the structural members

What is the difference in the design methodology of tension and compression members?

Compression members  
  • Structural Members subjected to axial compression/compressive forces  
  • Design governed by strength and buckling  
  • Columns are subjected to axial loads through the centroid.  
  • The stress in the column cross-section can be  calculated as  f = P/A where f is assumed to be uniform over the entire cross-section
  • The failure of the column depends upon its slenderness ratio.
Tension members
  • A tension member is defined as a structural member subjected to tensile force in the direction parallel to its longitudinal axis.
  • When a tension member is subjected to axial tensile force, then the distribution of stress over the cross-section is uniform. The complete net area of the member is effectively used at the maximum permissible uniform stress.
  • The various members used as tension members are not perfectly straight.
  • The initial crookedness and imperfections of the member result in a small eccentricity. The distribution of stress in the eccentrically loaded tension member is not uniform. The member is subjected to combined stress, (i.e., direct tension and bending). Besides this, a tension member may be subjected to axial tension and bending.
  • Two or more two members are used to form built-up members. When the single rolled steel sections cannot furnish the required area, then, the built-up sections are used. The built-up sections may be made more rigid and stiffer than the single structural shapes. The moment of inertia of the built-up sections may be increased.
  • A built-up section may be made of two channels placed back to back with a gusset plate in between them.
  • The heavy built-up tension members in the bridge truss girders are made of angles and plates. Such members can resist compression if the reversal of stresses occurs.
What are the loads to be considered in the design of Gantry Girders?
Type of Load on Gantry Gutter
  • Vertical Loads on Gantry Gutter.
  • Lateral Loads on Gantry Gutter.
  • Longitudinal Loads on Gantry Gutter.
  • Impact Loads on Gantry Gutter.

Write a brief note on the design of R.C. slabs.

One way slab
For design purposes, a strip (beam) of unit width along the supported edge in (i) or along the shorter span in (ii) is considered for analysis of bending moment and shear forces. 

Two-way slab
The design load on any point (i.e. on any common unit area crossed by two mutually perpendicular strips of unit width) is shared by both the spans. A two-way slab depending on the support conditions may be either of the following types :
  • Simply supported slab having no adequate provision to resist torsion at corners and to prevent corners from lifting
  • When the corners of a slab are prevented from lifting.
How will you design a footing?
The following steps are followed:
  • Step 1: Transfer of axial force at the base of the column 
  • Step 2: Size of the footing 
  • Step 3: Thickness of footing 
  • Step 4: Minimum reinforcement 
  • Step 5: Check for the gross base pressure 
---
The study material for AMIE/B Tech/Junior Engineer exams is available at https://amiestudycircle.com 

Comments

Popular posts from this blog

Mechanics of Fluids (Solved Numerical Problems)

Numerical The surface Tension of water in contact with air at 20°C is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm² greater than the outside pressure. Calculate the diameter of the droplet of water. (7 marks) (AMIE Summer 2023) Solution Surface tension, σ = 0.0725 N/m Pressure intensity, P = 0.02 N/m 2 P = 4σ/d Hence, the Diameter of the dropd = 4 x 0.0725/200 = 1.45 mm Numerical Find the surface tension in a soap bubble of 40 mm diameter when the inside pressure is 2.5 N/m² above atmospheric pressure. (7 marks) (AMIE Summer 2023) Answer: 0.0125 N/m Numerical The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if surface tension is given as 0.0725 N/m of water. (AMIE Summer 2023, 7 marks) Answer: 0.725 N/cm 2   Numerical An open lank contains water up to a depth of 2 m and above it an oil of specific gravity 0.9 for a depth of 1 m. Find the pressure intensity (i) at t...

Energy Systems (Solved Numerical Problems)

Wind at 1 standard atmospheric pressure and \({15^0}C\) has velocity of 15 m/s, calculate (i) the total power density in the wind stream (ii) the maximum obtainable power density (iii) a reasonably obtainable power density (iv) total power (v) torque and axial thrust Given: turbine diameter = 120 m, and turbine operating speed = 40 rpm at maximum efficiency. Propeller type wind turbine is considered. (AMIE Winter 2023) Solution For air, the value of gas constant is R = 0.287 kJ/kg.K 1 atm = 1.01325 x 105 Pa Air density \(\rho  = \frac{P}{{RT}} = \frac{{1.01325x{{10}^5}}}{{287}}(288) = 1.226\,kg/{m^3}\) Total Power \({P_{total}} = \rho A{V_1}^3/2\) Power density \(\begin{array}{l}\frac{{{P_{total}}}}{A} = \frac{1}{2}\rho {V_1}^3\\ = \frac{1}{2}(1.226){(15)^3}\\ = 2068.87{\mkern 1mu} W/{m^2}\end{array}\) Maximum power density \(\begin{array}{l}\frac{{{P_{\max }}}}{A} = \frac{8}{{27}}\rho A{V^3}_1\\ = \frac{8}{{27}}(1.226){(15)^3}\\ = 1226{\mkern 1mu} W/{m^2}\end{array}\) Assuming eff...

Design of Electrical Systems (Solved Numerical Problems)

Important note There is something wrong with this question paper. It seems that instead of "Design of Electrical Systems" the IEI has given problems from "Electrical Machines". You should raise a complaint to director_eea@ieindia.org in this regard. Numerical A 120 V DC shunt motor draws a current of 200A. The armature resistance is 0.02 ohms and the shunt field resistance is 30 ohms. Find back emf. If the lap wound armature has 90 slots with 4 conductors per slots, at what speed will the motor run when flux per pole is 0.04 Wb?​ (AMIE Summer 2023, 8 marks) Solution The back EMF (E b ) of a DC motor can be calculated using the formula: E b = V - I a R a   Given: V = 120 V I a = 200 A R a = 0.02 ohms Substituting the values into the formula: E b = 120 − 200 × 0.02 = 120 − 4​ = 116 V Now, let's calculate the speed (N) at which the motor will run using the given flux per pole (φ p ). The formula to calculate the speed of a DC motor is: N = 60×E b /(P×φ p ) Wh...