GATE CE Structural Engineering Formula Sheet 2027

The structural engineering formulas for GATE CE 2027 fit on a few pages, covering solid mechanics, structural analysis, plastic analysis and the constants of IS 456:2000 and IS 800:2007. This sheet groups them by topic, works each key one with made-up numbers, and names the trap that costs marks.

Key takeaways

• Structural Engineering is the largest technical section, averaging 15.1 marks across the sixteen counted papers.
• Most solid mechanics questions reduce to five relations: $\delta = PL/AE$, $\sigma = M/Z$, $T/J = \tau/r$, $q = VQ/I$ and $P_{cr} = \pi^2 EI/L_e^2$.
• Indeterminacy counting, fixed-end moments and member stiffness are cheap marks that need one formula and a careful count.
• IS 456:2000 and IS 800:2007 values are expected knowledge, not given data, so carry them in memory.
• A material partial safety factor always divides the strength; 0.87 belongs to concrete, not steel.
• Prestressed concrete, free vibration, the flexibility method and plate girders are off the 2027 syllabus, so they are not on this sheet.

The notation used on this sheet

$E$ is the modulus of elasticity and $I$ the second moment of area, so $EI$ is the flexural rigidity. $Z$ is the section modulus, which is $I$ divided by the distance to the extreme fibre.

In concrete, $f_{ck}$ is the characteristic cube strength and $f_y$ the yield strength of steel, both in N/mm². $x_u$ is the depth of the neutral axis from the compression face, and $d$ is the effective depth. In steel, $f_u$ is the ultimate stress and each $\gamma$ is a partial safety factor. One N/mm² equals one MPa.

Engineering mechanics: equilibrium and friction

Draw a free-body diagram and write $\sum F_x = 0$, $\sum F_y = 0$ and $\sum M = 0$. Limiting friction is $F = \mu N$, where $\mu$ is the coefficient of friction and $N$ the normal reaction.

• A block on a slope of angle $\theta$ slides without help only if $\tan\theta > \mu$.
• Three members at an unloaded joint, two of them collinear: the third carries zero force.
• Two non-collinear members at an unloaded joint: both carry zero force.

Solid mechanics: stress, bending, torsion and buckling

Axial, bending, torsion and shear flow

For $P = 100$ kN on a bar 2 m long with $A = 500$ mm² and $E = 200$ GPa, $\delta = (100 \times 10^3 \times 2000)/(500 \times 200 \times 10^3) = 2$ mm. For a beam 250 mm wide and 500 mm deep under 50 kN·m, $Z = 10.42 \times 10^6$ mm³ and $\sigma = 4.8$ MPa. A solid 80 mm shaft carrying 5 kN·m has $\tau_{\max} = 16 \times 5 \times 10^6/(\pi \times 80^3) = 49.74$ MPa.

Principal stresses from Mohr's circle

The circle is centred on the average normal stress, and its radius is the maximum in-plane shear stress:

For $\sigma_x = 80$, $\sigma_y = 20$ and $\tau_{xy} = 40$ MPa, the centre is 50 and the radius is $\sqrt{30^2 + 40^2} = 50$. So $\sigma_1 = 100$ MPa and $\sigma_2 = 0$.

Trap: Inside the root the term is $(\sigma_x - \sigma_y)/2$, not $\sigma_x - \sigma_y$. Check that $\sigma_1 + \sigma_2 = \sigma_x + \sigma_y$.

Combined stress and the kern

Under an eccentric load the extreme fibre stress is $\sigma = P/A \pm Pe/Z$. No tension develops while $e \le Z/A$.

Euler buckling

$P_{cr} = \pi^2 EI/L_e^2$, with the least $I$ governing, because a column buckles about its weak axis. Theoretical effective lengths are $L$ for both ends pinned, $0.5L$ both fixed, $2L$ fixed-free and $0.707L$ fixed-pinned.

For a pinned column 4 m long with $I = 5 \times 10^6$ mm⁴ and $E = 200$ GPa, $P_{cr} = \pi^2 \times 200 \times 10^3 \times 5 \times 10^6/4000^2 = 616.9$ kN.

Structural analysis: indeterminacy, deflections and stiffness

Counting indeterminacy

For a plane frame, the degree of static indeterminacy is

where $m$ is members, $r$ reactions and $j$ joints. Subtract one for each internal hinge joining two members. A two-bay single-storey frame with three fixed bases has $m = 5$, $r = 9$ and $j = 6$, so $D_s = 15 + 9 - 18 = 6$.

Kinematic indeterminacy counts independent joint displacements. A fixed-base single-bay portal, neglecting axial deformation, has two rotations and one sway, so $D_k = 3$.

Remember: A hinge releases only the moment. A hinge joining two members removes one unknown; one joining three members removes two.

Standard deflections and energy methods

By Castigliano's second theorem, $\delta = \partial U/\partial P$ with $U = \int M^2\,dx/(2EI)$. For $P = 20$ kN, $L = 2$ m and $EI = 20{,}000$ kN·m², the tip moves $160/60{,}000$ m $= 2.67$ mm.

Fixed-end moments and member stiffness

A 4 m fixed span under 30 kN/m has a fixed-end moment of $30 \times 16/12 = 40$ kN·m. With $EI = 12{,}000$ kN·m² and $L = 4$ m, the stiffness is 12,000 kN·m/rad with the far end fixed and 9,000 hinged.

Arches, cables and influence lines

A three-hinged parabolic arch under a full-span UDL has horizontal thrust $H = wL^2/(8h)$, where $h$ is the central rise, and zero bending moment everywhere. A cable takes the same $H$, constant along its length, with maximum tension at the supports, $T_{\max} = \sqrt{H^2 + V^2}$.

For a cable with $w = 12$ kN/m, $L = 30$ m and sag 3 m, $H = 450$ kN, $V = 180$ kN and $T_{\max} = 484.66$ kN. For a simply supported beam, the influence line for mid-span moment is a triangle with peak $L/4$.

This sheet is a selection. The book's last-minute sheet carries the formulae and code constants for all eight technical sections, each checked against a worked solution, and it is part of the GATE CE 2027 book.

Plastic analysis: shape factors and collapse loads

The shape factor is $Z_p/Z_e$, the plastic over the elastic section modulus.

Collapse loads come from external work equal to internal work. A simply supported beam with a central point load collapses at $4M_p/L$; a fixed-fixed beam at $8M_p/L$. With $M_p = 200$ kN·m and $L = 8$ m, those are 100 kN and 200 kN. Portal frames are now named too: 2026 Set 2, Q.49 asked for a combined mechanism by virtual work.

RCC to IS 456:2000: limit-state values

Partial safety factors and the stress block

Limiting neutral axis and limiting moment

$x_{u,\max}/d$ is 0.53 for Fe250, 0.48 for Fe415 and 0.46 for Fe500. The limiting moment of resistance is

The short forms are $0.148 f_{ck} b d^2$ for Fe250, $0.138 f_{ck} b d^2$ for Fe415 and $0.133 f_{ck} b d^2$ for Fe500. For $b = 250$ mm, $d = 450$ mm, M25 and Fe415, $M_{u,\lim} = 0.138 \times 25 \times 250 \times 450^2 = 174.7$ kN·m.

Development length and shear

Development length is $L_d = \phi\,\sigma_s/(4\tau_{bd})$, with $\sigma_s = 0.87 f_y$. The design bond stress $\tau_{bd}$ for plain bars in tension is 1.2 MPa for M20, 1.4 for M25 and 1.5 for M30. Raise it 60 per cent for deformed bars, and a further 25 per cent for bars in compression.

For a 20 mm deformed Fe500 bar in M25, $\tau_{bd} = 1.4 \times 1.6 = 2.24$ MPa and $L_d = 20 \times 435/(4 \times 2.24) = 971$ mm, about $48\phi$. For Fe415 in tension the book gives about $47\phi$ in M20, $40\phi$ in M25 and $38\phi$ in M30.

Vertical stirrups carry $V_{us} = 0.87 f_y A_{sv} d/s_v$. For $V_{us} = 120$ kN, $d = 400$ mm and two-legged 8 mm Fe500 stirrups, $A_{sv} = 100.5$ mm² and $s_v = 145.8$ mm. It is below the $0.75d$ and 300 mm caps, so it governs.

Trap: Forgetting the 60 per cent increase for deformed bars overstates $L_d$ by 60 per cent, and using one stirrup leg instead of two doubles the spacing.

In working stress design the modular ratio is $m = 280/(3\sigma_{cbc})$: 13.33 for M20, 10.98 for M25 and 9.33 for M30.

Steel to IS 800:2007: safety factors, tension members and welds

A 120 × 8 mm plate of Fe410 with $f_y = 250$ MPa gives $T_{dg} = 960 \times 250/1.10 = 218.2$ kN. If $A_n = 800$ mm², $T_{dn} = 0.9 \times 800 \times 410/1.25 = 236.2$ kN, so yielding governs.

An 8 mm site fillet weld 120 mm long has $t_t = 5.6$ mm and $f_{wd} = 410/(\sqrt{3} \times 1.50) = 157.8$ MPa. Its strength is $5.6 \times 120 \times 157.8 = 106.0$ kN.

IS 800:2007 effective lengths for compression members are more conservative than theory: $0.65L$ for both ends fixed and $0.80L$ for fixed-pinned.

In one line: IS 456 uses 1.15 and 1.5; IS 800 uses 1.10, 1.25 and 1.50, and never 0.87.

What this sheet leaves out, and why

Prestressed concrete, free vibration, the flexibility method and plate girders were all dropped for 2027. The GATE CE 2027 syllabus changes guide sets out each removal, and the GATE CE subject-wise weightage shows what the section is worth.

MSQs carry no partial credit under the marking scheme, which is common to every GATE paper. Practise these steps on the GATE virtual calculator, and confirm the current exam-day rules at gate2027.iitm.ac.in.

More formula sheets: all of GATE Civil · Environmental and Transportation · Fluid Mechanics and Hydrology · Geotechnical

Quick revision

1. $\delta = PL/AE$, $\sigma = M/Z$ with $Z = bd^2/6$, $\tau_{\max} = 16T/\pi d^3$, $q = VQ/I$.
2. $P_{cr} = \pi^2 EI/L_e^2$ with the least $I$; no tension while $e \le Z/A$.
3. $D_s = 3m + r - 3j$, minus one per hinge joining two members.
4. Fixed-end moment $wL^2/12$; stiffness $4EI/L$ or $3EI/L$; arch and cable thrust $wL^2/(8h)$.
5. Shape factors 1.50, 1.70, 2.00 and about 1.12 to 1.15; collapse loads $4M_p/L$ and $8M_p/L$.
6. IS 456: $0.87 f_y$, $0.446 f_{ck}$, $x_{u,\max}/d$ of 0.53, 0.48, 0.46, and $M_{u,\lim} = 0.138 f_{ck} b d^2$ for Fe415.
7. $L_d = \phi\,\sigma_s/(4\tau_{bd})$, $\tau_{bd}$ 1.2 MPa for M20, plus 60 per cent for deformed bars.
8. IS 800: $\gamma_{m0} = 1.10$, $\gamma_{m1} = 1.25$, welds 1.25 shop and 1.50 site, throat $0.7 \times$ size.