THEORY EXAMINATION (SEM–IV) 2016-17 STRUCTURAL ANALYSIS-I

Course: B.Tech (Civil Engineering)
Subject Code: CE401
Subject Title: Structural Analysis–I
Exam Type: Theory
Duration: 3 Hours
Maximum Marks: 100

SECTION – A (10 × 2 = 20 Marks)

Concept-based short questions assessing fundamental theory and formulas

SECTION – B (5 × 10 = 50 Marks)

Numerical and application-based questions, requiring analytical skill

(a) Influence Line for Shear Force

A simply supported beam (L = 25 m) — find shear at 10 m from left end.
Steps:

Draw influence line (ordinate changes linearly).

Compute shear for 5 kN point load + 2.4 kN/m UDL (5 m long).

Max shear occurs when load position covers critical region.

(b) UDL Moving Load on Beam

UDL = 40 kN/m over 3 m on 18 m span.
Find:

Bending moment at 4 m from left.

Absolute maximum bending moment.
→ Use influence lines and envelope concept.

(c) Three-Hinged Parabolic Arch

Span = 24 m, rise = 4 m.
Loads: 50 kN (18 m from left) + UDL 30 kN/m (left portion).
Find:

Normal thrust

Radial shear at 6 m section
→ Use equations:

y=4x(24−x)242y = \frac{4x(24 - x)}{24^2}y=2424x(24−x)​

and equilibrium equations for three-hinged arch.

(d) Cantilever Deflection (Moment-Area Method)

Find slope and deflection at free end; given: Moment of inertia of AC = 2 × inertia of BC.
→ Use moment area theorems:

Area under M/EI = change in slope.

Moment of that area = deflection.

(e) Beam ABCDE – Conjugate Beam Method

12 m long, load = 100 kN at C.
Moment of inertia:

AB and DE = I

BD = 2I
Compute deflections at B and C using conjugate beam equations.

(f) Cantilever Beam (Unit Load Method)

Span = 2 m, P = 20 kN at free end, section = 100 × 200 mm, E = 30 kN/mm².
Formulas:

θ=PL22EI,δ=PL33EIθ = \frac{PL^2}{2EI}, \quad δ = \frac{PL^3}{3EI}θ=2EIPL2​,δ=3EIPL3​

Compute slope and deflection at midspan using superposition.

(g) Castigliano’s Theorem (Proof)

Strain energy U=12∑PiδiU = \frac{1}{2}\sum P_iδ_iU=21​∑Pi​δi​
Differentiating: ∂U/∂Pi=δi∂U/∂P_i = δ_i∂U/∂Pi​=δi​.
→ Used for redundant and indeterminate structure analysis.

(h) Short Theory

(i) Fatigue – failure under repeated stress cycles.
(ii) Polar Moment of Inertia (J) – measure of torsional resistance.
(iii) Unsymmetrical Bending – occurs when loading plane not coincident with one principal axis.
(iv) Reasons: inclined loads, asymmetric cross-section, eccentric loading.

SECTION – C (2 × 15 = 30 Marks)

Complex numerical problems

Q3 (a) Variable Inertia Beam

Simply supported beam under UDL w kN/mw \, kN/mwkN/m.
Variable EI — find maximum deflection using:

δmax=5wL4384EIeqδ_{max} = \frac{5wL^4}{384EI_{eq}}δmax​=384EIeq​5wL4​

where EIeqEI_{eq}EIeq​ = equivalent flexural rigidity (weighted average).

Q3 (b) Beam Deflection via Conjugate Beam Method (See Figure on Page 2)

Beam: 9 m total, 80 kN load at 3 m, variable inertia (I, 2I).
Find slope at supports and deflection under load using conjugate beam relationships:

M/EI diagram of real beam → Shear/Moment diagram of conjugate beam.M/EI \text{ diagram of real beam → Shear/Moment diagram of conjugate beam.}M/EI diagram of real beam → Shear/Moment diagram of conjugate beam. 

Q4 (a) Truss Deflection Using Unit Load Method

(See diagram on page 2.)
Given: E=200 kN/mm2E = 200\,kN/mm^2E=200kN/mm2, A=1250 mm2A = 1250\,mm^2A=1250mm2.
Load W=25kNW = 25 kNW=25kN.
Steps:

Calculate member forces by joint method.

Apply unit load at deflection point.

Use:

Δ=∑PipiLiAEΔ = \sum \frac{P_i p_i L_i}{AE}Δ=∑AEPi​pi​Li​​

where PiP_iPi​ = real force, pip_ipi​ = unit-load force.

Q4 (b) Three-Hinged Parabolic Arch (See Figure on Page 3)

Given: UDL = 30 kN/m, span = 36 m, rise = 6 m, concentrated 60 kN at right.
Find:

Normal thrust, radial shear, bending moment at quarter span.

Draw Bending Moment Diagram (BMD).

Q5 Frame Analysis (See Figure on Page 3)

Frame with load 3.4 kN → find resultant stress at A & B.
Given: dimensions 48 mm × 18 mm, lever arm = 120 mm.
Use combined bending and direct stress:

σ=PA±MyIσ = \frac{P}{A} ± \frac{M y}{I}σ=AP​±IMy​ 

Key Formulas Summary