🔩 Design of Steel Structures 🔥 Day 1
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Steel Structures — Design of Structures

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Design of Steel Structures — Complete Study Notes
6 chapters · GATE, ESE (IES), SSC JE, State PSC, RRB JE
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Design of Steel Structures applies the analysis results from Theory of Structures to size real structural members as per IS 800 — from the Working Stress Method fundamentals, through structural fasteners (rivets, bolts, welds), tension and compression member design, beam design and lateral torsional buckling, plate girders and industrial roof systems, up to plastic (ultimate load) analysis. Every formula, IS code clause, diagram, solved example and exam-pattern table is included.

Learning Objectives

After studying this chapter you will be able to:

  • Design riveted, bolted (black bolt & HSFG) and welded connections, and check the strength of eccentric bolt/weld groups
  • Compute gross and net area of tension members, apply the shear-lag factor for single-angle connections, and design lug angles
  • Determine the permissible compressive stress from slenderness ratio and design laced/battened built-up compression members
  • Design steel beams for bending, shear and deflection, and identify when lateral torsional buckling governs the permissible bending stress
  • Proportion the web, flanges and stiffeners of a plate girder, and design purlins and crane girders for industrial roof structures
  • Apply plastic analysis — shape factor, plastic hinge formation, static and kinematic methods — to find the collapse load of indeterminate beams

Exam Relevance

ESE★★★★★
GATE★★★★☆
SSC JE★★★★☆
State PSC★★★☆☆
💡 Steel Structures is a high-weightage design subject in GATE Civil and ESE — rivet/bolt/weld strength, permissible stresses (WSM), slenderness ratio limits, and plastic collapse loads are asked almost every year.

Concept Map

IS 800 & Working Stress Method → Permissible Stresses
↓
Structural Fasteners (Rivets, Bolts, Welds) → Eccentric Connections
↓
Tension Members (Net Area, Shear Lag) ↔ Compression Members (Slenderness, Lacing/Battening)
↓
Beams (Bending, Shear, LTB) → Plate Girders & Industrial Roofs
↓
Plastic Analysis (Shape Factor, Collapse Load, Load Factor)

Where This Fits

Prerequisite: Theory of Structures (determinacy, analysis methods and BM/SF diagrams computed there are the direct inputs to member sizing here). Related: RCC/PCC Design covers the equivalent design process for reinforced/plain concrete members.

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Ch 1 · Structural Fasteners – Rivets, Bolts & Welds
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1.1 Introduction to IS 800 and WSM

Steel design in India is governed primarily by IS 800:2007 (Limit State Method) and the older IS 800:1984 (Working Stress Method). For competitive exams (GATE, ESE, SSC JE), both versions are relevant; traditional syllabi still emphasise WSM heavily.

ℹ️ Working Stress Method (WSM): Stresses in the structure are kept below permissible values which are a fraction of yield/ultimate strength. FOS is applied to material strength directly. Elastic analysis is used; no redistribution of forces. Key permissible stresses as per IS 800:1984 — \(\sigma_{at}\) = 150 MPa (tension), \(\sigma_{ac}\) = depends on slenderness, \(\tau_{va}\) = 100 MPa (shear).

1.2 Riveted Connections

Rivets are permanent fasteners driven hot and cooled in place, forming a head on the free end. Though largely replaced by bolts and welds in modern construction, they remain important for exam purposes.

Modes of Failure of a Riveted Joint

  • Shearing of rivet: Rivet shank shears across the interface
  • Crushing (bearing) of rivet: Rivet or plate material crushes under bearing pressure
  • Tearing of plate: Net section of plate fractures in tension
  • Shear-out (end shear) of plate: Plate shears out behind the end rivet
  • Splitting of plate: Plate splits at edge of hole (insufficient edge distance)

Strength of a Single Rivet (WSM)

Strength in single shear: \(P_s = (\dfrac{\pi}{4}) \times d^{2} \times \tau_{va}\). Strength in double shear: \(P_s = 2 \times (\dfrac{\pi}{4}) \times d^{2} \times \tau_{va}\). Strength in bearing: \(P_b = d \times t \times \sigma_{pb}\). Strength in tearing: \(P_t = (p - d) \times t \times \sigma_{at}\).

Where: d = gross diameter of rivet (nominal dia + 1.5 mm for hot-driven rivets, per IS 1929; use gross dia for shear, net for tearing); \(\tau_{va}\) = permissible shear stress in rivet = 100 MPa (IS 800); \(\sigma_{pb}\) = permissible bearing stress = 300 MPa (IS 800); \(\sigma_{at}\) = permissible axial tensile stress in plate = 150 MPa; p = pitch of rivets, t = thickness of plate. Rivet value \(R = \min(P_s, P_b)\) for a single rivet in a lap/butt joint.

Efficiency of a Riveted Joint

η = (Strength of joint per pitch) / (Strength of solid plate per pitch). For a single-riveted lap joint with one rivet per pitch: \(\eta_{tearing} = (p - d) / p\) and \(\eta_{shearing} = P_s / (p \times t \times \sigma_{at})\). A balanced joint has \(\eta_{tearing} = \eta_{shearing}\), giving the optimal pitch.

1.3 Rivet Spacing Rules (IS 1929 / IS 800)

DimensionMinimumMaximumIS 800 Clause
Pitch (p)3d (d = gross rivet dia)16t or 200 mm (in tension) / 12t or 200 mm (in compression)Cl. 10.2
Edge distance (e)1.5d (sheared edge), 1.25d (rolled/sawn)12t (12 × plate thickness)Cl. 10.2.4
Back pitch / gauge (g)3dSame as pitch max—

1.4 High-Strength Friction Grip (HSFG) Bolts

HSFG bolts (IS 3757) are tightened to a high preload (proof load), creating clamping friction between surfaces. Load transfer is by friction, not shear of the bolt shank — no slippage at working loads. Superior fatigue performance; used in bridges and dynamically loaded structures.

Slip resistance of HSFG bolt (per bolt per interface): \(P_{sf} = \mu \times T_f \times n_e\), where \(\mu\) = slip factor (0.45 for grit-blasted surfaces; 0.2 for as-rolled), \(T_f\) = proof load of bolt (from IS 1367 tables), \(n_e\) = number of effective interfaces (1 for single shear, 2 for double). For an M20 bolt Grade 8.8: \(T_f = 144\) kN; proof stress = 628 MPa.

1.5 Black Bolts vs HSFG Bolts

FeatureBlack Bolt (IS 1364)HSFG Bolt (IS 3757)
Load transferShear of bolt shank + bearing on holeFriction between clamped surfaces
Clearance hole2–3 mm larger than bolt1–2 mm larger; tight tolerances
Slip at working loadMay slip into bearing (not ideal for fatigue)No-slip; rigid connection
Grade designations4.6 (UTS 400 MPa, Fy 240); 8.8; 10.98.8 (most common)
UseOrdinary structures, non-fatigue loadsBridges, cranes, fatigue loading
TighteningSnug tight (hand-tight)Full proof load via torque wrench / turn-of-nut

1.6 Welded Connections

Welds are classified into butt welds (groove welds) and fillet welds. For competitive exams, fillet weld design is the most important.

Fillet Weld Design (WSM)

Effective throat thickness: \(t_e = 0.707 \times s\) (for 90° fillet welds; s = weld size = leg length). IS 816 uses effective throat \(= K \times s\), K = 0.7 for a 90° weld angle. Permissible shear stress in weld (on throat): \(\tau_w = 110\) MPa (IS 816, E41 electrode). Strength of fillet weld per unit length: \(q = t_e \times \tau_w = 0.707 \times s \times 110\) (N/mm per mm run of weld).

Weld size limits — minimum weld size depends on the thicker part being joined: up to 10 mm plate → 3 mm weld; 10–20 mm → 5 mm; 20–32 mm → 6 mm; > 32 mm → 8 mm. Maximum weld size: t − 1.5 mm for plates ≤ 6 mm; = t for plates > 6 mm (rounded corner). Minimum length of fillet weld: max(4s, 40 mm). Effective length = total length − 2s (deduct for starting/stopping craters).

Butt Welds

Full penetration butt weld (CJP): effective throat = thickness of thinner plate joined. Strength = throat × length × \(\sigma_{at}\) (permissible stress same as parent metal). There is no reduction for joint efficiency in full-penetration butt welds → 100% efficient.

1.7 Eccentric Connections — In-plane Eccentricity

When load P acts at eccentricity e from the CG of a bolt/weld group, the moment \(M = P \times e\). For a bolt group, the force on the most critical bolt has two components: Direct shear \(F_d = P / n\) (n = number of bolts) and Torsional shear \(F_t = M \times r_{max} / (\Sigma r^{2})\), where \(r_{max}\) = distance of the farthest bolt from the CG and \(\Sigma r^2\) = sum of squared distances. Resultant on the critical bolt: \(F_R = \sqrt{F_d^{2} + F_t^{2} + 2\cdot F_d\cdot F_t\cdot \cos\theta}\), where \(\theta\) = angle between \(F_d\) and \(F_t\) directions at the critical bolt.

1.8 Beam-Column Connections

Connection TypeMoment TransferRotationUse
Simple (pin) connectionNo moment (shear only)Free rotationNon-moment frames; beam-to-girder web
Semi-rigid connectionPartial momentPartial rotationComposite frames; web cleats + top cleat
Rigid (fixed) connectionFull momentNo relative rotationPortal frames; moment-resisting frames
📝 GATE Tip: For WSM, rivet value R = min(shear strength, bearing strength). Efficiency η = R / (p × t × σ_at) × 100%. Maximum efficiency for a single rivet lap joint is ~100% theoretically but practically 70–80% due to minimum pitch constraints.
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Ch 1 · Formulas
Rivet Strength (Single Shear / Bearing)
$$ P_s = \frac{\pi}{4}d^2\tau_{va},\quad P_b = d\cdot t\cdot\sigma_{pb} $$
Double shear: \(P_s\) × 2. Rivet value \(R=\min(P_s,P_b)\)
N
Joint Efficiency
$$ \eta = \frac{\min(P_s,P_b,P_t)}{p\times t\times\sigma_{at}}\times100\% $$
Balanced joint: \(\eta_{tearing}=\eta_{shearing}\)
%
HSFG Bolt Slip Resistance
$$ P_{sf} = \mu\times T_f\times n_e $$
μ = 0.45 (grit-blasted), 0.2 (as-rolled)
N
Fillet Weld Strength per Unit Length
$$ q = t_e\times\tau_w = 0.707\,s\times110 $$
Min length = max(4s, 40mm); effective length = total − 2s
N/mm
Resultant Force on Critical Bolt
$$ F_R=\sqrt{F_d^2+F_t^2+2F_dF_t\cos\theta} $$
\(F_t=M\cdot \dfrac{r_{max}}{\Sigma} r^2\); \(F_d=\dfrac{P}{n}\)
N
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Ch 1 · Solved Example
GATE CE / ESE level
Example 1 Medium
A power-driven rivet of gross diameter 20 mm connects two plates each 12 mm thick in single shear. Given \(\tau_{va} = 100\) MPa and \(\sigma_{pb} = 300\) MPa, find the rivet value.
Approach hint
Compute shear strength \(P_s\) and bearing strength \(P_b\) separately, then take the rivet value as the minimum of the two.
  1. \(P_s = (\dfrac{\pi}{4})\times20^2\times100 = 31{,}416\) N ≈ 31.4 kN
  2. \(P_b = d\times t\times\sigma_{pb} = 20\times12\times300 = 72{,}000\) N = 72.0 kN
  3. \(R = \min(P_s, P_b) = \min(31.4, 72.0) = 31.4\) kN
✓ Rivet value R = 31.4 kN (shear governs)
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Ch 1 · Diagrams
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Ch 1 · Quick Quiz
3 questions · tap an option to reveal answer
Question 1
Load transfer in an HSFG bolted connection occurs primarily by:
HSFG bolts are pre-tensioned to a high proof load, clamping the plates together so the load transfers by friction — not by shear or bearing on the bolt ✓
Question 2
The effective throat thickness of a 90° fillet weld of leg size s is:
For a 90° fillet weld, the throat thickness is the perpendicular distance from the root to the hypotenuse: \(t_e = 0.707s\) (K = 0.7 per IS 816) ✓
Question 3
Which of the following is NOT a mode of failure of a riveted joint?
Lateral torsional buckling is a beam failure mode (Ch 4), not a riveted-joint failure mode. The five joint modes are shearing, crushing/bearing, tearing, shear-out, and splitting of plate ✓
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Ch 2 · Tension Members
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2.1 Introduction

Tension members are structural elements carrying axial tensile forces — bottom chords of trusses, hangers, cable stays, tie rods, bracing members. Their design is governed by avoiding yielding of the gross section and fracture at the net section.

2.2 Gross Area and Net Area

Gross area \(A_g\) = full cross-sectional area (ignoring holes). Net area at a bolt/rivet hole section: \(A_n = A_g - n \times d_h \times t\), where \(n\) = number of holes at the critical section, \(d_h\) = hole dia = rivet dia + 1.5 mm (IS 800), \(t\) = thickness of element.

For staggered holes — net width (Cochrane formula): \(w_n = w_g - \Sigma d_h + \Sigma(\dfrac{s^{2}}{4g})\), where \(s\) = stagger (longitudinal spacing between holes), \(g\) = gauge (transverse spacing); \(A_n = w_n \times t\). Apply this to all possible failure paths and use the minimum net area.

2.3 Permissible Tensile Stress (WSM — IS 800:1984)

Permissible axial tensile stress: \(\sigma_{at} = 0.6 \times f_y\). For Fe250 (mild steel, \(f_y\) = 250 MPa): \(\sigma_{at} = 150\) MPa. Design tensile capacity: \(T_{dg} = A_g \times \sigma_{at}\) (based on gross area yielding); \(T_{dn} = A_n \times 0.9 \times f_u / \gamma_{m1}\) (LSD approach for fracture at net section). WSM: \(T_{allowed} = A_n \times \sigma_{at}\) (net area governs when holes are present).

2.4 Shear Lag and Net Section Efficiency

When only one leg of an angle is connected, the outstanding (unconnected) leg does not carry full stress — this is shear lag. The net effective area is reduced by a shear lag factor.

Net effective area for outstanding leg (IS 800:1984): \(A_{eff} = A_1 + A_2 \times k\), where \(k = 3A_1 / (3A_1 + A_2)\); \(A_1\) = net area of connected leg (after deducting hole area), \(A_2\) = gross area of unconnected leg. Alternatively per IS 800:2007, shear lag factor \(\beta = 1 - \dfrac{w_s}{L_c}\) (for connection length \(L_c\), shear lag width \(w_s\)).

2.5 Types of Tension Members

TypeTypical SectionAdvantageLimitation
Single angleISA (unequal/equal)Simple; easy to connectEccentric connection; shear lag; unsymmetric
Double angle (back-to-back)2×ISA with gussetSymmetric; greater areaNeed lacing/batten plates for compound section
T-sectionIST (rolled T)Direct connection to gussetLimited sections available
ChannelsISMC, ISSCLarger area; stiff laterallySingle-sided connection eccentricity
Built-up box section2 channels + platesLarge area; symmetric; good torsional stiffnessMore fabrication; expensive
Wire ropes / rodsCircularVery high tension capacity; flexibleNo compression capacity; needs turnbuckles

2.6 Lug Angles

A lug angle is a short angle cleat welded or bolted to the outstanding leg of a main angle member near the connection point, to transfer part of the load directly, reducing the shear lag effect and the required connection length.

With lug angles, the effective net area = full gross area (shear lag eliminated). Lug angle force = \((A_2 / A_{total}) \times P\), where \(A_2\) = unconnected leg area. The lug angle connection must carry ≥ 120% of the force in the attached leg (IS 800).

📝 GATE Tip: For a single angle tension member connected through one leg: \(A_{eff}=A_1+k\cdot A_2\) where \(k=\dfrac{3A_1}{3A_1+A_2}\). If both legs are connected (via gusset both sides), \(A_{eff}=A_g\) (full gross area). Memorise the k formula — it appears every 2–3 years in GATE.
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Ch 2 · Formulas
Net Area (No Stagger)
$$ A_n = A_g - n\times d_h\times t $$
\(d_h\) = rivet/bolt dia + 1.5 mm
mm²
Net Width — Staggered Holes (Cochrane)
$$ w_n = w_g - \Sigma d_h + \Sigma\frac{s^2}{4g} $$
Check all failure paths; take minimum \(A_n=w_n\times t\)
mm
Permissible Tensile Stress
$$ \sigma_{at} = 0.6f_y = 150\ \text{MPa (Fe250)} $$
WSM design capacity: \(T_{allowed}=A_n\times\sigma_{at}\)
MPa
Shear Lag Factor (Single Angle)
$$ A_{eff}=A_1+kA_2,\quad k=\frac{3A_1}{3A_1+A_2} $$
Both legs connected → \(A_{eff}=A_g\)
mm²
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Ch 2 · Solved Example
GATE CE / ESE level
Example 1 Medium
A single equal angle ISA 100×100×10 mm is connected through one leg only. The connected leg has net area \(A_1 = 850\) mm² and the unconnected leg has gross area \(A_2 = 950\) mm². Find the effective net area \(A_{eff}\).
Approach hint
Compute the shear-lag factor \(k=\dfrac{3A_1}{3A_1+A_2}\) first, then apply \(A_{eff}=A_1+kA_2\).
  1. \(k = 3\dfrac{\times850}{3\times850+950} = \dfrac{2550}{3500} = 0.729\)
  2. \(A_{eff} = A_1 + kA_2 = 850 + 0.729\times950 = 850+692 = 1542\) mm²
✓ \(A_{eff} \approx 1542\) mm² (compare to full \(A_g \approx 1800\) mm² — shear lag reduces capacity by about 14%)
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Ch 2 · Diagrams
Click any diagram to zoom
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Ch 2 · Quick Quiz
3 questions · tap an option to reveal answer
Question 1
For a plate with staggered bolt holes, the correct approach to find the design net area is to:
Since fracture will occur along whichever path is weakest, every possible zig-zag/straight path must be checked using the Cochrane formula, and the minimum net area governs design ✓
Question 2
Shear lag reduces the effective area of a tension member when:
When only one leg is connected, the unconnected (outstanding) leg lags behind in picking up stress near the connection — this shear lag effect reduces the effective area below the full net area ✓
Question 3
The primary purpose of a lug angle is to:
A lug angle picks up load directly from the outstanding leg near the connection, effectively eliminating shear lag so the full gross area can be used in design ✓
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Ch 3 · Compression Members
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3.1 Introduction

Compression members (struts, columns, top chords of trusses, stanchions) must resist yielding of the cross-section and flexural buckling (Euler-type instability). As slenderness ratio increases, buckling becomes the governing criterion.

3.2 Permissible Axial Compressive Stress (IS 800:1984)

Permissible compressive stress \(\sigma_{ac}\) depends on slenderness ratio \(\lambda = \dfrac{l}{r}\). From IS 800 Table 5.1 (for \(f_y = 250\) MPa):

\(\lambda\)050100150180200
\(\sigma_{ac}\) (MPa)150.0139.0107.564.446.538.4

At \(\lambda = 0\), \(\sigma_{ac} = \sigma_{at} = 150\) MPa since the gross section simply yields. Maximum permitted \(\lambda\) for main members is 180 (\(\lambda=200\) is only reached for secondary/bracing members). Maximum slenderness ratio (IS 800:1984): main compression members \(\lambda_{max} = 180\); secondary members (bracing) \(\lambda_{max} = 200\). Design: \(P_{allowed} = A_g \times \sigma_{ac}(\lambda)\).

3.3 Effective Length of Compression Members

End ConditionEffective Length \(L_{eff}\)IS 800 Factor
Both ends pin (pin-pin)L1.0 L
Both ends fixed0.5 L0.5 L
One fixed, one pin0.7 L (≈ \(\dfrac{L}{\sqrt{2}}\))0.7 L
One fixed, one free (flagpole)2.0 L2.0 L
Fixed-pin with restraint against sway0.85 L0.85 L
Both fixed with sway possible1.2 L1.2 L

3.4 Lacing and Batten Plate Design

Lacing

Lacing is designed for a transverse shear force \(V\) = 2.5% of \(P\) (axial column load), per IS 800 Cl. 7.6.6.1 — this shear is assumed to act on the whole column. For N-lacing (single), force in each lacing bar: \(F_{lace} = \dfrac{V}{2\sin\theta}\), where \(\theta\) = angle of the lacing bar with the axis perpendicular to the member axis.

Lacing bar slenderness: \(\dfrac{l}{r_{min}} \le 145\) (single lacing). Lacing angle to longitudinal axis: 40° to 70° (optimal ~50°–60°). Minimum width of flat lacing: l/40 but not < 25 mm. Minimum thickness: l/60 for single; l/40 for double (l = lacing length).

Batten Plates

Battens are designed for longitudinal shear \(V_L = V \times C / (2 \times n_b \times S)\) and moment in batten \(M = V \times C / (2 \times n_b)\), where \(C\) = distance between centroids of connections on each main member, \(n_b\) = number of bays, \(S\) = distance between panel points.

Batten dimensions (IS 800): depth ≥ 0.75b (b = distance between inner edges of main members); thickness ≥ 1/50 × distance between inner edges, min 6 mm; spacing not more than 0.7 times the min radius of gyration of the main member × 50.

3.5 Splices in Compression Members

A splice is a joint connecting two lengths of the same member. Compression splices must transfer the full compressive force plus any accidental bending. If ends are machined flat (milled ends): 50% of load is transferred by direct bearing, 50% by fasteners. If ends are not milled: full load transfer through fasteners. Splice plate design: sized to carry the full design load in compression. IS 800 requires the splice to carry at least 50% of \(P\) (even if milled ends are assumed).

📝 GATE Tip: Maximum slenderness ratio for main compression members = 180; for secondary/bracing members = 200. Lacing angle optimum = 40°–70°. Transverse shear for lacing design = 2.5% of P. These three values appear repeatedly in GATE and ESE.

3.6 Column Bases (Base Plates)

A column base transfers the column load to the concrete pedestal/footing over an area large enough to keep the bearing pressure within the permissible value for concrete. Two principal types:

TypeDescriptionLoad type
Slab baseThick plate directly under the column, machined for bearing; column may be faced for direct bearingAxial (concentric) load
Gusseted baseBase plate + gusset plates/angles that increase the load-transfer area and stiffnessHeavier axial loads
Moment / pocket baseBase with anchor bolts resisting uplift/moment (or column grouted into a pocket)Axial load + moment

Slab base design (IS 800 Cl. 7.4): the plate thickness is governed by the cantilever bending of the plate projection under the uniform bearing pressure \(w\):

$$ t_s = \sqrt{\frac{2.5\,w\,(a^2 - 0.3b^2)\,\gamma_{m0}}{f_y}} $$

where \(a, b\) = larger and smaller projections of the plate beyond the column, \(w\) = uniform pressure under the plate. Anchor bolts (holding-down bolts) connect the base to the foundation and resist uplift and shear; a minimum of four bolts is provided for stability. For bases carrying moment, the plate is designed for the resulting non-uniform (trapezoidal or triangular, with anchor-bolt tension) pressure distribution.

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Ch 3 · Formulas
Slenderness Ratio & Design Load
$$ \lambda = \frac{L_{eff}}{r},\quad P_{allowed}=A_g\cdot\sigma_{ac}(\lambda) $$
\(\sigma_{ac}\) read from IS 800 Table 5.1
—
Maximum Slenderness Ratios
$$ \lambda_{main}\le180,\quad \lambda_{bracing}\le200 $$
IS 800:1984 limits
—
Lacing Bar Force
$$ F_{lace} = \frac{V}{2\sin\theta},\quad V=0.025P $$
θ = lacing angle, typically 40°–70°
N
Batten Longitudinal Shear & Moment
$$ V_L=\frac{VC}{2n_bS},\quad M=\frac{VC}{2n_b} $$
C = centroid spacing, \(n_b\) = bays, S = panel spacing
N, N·mm
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Ch 3 · Solved Example
GATE CE / ESE level
Example 1 Easy
A built-up laced column carries an axial load P = 800 kN. Find the transverse shear force for which the lacing system must be designed, and the force in a single lacing bar inclined at 50° if this shear splits equally between two lacing planes.
Approach hint
The transverse shear for lacing design is always taken as 2.5% of the axial load P (IS 800 Cl. 7.6.6.1), regardless of the actual computed transverse shear. Divide between the two lacing planes, then resolve using \(F_{lace}=\dfrac{V}{2\sin\theta}\) for each plane's share.
  1. Total transverse shear: \(V = 0.025 \times P = 0.025\times800 = 20\) kN
  2. Shear per lacing plane (2 parallel lacing systems): \(V_{plane} = \dfrac{20}{2} = 10\) kN
  3. Force in one lacing bar: \(F_{lace} = \dfrac{V_{plane}}{2\sin50°} = \dfrac{10}{2\times0.766} = 6.53\) kN
✓ Design transverse shear V = 20 kN; force per lacing bar ≈ 6.53 kN
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Ch 3 · Diagrams
Click any diagram to zoom
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Ch 3 · Quick Quiz
3 questions · tap an option to reveal answer
Question 1
The permissible compressive stress \(\sigma_{ac}\) for a compression member:
As λ increases, buckling instability governs over material yielding, so IS 800 Table 5.1 gives progressively lower permissible stress — from 150 MPa at λ=0 down to 38.4 MPa at λ=200 ✓
Question 2
The effective length of a column fixed at both ends is:
Both ends fixed gives the shortest effective length, 0.5L — the stiffest end condition (compare: pin-pin = L, fixed-free = 2L) ✓
Question 3
Per IS 800 Cl. 7.6.6.1, the transverse shear force used to design lacing is taken as:
IS 800 conservatively assumes a nominal transverse shear of 2.5% of the axial load P acts on the whole laced/battened column, regardless of actual computed shear ✓
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Ch 4 · Beams – Design of Steel Beams
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4.1 Classification of Steel Beams

TypeSectionSpan RangeApplication
Simple rolled beamISMB, ISWB, ISHBUp to 8–10 mFloors, bridges, crane girders
Compound beamRolled section + cover plates10–15 mWhen standard sections insufficient
Built-up girderPlate girder (web + flanges)15–30 mHeavy loads, longer spans
Composite beamSteel beam + RC slab8–25 mModern buildings; bridges
Castellated beamISMB with hexagonal web cut-outs10–20 mLight loads; service passage through web

4.2 Permissible Bending Stress (WSM — IS 800:1984)

For compact sections with full lateral support: \(\sigma_{bt} = \sigma_{bc} = 0.66 \times f_y = 165\) MPa (for Fe250, \(f_y\) = 250 MPa). Required section modulus: \(Z_{required} = M_{max} / \sigma_{bc}\).

Shear stress check: average shear stress \(\tau_{va} = V / (d \times t_w) \leq 100\) MPa; maximum shear stress \(\tau_{vm} = 1.5 \times \tau_{va} \leq 115\) MPa (for thin webs). Deflection check: \(\delta_{max} \le \dfrac{L}{325}\) (IS 800) for beams carrying plaster or sensitive finishes; \(\delta_{max} \le \dfrac{L}{360}\) for roofs (without plaster).

4.3 Lateral Torsional Buckling (LTB)

When a beam's compression flange is not adequately restrained laterally, it may buckle sideways (lateral torsional buckling) at a load below the full plastic/elastic capacity. The effective length for LTB governs the permissible bending stress.

Permissible bending compressive stress \(\sigma_{bc}\) (IS 800:1984) depends on: (1) \(\dfrac{D}{T}\) (depth-to-flange thickness ratio); (2) \(\dfrac{l}{r_y}\) (slenderness for LTB; \(l\) = effective length between lateral supports). From IS 800 Table 6.1B: \(\sigma_{bc}\) decreases as \(\dfrac{l}{r_y}\) or \(\dfrac{D}{T}\) increases. Full plastic moment is available only when \(\dfrac{l}{r_y} \le\) the permissible limit; otherwise a reduced \(\sigma_{bc}\) from the code tables applies.

Effective length for LTB (IS 800 Table 6.3): compression flange fully restrained (both ends) → \(l = 0.7 \times L\) (L = span); compression flange partially restrained → \(l = 0.85 \times L\); compression flange unrestrained → \(l = 1.0 \times L\).

4.4 Web Buckling and Web Crippling

Web Buckling (under concentrated load or reaction): dispersal angle through flange = 45° (IS 800); bearing length at web \(b_1\) = load dispersal width; load-bearing capacity \(P_{wb} = (b_1 + n_1) \times t_w \times \sigma_{ac}(\lambda)\), where \(\lambda = 2.5 \times d / t_w\) (slenderness of web in bearing).

Web Crippling (local yielding at support/point load): \(P_{wc} = (b + 2.5 \times T_f) \times t_w \times f_{yw}\), where \(b\) = bearing length, \(T_f\) = flange thickness, \(t_w\) = web thickness, \(f_{yw}\) = yield stress of web.

4.5 Compound and Cover-Plated Beams

When the required Z exceeds available rolled sections, add cover plates to the flanges: \(Z_{compound} = Z_{rolled} + A_{plate} \times \bar y_{plate}\). Horizontal shear per unit length at the plate-flange interface (shear flow): \(q = V \times Q / I\), where \(Q = A_{plate} \times \bar y_{plate}\) (first moment of the cover plate about the NA). Weld/rivet spacing to resist shear flow: \(spacing = R / q\), where R = weld strength per unit length or rivet value.

4.6 Beam-Column Design (Combined Axial + Bending)

Interaction equation (IS 800:1984 WSM): \(\dfrac{P}{P_a} + \dfrac{M_x}{M_{ax}} + \dfrac{M_y}{M_{ay}} \le 1.0\), where \(P\) = applied axial load, \(P_a\) = axial capacity \(= A_g \times \sigma_{ac}\); \(M_x\) = bending moment about the major axis, \(M_{ax} = Z_x \times \sigma_{bc}\); \(M_y\) = bending moment about the minor axis, \(M_{ay} = Z_y \times \sigma_{bc}\).

Additional check for yielding at the extreme fibre (interaction): \(\dfrac{P}{P_e} + C_m \times M_x / [(1 - \dfrac{P}{P_e}) \times M_{ax}] \le 1.0\), where \(P_e = \pi^{2}EI_x / (L_{eff})^{2}\) (Euler load for the major axis).

📝 GATE Tip: Permissible bending stress for laterally supported beams = 0.66 f_y = 165 MPa (Fe250). For unsupported beams, σ_bc is reduced — read from IS 800 Table 6.1B based on D/T and l/r_y. The shear flow formula q = VQ/I is used for connecting cover plates and also for composite beams.
📏
Ch 4 · Formulas
Permissible Bending Stress & Section Modulus
$$ \sigma_{bc}=0.66f_y=165\ \text{MPa},\quad Z_{req}=\frac{M_{max}}{\sigma_{bc}} $$
For laterally-supported compact section, Fe250
MPa, mm³
Shear & Deflection Limits
$$ \tau_{va}=\frac{V}{d\,t_w}\le100\ \text{MPa},\quad \delta\le\frac{L}{325} $$
L/360 for roofs without plaster
—
Effective Length for LTB
$$ l=0.7L\ \text{to}\ 1.0L $$
0.7L fully restrained; 0.85L partial; 1.0L unrestrained
—
Web Crippling
$$ P_{wc}=(b+2.5T_f)\,t_w\,f_{yw} $$
Local yielding under concentrated load/reaction
N
Beam-Column Interaction
$$ \frac{P}{P_a}+\frac{M_x}{M_{ax}}+\frac{M_y}{M_{ay}}\le1.0 $$
\(P_a=A_g\sigma_{ac}\); \(M_{ax}=Z_x\sigma_{bc}\)
—
📏
Ch 4 · Solved Example
GATE CE / ESE level
Example 1 Easy
A simply supported, laterally-supported steel beam of Fe250 grade carries a maximum bending moment of 82.5 kN·m. Find the required section modulus.
Approach hint
Since the beam is laterally supported (compact section, no LTB reduction), use \(\sigma_{bc}=165\) MPa directly in \(Z_{req}=\dfrac{M_{max}}{\sigma_{bc}}\).
  1. \(M_{max} = 82.5\) kN·m \(= 82.5\times10^6\) N·mm
  2. \(Z_{req} = \dfrac{M_{max}}{\sigma_{bc}} = 82.5\dfrac{\times10^6}{165} = 500{,}000\) mm³ \(= 500\) cm³
✓ \(Z_{req} = 500\) cm³ — select a rolled section (e.g. ISMB) with \(Z_x \ge 500\) cm³
📏
Ch 4 · Diagrams
Click any diagram to zoom
📏
Ch 4 · Quick Quiz
3 questions · tap an option to reveal answer
Question 1
Lateral torsional buckling of a steel beam is prevented primarily by:
LTB occurs because the unrestrained compression flange buckles sideways; providing lateral (and torsional) restraint at intervals directly prevents this failure mode ✓
Question 2
For a laterally-supported compact steel beam in Fe250, the permissible bending stress is:
\(\sigma_{bc}=0.66f_y=0.66\times250=165\) MPa for a compact, laterally-supported Fe250 section ✓
Question 3
The shear flow formula q = VQ/I is used to design:
The horizontal shear flow q = VQ/I gives the force per unit length that the connectors (welds/rivets) between the cover plate and the rolled section must resist ✓
🏭
Ch 5 · Plate Girders & Industrial Roofs
○ Not Started

5.1 Plate Girder — Introduction

A plate girder is a built-up flexural member fabricated by welding (or riveting) flat plates to form an I-shape — a web plate and two flange plates. Used when spans exceed the economical rolled-section range (15–100 m) or loads are too heavy for standard sections.

5.2 Design of Plate Girder Web

Web depth d (economic depth — minimises total steel weight): \(d_{eco} = \dfrac{M}{\sigma_{bc}} \times \left(\dfrac{12}{t_w}\right)^{1/3}\) [approximate; varies with flange assumptions].

Web thickness \(t_w\) limits (IS 800): without horizontal stiffeners \(\dfrac{d}{t_w} \le 200\); with horizontal stiffeners \(\dfrac{d}{t_w} \le 250\). Average web shear stress: \(\tau = \dfrac{V}{d \times t_w} \le \tau_{va} = 100\) MPa. For slender webs (\(\dfrac{d}{t_w} > 85\)): tension field action may be considered.

5.3 Flange Design

Flange area required (approximate — web carries moment contribution ≈ 1/6): \(A_f = M / (\sigma_{bc} \times d) - t_w \times d / 6\). OR more precisely: \(I_{required} = M \times y_{max} / \sigma_{bc}\) (full section MI from extreme fibre stress); \(I_{web} = t_w \times d^{3} / 12\); \(I_{flange,needed} = I_{required} - I_{web}\); \(A_f = I_{flange,needed} / (\dfrac{d}{2})^2\) [approx, flange as thin plate at \(\dfrac{d}{2}\) from NA].

Flange width \(b_f\) limits (outstand criterion to prevent local buckling): \(\dfrac{b}{t} \le \dfrac{256}{\sqrt{f_y}}\) (plate outstand from web face; IS 800 WSM). For \(f_y = 250\) MPa: \(\dfrac{b}{t} \le \dfrac{256}{\sqrt{250}} \approx 16.2\).

5.4 Stiffeners in Plate Girders

Stiffener TypeLocationPurposeDesign For
Bearing stiffener (load-bearing)Over supports and under concentrated loadsPrevent web crippling; transfer reactionAxial compression from reaction/point load
Intermediate transverseAt regular intervals along webPrevent shear buckling of webShear buckling (d/t_w, panel aspect ratio)
Longitudinal stiffenerHorizontal, in compression zone of webPrevent bending buckling of webd/t_w > 200 (prevents need for thicker web)
Flange stiffenerAttached to flange (plates)Reduce flange outstandLocal flange buckling control

Transverse stiffener spacing (IS 800 — to prevent shear buckling): \(\dfrac{c}{d} \le 1.5\) (aspect ratio of each panel). Without diagonal tension, shear stress \(\tau \le \tau_{cr}\) (critical shear buckling stress): \(\tau_{cr} = k_s \times \pi^{2}E / (12(1-\nu^{2})) \times (\dfrac{t_w}{d})^{2}\), where \(k_s\) = shear buckling coefficient \(\approx 5.35 + \dfrac{4.0}{(c/d)^2}\) for intermediate panels.

5.5 Flange-to-Web Connection

Horizontal shear flow at the flange-web interface: \(q_{fw} = V \times A_f \times \bar{y}_f / I\). Weld size required: \(s_w = q_{fw} / (2 \times 0.707 \times \tau_w) = q_{fw} / (2 \times 0.707 \times 110)\) [two fillet welds, one each side of web; E41 electrode, \(\tau_w = 110\) MPa].

5.6 Industrial Roof Structures

Industrial buildings use steel roof trusses (spanning 10–30 m) or lattice girders for large clear spans. The roof system must carry dead loads (sheeting, purlins), live loads, wind, and snow/maintenance loads.

Roof Truss Loads

Load TypeSourceTypical ValueApplication
Dead load (roofing)Sheeting, purlins, truss self-weight0.4–1.0 kN/m² (horizontal)Permanent; full span
Live load (roof)Maintenance, minor snow0.75 kN/m² (IS 875 Part 2)Reduced for slopes > 10°
Wind loadExternal pressure / suctionPer IS 875 Part 3; Cp valuesMay cause uplift on windward slope
Snow loadAccumulation (applicable zones)0.25–2.5 kN/m² (IS 875 Part 4)Hilly/cold regions; Jammu, NE India
Crane girder loadsOverhead traveling cranesWheel loads per crane capacityIndustrial buildings with cranes

Purlins

Purlins are secondary beams spanning between roof trusses, supporting the roof covering (sheeting). They are typically angle sections, channel sections, or Z-sections.

Purlin load: \(w = (\text{dead} + \text{live} + \text{wind}) \times \text{spacing of purlins} \times \cos\theta\) (the component perpendicular to the roof surface governs bending). Biaxial bending check: \(\dfrac{M_u}{Z_{xx}} + \dfrac{M_v}{Z_{yy}} \le \sigma_{bc}\), where \(M_u = \dfrac{wL^2}{8} \times \cos\alpha\) (in plane of roof) and \(M_v = \dfrac{wL^2}{8} \times \sin\alpha\) (perpendicular to roof); \(\alpha\) = roof slope angle. Sag rods reduce \(M_v\) by providing intermediate support in the minor axis.

5.7 Crane Girder Design

Crane girder must be designed for: (1) Vertical loads — wheel loads with impact factor (IS 875 Part 2): \(P_{design} = P_{wheel} \times\) (1 + impact factor); impact factor = 10–25% of the lifted load (depending on class of crane). (2) Horizontal lateral loads (surge): 10% of wheel load (side thrust from crane travel). (3) Fatigue considerations for frequently operated cranes.

Deflection limit for crane girder: \(\delta \le \dfrac{L}{750}\) (running clearance requirement).

📝 GATE Tip: Economic depth of plate girder \(d_{eco} \approx 1.1 \times (\dfrac{M}{\sigma_{bc}})^{0.5} \times (\dfrac{1}{t_w})^{1/3}\). Web d/t_w ≤ 200 (without horizontal stiffener). Flange b/t ≤ 16 (for Fe250). North light truss gives diffuse daylight — preferred for textile industries. These facts are classic MCQ fodder.
🏭
Ch 5 · Formulas
Web Thickness Limits
$$ \frac{d}{t_w}\le200\ (\text{no stiff.}),\quad \le250\ (\text{with stiff.}) $$
Average shear \(\tau=\dfrac{V}{d\,t_w}\le100\) MPa
—
Flange Outstand Limit
$$ \frac{b}{t}\le\frac{256}{\sqrt{f_y}}\approx16.2\ (\text{Fe250}) $$
Prevents local flange buckling
—
Critical Shear Buckling Stress
$$ \tau_{cr}=k_s\frac{\pi^2E}{12(1-\nu^2)}\left(\frac{t_w}{d}\right)^2 $$
\(k_s\approx5.35+\dfrac{4.0}{(c/d)^2}\); stiffener spacing \(\dfrac{c}{d}\le1.5\)
MPa
Flange-Web Weld Size
$$ s_w=\frac{q_{fw}}{2\times0.707\times110} $$
\(q_{fw}=VA_f\bar y_f/I\); two fillet welds
mm
Crane Girder Deflection Limit
$$ \delta \le \frac{L}{750} $$
Running clearance requirement (stricter than L/325)
—
🏭
Ch 5 · Solved Example
GATE CE / ESE level
Example 1 Medium
A plate girder web has depth d = 1800 mm and no horizontal stiffeners are provided. Find the minimum permissible web thickness \(t_w\), and the maximum shear force it can carry at \(\tau_{va}=100\) MPa.
Approach hint
Without horizontal stiffeners, the limit is \(\dfrac{d}{t_w} \le 200\); use this to find the minimum \(t_w\), then apply \(\tau=\dfrac{V}{d\,t_w}\) to find the shear capacity.
  1. \(t_{w,min} = \dfrac{d}{200} = \dfrac{1800}{200} = 9\) mm
  2. \(V_{max} = \tau_{va} \times d \times t_w = 100\times1800\times9 = 1{,}620{,}000\) N = 1620 kN
✓ \(t_{w,min} = 9\) mm; maximum shear capacity ≈ 1620 kN
🏭
Ch 5 · Diagrams
Click any diagram to zoom
🏭
Ch 5 · Quick Quiz
3 questions · tap an option to reveal answer
Question 1
The maximum permissible d/t_w ratio for a plate girder web WITHOUT horizontal stiffeners is:
IS 800 limits d/t_w ≤ 200 without horizontal stiffeners; adding a horizontal stiffener raises the limit to 250 ✓
Question 2
A North Light (sawtooth) roof truss is preferred in textile mills primarily because it:
The steep glazed face of a North Light truss faces away from direct sun (north in the northern hemisphere), giving even, glare-free natural light — ideal for textile/precision work ✓
Question 3
The deflection limit for a crane girder (running clearance requirement) is stricter than for an ordinary floor beam because it must satisfy:
Crane girders must limit deflection to L/750 — much stricter than L/325 (plaster) or L/360 (roofs) — to maintain running clearance for the crane wheels ✓
🔓
Ch 6 · Plastic Analysis (Ultimate Load Design)
○ Not Started

6.1 Introduction — Plastic vs Elastic Design

Elastic design assumes the structure fails when the most stressed fibre yields. Plastic (ultimate load) design recognises that after first yield, load redistribution continues until a plastic collapse mechanism forms. This allows more economical design because unused capacity of redundant structures is utilised.

⭐ Key advantage of plastic design: Ductile steel structures can sustain loads well beyond first yield. A fixed-end beam can carry 1.5× the load that caused first yield, before collapse — this extra 50% is the benefit of plastic design over elastic design for a fixed-fixed beam.

6.2 Plastic Bending — Key Concepts

Elastic limit: \(\sigma_y\) at extreme fibre; NA at geometric centroid. At plastic limit (full section yielded): Plastic moment \(M_p = f_y \times Z_p\); elastic moment \(M_y = f_y \times Z_e\) (\(Z_e\) = elastic section modulus = I/y).

Plastic section modulus \(Z_p\) = first moment of area about the equal-area axis: \(Z_p = \dfrac{A}{2} \times (\bar y_{top} + \bar y_{bot})\), where \(\bar y\) = centroid of the top/bottom half from the plastic neutral axis. Shape factor (form factor): \(f = M_p / M_y = Z_p / Z_e\). Rectangle: f = 1.5. Solid circle: f = 1.698 ≈ 1.7. Diamond: f = 2.0. I-section: f ≈ 1.12–1.15 (close to 1.0 for deep, slender I). Hollow circle: f ≈ 1.27 (thin-walled).

6.3 Plastic Hinge

A plastic hinge forms at a cross-section where the full plastic moment \(M_p\) has been reached. Unlike a real hinge (no moment), a plastic hinge can rotate freely while maintaining \(M_p\). After formation, it acts as a pin in further load analysis — i.e., it transmits constant \(M_p\) while allowing rotation to continue.

Number of plastic hinges needed for collapse (mechanism): \(n_h = DSI + 1\) (for a one-degree mechanism), where DSI = degree of static indeterminacy. Simply supported beam (DSI=0): 1 hinge for collapse. Propped cantilever (DSI=1): 2 hinges for collapse. Fixed-fixed beam under one load: needs \(DSI_{bending} + 1 = 2+1 = 3\) hinges ✓ (3 bending redundants → need 3 hinges for a single-load mechanism).

6.4 Collapse Load — Standard Cases

Beam TypeLoadingCollapse LoadElastic LoadLoad Factor f
SS beamCentral P4M_p/L4M_y/Lf = M_p/M_y = shape factor
SS beamUDL w8M_p/L²8M_y/L²f = shape factor
Propped cantileverCentral P6M_p/L5.03M_y/L (first yield at fixed end)~1.19×M_p/M_y
Propped cantileverUDL w11.66M_p/L²8M_y/L² (fixed end)~1.46×M_p/M_y
Fixed-fixed beamCentral P8M_p/L6M_y/L\(\frac{4}{3}\times(\dfrac{M_p}{M_y})\)
Fixed-fixed beamUDL w16M_p/L²12M_y/L² (ends)\(\frac{4}{3}\times(\dfrac{M_p}{M_y})\)

6.5 Methods of Plastic Analysis

Static (Lower Bound) Method

Rules (Lower Bound Theorem): (1) Equilibrium — the moment diagram must be in equilibrium with the applied loads; (2) Yield — \(|M| \le M_p\) at all sections; (3) The collapse load so determined is a lower bound (safe side). Procedure: assume plastic hinge locations → equilibrium gives \(P_u\) → check \(|M| \le M_p\) everywhere.

Kinematic (Upper Bound / Mechanism) Method

Rules (Upper Bound Theorem): (1) Mechanism — assume a collapse mechanism (sufficient hinges to turn the structure into a mechanism); (2) Energy balance — external work done = internal energy dissipated at plastic hinges; (3) The collapse load so found is ≥ the true collapse load (unsafe side unless the exact mechanism is found). Energy method: \(\Sigma(P_i \times \delta_i) = \Sigma(M_p \times \theta_j)\), where \(P_i\) = applied loads, \(\delta_i\) = displacements, \(\theta_j\) = hinge rotation.

6.6 Worked Example — Propped Cantilever with UDL

Fixed-pinned beam, UDL w over full span L: a plastic hinge forms first at the fixed end (max hogging BM). After the first hinge, the system becomes an SS beam (the propped cantilever rotates at the fixed end); a second hinge forms at the location of max +ve BM.

Energy method for the mechanism (2 hinges): let hinge 1 be at fixed end A (rotation \(\theta_1\)), hinge 2 at \(x = x_0\) from A (rotation \(\theta_1 + \theta_2\)). At B (roller), \(\delta_B = 0\), so compatibility gives \(x_0\theta_1 = (L-x_0)\theta_2 \to \theta_2 = x_0\dfrac{\theta_1}{L-x_0}\). Internal work = \(M_p \times \theta_1 + M_p \times (\theta_1 + \theta_2) = M_p \times (2\theta_1 + \theta_2)\). Minimising \(w_u\) with respect to \(x_0\): \(\dfrac{dw_u}{dx_{0}} = 0 \to x_{0} = 0.414L\), giving the exact result \(w_u = 11.66\,M_p / L^{2}\).

6.7 Load Factor in Plastic Design

Load factor = Collapse load / Working load = \(P_u / P_{working}\). Minimum load factor (IS 800 plastic design): \(\lambda_{min} = 1.85\) (dead + live). Design approach: \(P_u = \lambda_{min} \times P_{working}\); determine \(M_p\) required for each member from collapse analysis; select a section with \(Z_p \ge \dfrac{M_p}{f_y}\).

Requirements for plastic design (IS 800 Cl. 9): (1) Material — \(f_y \le 450\) MPa, elongation ≥ 15%, \(\dfrac{f_u}{f_y} \ge 1.2\); (2) Section — compact section (\(\dfrac{b}{t} \leq 8.9\sqrt{\dfrac{250}{f_y}}\) for flanges); (3) No LTB — adequate lateral support at plastic hinge locations; (4) Deflection — check under service loads (elastic analysis).

6.8 Shape Factors and Z_p Values

SectionZ_eZ_pShape Factor f
Rectangle (b × d)bd²/6bd²/41.5
Solid circle (D)πD³/32D³/616/(3π) ≈ 1.698
Thin circular tube (D, t)πD²t/4D²t4/π ≈ 1.27
Diamond (diag. 2a × 2b)ab²/3ab²/22.0
I-section (approx)I/(d/2)b_f·t_f·(d−t_f)+t_w·d_w²/41.12–1.15
Triangle (b, h)bh²/24bh²/122.0 (about apex); varies
⚠️ Plastic design is only applicable to structures made of ductile steel with adequate rotation capacity. It must NOT be used for structures subject to repeated loading (fatigue), structures where deformations are critical, or members where LTB or local buckling would limit rotation at hinges.
🔓
Ch 6 · Formulas
Plastic & Elastic Moment
$$ M_p=f_yZ_p,\quad M_y=f_yZ_e $$
Shape factor \(f=\dfrac{M_p}{M_y}=\dfrac{Z_p}{Z_e}\)
N·mm
Hinges Needed for Collapse Mechanism
$$ n_h = DSI + 1 $$
Per independent (one-degree) mechanism
—
Energy (Virtual Work) Method
$$ \Sigma(P_i\delta_i) = \Sigma(M_p\theta_j) $$
External work = internal energy dissipated at hinges
—
Collapse Loads — Standard Cases
$$ P_u=\frac{4M_p}{L}\ (\text{SS}),\quad \frac{8M_p}{L}\ (\text{fixed-fixed}) $$
Propped cantilever central P: \(\dfrac{6M_p}{L}\)
N
Load Factor
$$ \lambda = \frac{P_u}{P_{working}} \ge 1.85 $$
Minimum per IS 800 plastic design
—
🔓
Ch 6 · Solved Example
GATE CE / ESE level
Example 1 Medium
A fixed-fixed steel beam of span 6 m has plastic moment capacity \(M_p = 120\) kN·m. Find the collapse load \(P_u\) for a central point load, and compare it to the elastic first-yield load \(P_y = \dfrac{6M_p}{L}\).
Approach hint
Use the standard fixed-fixed-beam-with-central-load collapse formula \(P_u=\dfrac{8M_p}{L}\), then compute \(P_y\) using the given elastic formula and find the ratio.
  1. \(P_u = \dfrac{8M_p}{L} = 8\dfrac{\times120}{6} = 160\) kN
  2. \(P_y = \dfrac{6M_p}{L} = 6\dfrac{\times120}{6} = 120\) kN
  3. Ratio \(\dfrac{P_u}{P_y} = \dfrac{160}{120} = \dfrac{4}{3} \approx 1.33\)
✓ \(P_u = 160\) kN — the beam carries 33% more load than at first yield before full collapse, consistent with the standard 4/3 ratio for fixed-fixed beams
🔓
Ch 6 · Diagrams
Click any diagram to zoom
🔓
Ch 6 · Quick Quiz
3 questions · tap an option to reveal answer
Question 1
The shape factor of a rectangular section is:
For a rectangle, \(Z_p=\dfrac{bd^2}{4}\) and \(Z_e=\dfrac{bd^2}{6}\), so \(f=\dfrac{Z_p}{Z_e}=1.5\) — a standard, frequently tested value ✓
Question 2
A propped cantilever (DSI = 1) requires how many plastic hinges to form a collapse mechanism?
\(n_h=DSI+1=1+1=2\) hinges — one at the fixed end, one at the point of maximum span moment ✓
Question 3
The Kinematic (Upper Bound) method of plastic analysis gives a collapse load that is:
The Upper Bound theorem states that any assumed mechanism gives a collapse load ≥ the true value — only the correct mechanism gives equality, so an incorrectly guessed mechanism is unsafe (overestimates capacity) ✓
📚
Ch 7 · Quick Revision — All Formulae, Tables & Mnemonics
Cross-chapter recall sheet, mnemonics, exam-angle comparison, mini quiz and practice questions
○ Not Started

All Key Formulae — Design of Steel Structures

IS 800:1984 Permissible Stresses (Fe250)
$$ \sigma_{at}=0.6f_y=150,\quad \sigma_{bc}=0.66f_y=165 $$
MPa. Shear in rivet τ_va=100; max τ_vm=115; bearing σ_pb=300; weld τ_w=110
MPa
Rivet and Bolt Design
$$ R=\min(P_s,P_b),\quad P_s=\frac{\pi}{4}d^2\tau_{va} $$
Gross dia = nominal + 1.5mm; pitch min 3d, max 16t/200 (tension) 12t/200 (compression); edge min 1.5d sheared, 1.25d rolled
—
Fillet Weld
$$ t_e=0.707s,\quad q=0.707s\times110 $$
Min length=max(4s,40mm); effective length=total−2s; min weld 3mm(≤10mm)→8mm(>32mm)
N/mm
Tension Members
$$ A_n=A_g-n\,d_h\,t,\quad A_{eff}=A_1+kA_2 $$
Staggered: \(w_n=w-\Sigma d_h+\Sigma(\dfrac{s^2}{4g})\); \(k=\dfrac{3A_1}{3A_1+A_2}\); \(T_{allow}=A_{eff}\times150\)
mm²
Compression Members
$$ \lambda=\frac{L_{eff}}{r},\quad \lambda_{main}\le180,\ \lambda_{brace}\le200 $$
Eff. lengths: PP=L, FF=0.5L, FP=0.7L, F-Free=2L. Lacing shear=2.5%P; angle 40–70°
—
Beams
$$ Z_{req}=\frac{M_{max}}{\sigma_{bc}},\quad \tau_{va}=\frac{V}{d\,t_w}\le100 $$
Deflection: δ≤L/325 (plaster), L/360 (roofs). LTB eff. length 0.7L–1.0L. Shear flow q=VQ/I
—
Plate Girder
$$ \frac{d}{t_w}\le200\ (\text{or }250),\quad \frac{b}{t}\le\frac{256}{\sqrt{f_y}} $$
Stiffener spacing c/d≤1.5. Flange-web weld: \(q_{fw}=VA_f\bar y_f/I\). Crane girder δ≤L/750
—
Plastic Analysis
$$ M_p=f_yZ_p,\quad n_h=DSI+1 $$
Shape factor: Rect=1.5, Circle=1.698, I≈1.12, Diamond=2.0. Collapse: SS-P→4M_p/L, Fixed-P→8M_p/L. Load factor min=1.85
—

Key Mnemonics

💡 Rivet failure modes: "SCTSE"
Shearing of rivet | Crushing (bearing) | Tearing of plate | Shear-out | Edge split

Rivet pitch limits: "3d to 16t or 200 (tension) / 12t or 200 (compression)"

Slenderness ratio limits: "180 – 200"
Main members ≤ 180; Bracing/secondary ≤ 200
"Main members May (180); Bracing can Be a bit more (200)"

Effective lengths: "2 – 1 – 0.7 – 0.5"
Fixed-Free: 2L | Pin-Pin: 1L | Fixed-Pin: 0.7L | Fixed-Fixed: 0.5L
"Worst to best: 2, 1, 0.7, 0.5"

Shape factors (descending order):
Diamond 2.0 → Triangle ≈ 2.0 → Circle 1.70 → Hollow Circle 1.27 → Rectangle 1.50 → I-section 1.12
"Diamonds are Round; Rectangles are Half-way; I-sections barely Improve"

Fillet weld throat: "0.707 × size"
"707 — like Boeing 707, always at 45°"

Plate girder web limit: "200/250"
Without H-stiffener: d/t ≤ 200 | With H-stiffener: d/t ≤ 250
"Add a Horizontal stiffener to get 250 — just 50 more"

Plastic collapse: "Hinges = DSI + 1"
"Need one more hinge than redundants to make a mechanism (one more move to checkmate)"

Exam-Angle Comparison Table

TopicGATE FocusESE FocusSSC JE Focus
FastenersRivet value, efficiency, fillet weld strength per unit lengthHSFG bolt design; eccentric bolt/weld group; combined shear + tensionPitch/edge distance rules; failure modes
Tension MembersNet area; shear lag k formula; A_effLug angle; staggered holes detailed calc; block shearGross vs net area concept
Compression Membersσ_ac from table; lacing design; buckling loadBatten plate design; stability of laced column; load eccentricityMax λ = 180; effective length; types
BeamsZ_req, σ_bc, LTB concept; shear checkBeam-column interaction; compound beam design; crane girderZ_req formula; deflection limits
Plate GirderWeb d/t limits; flange b/t; stiffener spacingFull plate girder design from scratch; tension field actionComponent names; web limit
Industrial RoofsRoof truss types; purlin design in biaxial bendingComplete purlin, rafter, truss design; wind load Cp valuesTruss type identification; load types
Plastic AnalysisShape factor; collapse load for standard cases; number of hingesVirtual work method; multi-span frame collapse; load factorShape factor values; plastic hinge concept

Mini Quiz

Q1
A rivet of gross diameter 22 mm connects plates 10 mm thick in double shear, with \(\tau_{va}=100\) MPa. Find the shear strength.
Single shear \(P_s=(\dfrac{\pi}{4})\times22^2\times100=38{,}013\) N ≈ 38.0 kN; double shear = 2×38.0 = 76.0 kN ✓
Q2
Which of the following governs the design of a single-angle tension member connected through one leg only?
Since only one leg is connected, shear lag reduces the effective area: \(A_{eff}=A_1+kA_2\), \(k=\dfrac{3A_1}{3A_1+A_2}\) ✓
Q3
A compression member has slenderness ratio λ = 180. This is the maximum permitted value for:
λ_max = 180 applies to main compression members; secondary/bracing members are allowed up to λ_max = 200 ✓
Q4
A simply supported beam has plastic moment capacity \(M_p = 90\) kN·m and span 4.5 m. Find the collapse load for a central point load.
\(P_u = \dfrac{4M_p}{L} = 4\dfrac{\times90}{4.5} = 80\) kN ✓
Q5
The effective throat thickness of a fillet weld with leg size 8 mm is approximately:
\(t_e=0.707\times8=5.66\) mm ✓

Practice Questions

EasyList the five modes of failure of a riveted joint and state which two determine the "rivet value" R.
The five modes are: shearing of the rivet, crushing (bearing) of the rivet, tearing of the plate, shear-out of the plate, and splitting of the plate at the edge. The rivet value \(R=\min(P_s,P_b)\) is governed by shearing and bearing strength only — tearing and shear-out are checked separately against the plate's net-section and edge-distance capacity.
EasyWhy does an HSFG bolted connection have superior fatigue performance compared to an ordinary black-bolted connection?
HSFG bolts are pre-tensioned to a high proof load, clamping the plates together so friction — not bolt shear — transfers the working load, meaning there is no slip and no cyclic bearing stress concentration at the bolt hole. Black bolts can slip into bearing under load reversal, causing repeated stress concentration at the hole edge that initiates fatigue cracks — a mechanism HSFG connections avoid as long as the friction capacity is not exceeded.
MediumA tension member carries a factored load requiring a net area of 1600 mm². If shear lag reduces a candidate single-angle section's effective area to 82% of its net area, what net area must the selected angle actually provide?
Required \(A_{eff}=1600\) mm² and \(A_{eff}=0.82\times A_1\) (approximately, treating the reduction factor as an overall efficiency), so \(A_1 = \dfrac{1600}{0.82} \approx 1951\) mm² net area must be provided by the connected leg region before the shear-lag reduction is applied.
MediumExplain why plate girders use transverse stiffeners at intervals rather than simply specifying a thicker web throughout.
A thick web adequate to prevent shear buckling everywhere would add substantial self-weight and cost across the full girder length, even though shear demand is often concentrated near supports. Transverse stiffeners instead subdivide a thin, lighter web into smaller panels (aspect ratio c/d ≤ 1.5), raising the critical shear buckling stress τ_cr in each panel without adding steel along the full span — a far more material-efficient solution, especially when tension field action is exploited post-buckling.
HardA fixed-fixed beam under a central point load has a load factor of 4/3 (M_p/M_y ratio aside), while an SS beam under the same loading has a load factor equal to its shape factor alone. Explain the source of this extra 4/3 factor for the fixed-fixed case.
For an SS beam (DSI = 0), only one plastic hinge (at midspan) is needed to form a mechanism, so the ratio of collapse load to first-yield load depends purely on the cross-section's shape factor — the redistribution benefit comes entirely from the plastic reserve within one section. For a fixed-fixed beam under a central load, three hinges are needed (2 ends + midspan). First yield occurs at the ends, where the elastic moment is largest (\(P_y=\dfrac{6M_p}{L}\)), but the beam does not collapse there — it keeps carrying more load as the end sections plasticize while the midspan moment keeps rising elastically, until a third hinge finally forms at midspan and turns the beam into a mechanism at \(P_u=\dfrac{8M_p}{L}\). The ratio \(\dfrac{P_u}{P_y}=\dfrac{8}{6}=\dfrac{4}{3}\) captures this additional structural-redundancy benefit — indeterminate structures redistribute moment among multiple critical sections before collapse, on top of whatever cross-sectional reserve (shape factor) each section individually has.
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