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
