UPSC ESE Civil Engineering Civil Engineering Flashcards

24 sample cards from the Civil Engineering deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

1. What are the assumptions made in the analysis of trusses?

   Members are straight and connected by frictionless pins; loads and reactions act only at joints; members carry only axial force (tension or compression); self-weight is neglected or lumped at joints.

2. Differentiate the method of joints and the method of sections for truss analysis.

   Method of joints applies $\sum F_x=0$ and $\sum F_y=0$ at each joint (best for finding all member forces). Method of sections cuts the truss and applies all three equilibrium equations to a portion (best for a few specific members).

3. What is the carry-over factor and distribution factor in the moment distribution method?

   Carry-over factor = ratio of moment induced at the far end to that applied at the near end; equals $\frac{1}{2}$ for a prismatic member with a fixed far end. Distribution factor $DF=\frac{K}{\sum K}$, the fraction of unbalanced moment a member takes, proportional to its stiffness $K$.

4. Give the stiffness $K$ of a prismatic member with the far end fixed and with the far end pinned.

   Far end fixed: $K=\frac{4EI}{L}$. Far end pinned (hinged): $K=\frac{3EI}{L}$ (modified/reduced stiffness).

5. What is a fixed-end moment, and give the FEM for a fixed beam with central point load $W$ and with UDL $w$.

   A fixed-end moment is the reactive moment at the ends of a fixed beam under load. Central point load: $M_{FEM}=\frac{WL}{8}$. UDL: $M_{FEM}=\frac{wL^{2}}{12}$.

6. Write the slope-deflection equation for member AB.

   $M_{AB}=M_{FAB}+\frac{2EI}{L}\left(2\theta_A+\theta_B-3\frac{\Delta}{L}\right)$, where $M_{FAB}$ is the fixed-end moment, $\theta_A,\theta_B$ are end rotations and $\Delta$ is the relative settlement (sway).

7. Compare the slope-deflection method and the moment-distribution method.

   Slope-deflection is a displacement method that solves simultaneous equilibrium equations for unknown rotations/sway exactly. Moment distribution is an iterative displacement method that locks/releases joints successively, avoiding simultaneous equations but converging approximately.

8. In matrix methods, distinguish the stiffness (displacement) method from the flexibility (force) method.

   Stiffness method: unknowns are displacements; solve $[K]\{D\}=\{F\}$ where $[K]$ is the stiffness matrix — suited to computer/kinematically indeterminate problems. Flexibility method: unknowns are redundant forces; solve using the flexibility matrix $[\delta]$ — suited to statically indeterminate structures.

9. What is the element stiffness matrix of an axial bar element of area $A$, length $L$, modulus $E$?

   $[k]=\frac{AE}{L}\begin{pmatrix}1 & -1\\ -1 & 1\end{pmatrix}$, relating the two nodal axial forces to the two nodal axial displacements.

10. Name the types of bolted/welded connections in steel structures.

    Connections are classified as bolted (riveted) connections — lap, butt joints in shear/bearing — and welded connections — fillet welds and butt (groove) welds. By behaviour: shear, bearing, and tension/friction-grip connections.

11. Give the strength of a fillet weld in terms of throat thickness and effective length (limit state).

    Design strength $P=\frac{0.7s\,L_w\,f_u}{\sqrt{3}\,\gamma_{mw}}$, where throat thickness $t=0.7s$ ($s$ = weld size), $L_w$ = effective length, $f_u$ = ultimate stress, $\gamma_{mw}$ = weld partial safety factor ($1.25$ shop, $1.5$ site).

12. What is the difference between a lap joint and a butt joint in connections?

    In a lap joint the two plates overlap and the fasteners are in single shear (eccentric, causes bending). In a butt joint the plates meet end-to-end with cover plate(s); fasteners are typically in double shear and load is concentric.

13. How is the net section area of a tension member with bolt holes computed?

    $A_{net}=A_{gross}-n\,d\,t+\sum\frac{p^{2}}{4g}t$, where $d$ = hole diameter, $t$ = thickness, $n$ = holes in the failure line, and the staggered term $\frac{p^{2}}{4g}$ accounts for pitch $p$ and gauge $g$.

14. List the three design strengths to be checked for a tension member (limit state).

    Yielding of gross section: $T_{dg}=\frac{A_g f_y}{\gamma_{m0}}$; Rupture of net section: $T_{dn}=\frac{0.9 A_{nc} f_u}{\gamma_{m1}}+\beta\frac{A_{go}f_y}{\gamma_{m0}}$; and Block shear failure. Design strength is the least of these.

15. State Euler's buckling load for a column and define effective length.

    $P_{cr}=\frac{\pi^{2}EI}{L_e^{2}}$, where $L_e$ is the effective length = $KL$. Effective length accounts for end conditions: both ends pinned $L_e=L$; both fixed $L_e=0.5L$; one fixed one free $L_e=2L$; one fixed one pinned $L_e=0.7L$.

16. Define slenderness ratio and explain its role in column design.

    Slenderness ratio $\lambda=\frac{L_e}{r}$, where $r=\sqrt{I/A}$ is the radius of gyration. High slenderness means the column fails by buckling (long column); low slenderness means crushing (short column).

17. What is a plate girder and why are intermediate and bearing stiffeners provided?

    A plate girder is a deep built-up I-section (web + flange plates) used for large spans/heavy loads. Intermediate (transverse) stiffeners prevent web shear buckling; bearing (load) stiffeners prevent web crippling under concentrated loads/reactions.

18. Compare working stress method (WSM) and limit state method (LSM) of design.

    WSM keeps stresses within permissible limits using a single factor of safety on material strength, assuming elastic behaviour. LSM uses partial safety factors on both loads and material strengths and checks limit states of collapse (strength) and serviceability (deflection, cracking), giving more rational, economical designs.

19. State the partial safety factors for loads and materials in IS 456 limit state design.

    Load factor for DL+LL = $1.5$. Material partial safety factors: concrete $\gamma_c=1.5$, steel $\gamma_s=1.15$. Hence design strengths use $0.446f_{ck}$ for concrete and $0.87f_y$ for steel.

20. Give the limiting depth of neutral axis ratio $x_{u,max}/d$ for Fe415 and Fe500 steel (IS 456 LSM).

    $\frac{x_{u,max}}{d}=0.48$ for Fe415, $0.46$ for Fe500, and $0.53$ for Fe250 (mild steel). These limit balanced/over-reinforced sections.

21. Write the limiting moment of resistance for a singly reinforced rectangular beam (IS 456).

    $M_{u,lim}=0.36\,f_{ck}\,b\,x_{u,max}\left(d-0.42\,x_{u,max}\right)$; e.g. for Fe415, $M_{u,lim}=0.138\,f_{ck}\,b\,d^{2}$.

22. How are one-way and two-way slabs distinguished, and what governs the distinction?

    A slab is one-way if the ratio of longer span to shorter span $\frac{L_y}{L_x}>2$ (it bends mainly in the short direction). It is two-way if $\frac{L_y}{L_x}\leq2$, bending in both directions and carrying load to all four supports.

23. State the design axial load formula for a short axially loaded RC column (IS 456) and the minimum/maximum steel limits.

    $P_u=0.4\,f_{ck}\,A_c+0.67\,f_y\,A_{sc}$. Longitudinal steel: minimum $0.8\%$ and maximum $6\%$ of gross area; minimum 4 bars (rectangular) or 6 bars (circular).

24. What are the two main types of column footings and when is each used?

    Isolated (independent) footing supports a single column on firm soil with adequate bearing capacity. Combined footing supports two or more columns when they are close together or near a property line, or when soil bearing capacity is low and isolated footings would overlap.

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