(SEM VII) THEORY EXAMINATION 2022-23 DESIGN OF STEEL STRUCTURES

DESIGN OF STEEL STRUCTURES (KCE075)

B.Tech SEM VII – Complete Solved Question Paper (2022–23)
                                                                                                                ⏱ Time: 3 Hours | 📊 Marks: 100
                                                                                                            IS 800:2007 & Steel Tables Allowed

SECTION A

Attempt all questions in brief (2 × 10 = 20 Marks)

(a) Five types of steel structures                                         Industrial buildings

Truss structures                                                                   Steel bridges

Transmission towers                                                           Multi-storey steel buildings 

(b) How will you calculate wind load?

Wind load is calculated using IS 875 (Part 3):

Vz=Vb×k1×k2×k3V_z = V_b \times k_1 \times k_2 \times k_3Vz​=Vb​×k1​×k2​×k3​

Design wind pressure:

Pz=0.6Vz2P_z = 0.6 V_z^2Pz​=0.6Vz2​

Where VbV_bVb​ is basic wind speed and k1,k2,k3k_1, k_2, k_3k1​,k2​,k3​ are risk, terrain, and topography factors 

(c) Pattern of riveted joint

Common rivet patterns are chain riveting and zig-zag riveting.
(In exam, draw neat sketch showing rivets arrangement) 

(d) Assumptions of bearing type connections       Load is transferred by bearing of bolt/rivet

Friction between plates is neglected                      Load is uniformly distributed among fasteners

Failure occurs by bearing, shear, or tearing 

(e) Which angle sections are used in roof truss and why?

Single or double angle sections are used because they are lightweight, economical, easy to connect, and efficient in carrying axial forces 

(f) How efficiency can be increased in a tension member?

Efficiency can be increased by:                                Reducing number of bolt holes

Using higher grade steel                                         Proper detailing to reduce stress concentration

Using lug angles where required 

(g) Assumptions made while designing a compression member

Load acts axially                                                       Member is initially straight

Material is homogeneous and isotropic                  Failure occurs by buckling or crushing 

(h) Define lattice column

A lattice column consists of two or more main members connected by lacing bars to act as a single unit.
(Draw neat sketch in exam) 

(i) Slender cross section

A slender cross section is one in which width-to-thickness ratio exceeds limiting values, causing local buckling before yielding 

(j) Define rafter

A rafter is an inclined structural member supporting roof covering and transferring loads to columns or walls 

SECTION B

Attempt any THREE (10 × 1 = 30 Marks)

(a) Load combinations on frame

Given:                                                                             Dead load = 5 kN/m
Imposed load = 15 kN/m                                               Wind load = 10 kN/m

(i) DL + IL

=5+15=20 kN/m= 5 + 15 = 20 \text{ kN/m}=5+15=20 kN/m

(ii) DL + WL

=5+10=15 kN/m= 5 + 10 = 15 \text{ kN/m}=5+10=15 kN/m

The maximum design load = 20 kN/m 

(b) Types of butt weld and fillet weld

Types of butt weld:                                                      Square butt

Single V butt                                                                  Double V butt

Single U butt                                                                  Double U butt

Fillet weld: Triangular weld used in lap and T-joints.
(Draw neat sketches) 

(c) Design of single angle tension member (Outline)

Factored load = 400 kN                                                 fy = 250 MPa, fu = 410 MPa

Design steps:                                                                  Calculate required net area

Select suitable angle section                                          Check yielding, rupture, and block shear

Design gusset plate and lug angle connections

( Write formulas clearly in exam) 

(d) Design load on column ISMB 450

Given:                                                                           Height = 4 m, pin-ended
fy = 250 MPa

Steps:                                                                            Calculate effective length

Find slenderness ratio                                                  Obtain compressive stress from IS 800

Calculate design compressive strength

( Numerical steps expected) 

(e) Recommended position of purlins

Purlins are placed perpendicular to rafters and parallel to ridge line to support roof sheets efficiently and reduce bending moments.
(Draw roof truss sketch) 

SECTION C

Q3

(a) Short notes

(i) Notch toughness: Ability of steel to resist brittle fracture
(ii) Fatigue strength: Resistance to failure under repeated loading
(iii) Corrosion resistance: Ability to resist rusting and chemical attack 

(b) Tension bar numerical

Area = 100 × 10 = 1000 mm²                               Test specimen area = 800 mm²
Ultimate strength:

=400×103800=500 MPa= \frac{400 \times 10^3}{800} = 500 \text{ MPa}=800400×103​=500 MPa

Factor of safety:

=500150≈3.33= \frac{500}{150} \approx 3.33=150500​≈3.33

Gauge length calculated using IS provisions 

Q4

(a) Failure of bolted joint

Types of failures:                                                     Shear failure of bolt

Bearing failure                                                        Tearing of plate

Block shear failure
( Draw sketches) 

(b) Design of bolted truss joint

Using M16 black bolts (grade 4.6):                         Steps include:

Calculate shear capacity                                         Calculate bearing capacity

Determine number of bolts                                    Check spacing and edge distance 

Q5

(a) Block shear strength

Block shear strength is calculated considering tension and shear planes using IS 800 equations.
(Draw block shear failure pattern) 

(b) Design of tension member (210 kN)

Steps:                                                                      Calculate required net area

Select suitable section                                            Check yielding and rupture

Design bolted connection 

Q6

(a) Elastic buckling of slender column

Euler’s formula:

Pcr=π2EI(Leff)2P_{cr} = \frac{\pi^2 E I}{(L_{eff})^2}Pcr​=(Leff​)2π2EI​

Used for long columns 

(b) Two channels toe-to-toe column

Given load = 1500 kN
Steps include:                                                             Select channel section

Calculate effective length                                           Check compressive strength

Ensure safety against buckling 

Q7

(a) Design of laterally supported beam                      Span = 4 m, load = 400 kN

Steps:                                                                          Calculate bending moment

Check section modulus                                               Check deflection using ISMB 400

Verify safety 

(b) Laterally unsupported beam strength

ISLB 450 section:                                                 Design strength reduced due to lateral-torsional buckling.
Calculate using IS 800 provisions for unsupported compression flange