TGPSC HMWSSB Manager Strength of Materials & Structural Analysis previous year questions with answers
Every Strength of Materials & Structural Analysis question asked in the TGPSC HMWSSB Manager (Engineering) papers on this site, grouped by topic. Answers come from TSPSC's final key; each has a short explanation.
Practise Strength of Materials & Structural Analysis onlineStrength of Materials
A rectangular box made with thin uniform plate measures 2000mm x 1000mm x 1000mm. When the box is subjected to certain internal pressure, the dimensions in respective directions have changed by +2.0mm, -1mm and +1mm. The change in the volume of the box is ______.
- 2.0 × 10⁶ mm³
- 1.0 × 10⁶ mm³
- 2.0 × 10³ mm³
- 2.0 × 10⁵ mm³
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Answer: (1) 2.0 × 10⁶ mm³
Volumetric strain = 2/2000 − 1/1000 + 1/1000 = 0.001, and V = 2000 × 1000 × 1000 = 2×10⁹ mm³, so ΔV = 0.001 × 2×10⁹ = 2.0×10⁶ mm³.
A rectangular section of a beam is acted upon by certain amounts of shear force and bending moment. Whereas the shear stress varies ______, the variation of bending stress is ______ along the depth.
- linear with zero value at centroid, linear with zero value at centroid
- parabolic with zero value at centroid, linear with zero value at centroid
- parabolic, with zero value at top & bottom, linear with zero value at middle
- linear with zero value at centroid, parabolic with zero value at top & bottom
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Answer: (3) parabolic, with zero value at top & bottom, linear with zero value at middle
In a rectangular section the shear stress τ = (F/2I)(d²/4 − y²) is parabolic, zero at the top and bottom and maximum (1.5 times average) at the neutral axis, while bending stress σ = My/I is linear and zero at the neutral axis (middle).
A simply supported beam of span L, carries a UVL with intensity varying from w/unit downward at left support to w/unit upward at right support. The reactions at left and right supports are respectively:
- wL/4 upward and wL/4 downward
- wL/4 downward and wL/4 upward
- wL/6 upward and wL/6 downward
- wL/8 upward and wL/8 downward
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Answer: (3) wL/6 upward and wL/6 downward
The load is a downward triangle wL/4 acting at L/6 from A and an upward triangle wL/4 acting at 5L/6 from A, so the net load is zero; moments about A give R_B × L = (wL/4)(L/6) − (wL/4)(5L/6) = −wL²/6, so R_B = wL/6 downward and R_A = wL/6 upward.
A cantilever beam of span L, uniform flexural rigidity EI is subjected to a unit couple at its free end. The deflection at the centre of the beam is:
- L²/2EI
- L²/8EI
- L²/4EI
- L²/16EI
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Answer: (2) L²/8EI
A couple M at the free end gives constant bending moment, so the deflection at distance x from the fixed end is Mx²/(2EI); with M = 1 and x = L/2 it is (L/2)²/(2EI) = L²/8EI (L²/2EI is the tip deflection).
A simply supported beam of span L, carries two couples of magnitude M each acting at both middle third locations of the beam. While one of them is acting clockwise the other is acting counter clockwise. Magnitude of the maximum shear force acting in the beam ______.
- 2M/L
- M/L
- 1.5M/L
- 0
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Answer: (4) 0
The two equal and opposite couples cancel, so both support reactions are zero; with no transverse load and no reactions the shear force is zero everywhere (the bending moment is M between the couples, but there is no shear).
A cantilever beam of uniform flexural rigidity with span L and depth D is subjected to temperatures T_1 on the upper face and T_2 on the lower face. If T_1 < T_2 and α is the coefficient of linear expansion for the material, the deflection at the free end of the beam is ______.
- α (T_2 - T_1) L²/2D (upward)
- α (T_2 - T_1) L²/D (upward)
- α (T_2 - T_1) L/2D (upward)
- α (T_2 - T_1) L²/2D (downward)
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Answer: (1) α (T_2 - T_1) L²/2D (upward)
The temperature difference across the depth gives a uniform curvature κ = α(T_2 − T_1)/D; the hotter lower face expands more, so the cantilever curls upward. Free-end deflection = κL²/2 = α(T_2 − T_1)L²/(2D), upward.
At a point in elastic medium normal stresses in two mutually perpendicular directions are 120 MPa, 40 MPa (both tensile) associated with a tangential stress of 30 MPa. The principal stresses at the location are ______.
- 120 MPa, 40 MPa (both tensile)
- 130 MPa, 30 MPa (both tensile)
- 130 MPa, 30 MPa (both compressive)
- 130 MPa (tensile), 30 MPa (compressive)
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Answer: (2) 130 MPa, 30 MPa (both tensile)
Centre of Mohr's circle = (120 + 40)/2 = 80 MPa and radius = √(40² + 30²) = 50 MPa, so σ1 = 80 + 50 = 130 MPa and σ2 = 80 − 50 = 30 MPa, both tensile.
A weight W falls freely on a body from a height of h. If the instantaneous deformation of the body in the direction of weight is δ, the work done by the force is ______.
- ½ W (h + δ)
- W (h + δ)
- ½ W h + W δ
- W h + ½ W δ
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Answer: (2) W (h + δ)
The weight falls through h and then moves a further δ while the body deforms, so the work done by the constant force W is W(h + δ), which equals the strain energy stored at maximum deformation. The factor ½ applies to a gradually applied load, not to a falling weight.
In a beam, where the variation of shear force is a second degree parabola and the variation of loading is ______, the variation of bending moment is ______.
- constant, cubic parabola
- linear, square parabola
- cubic parabola, square parabola
- linear, cubic parabola
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Answer: (4) linear, cubic parabola
Since dV/dx = −w and dM/dx = V, a shear force varying as a second-degree parabola means the load varies linearly (one degree lower) and the bending moment varies as a cubic parabola (one degree higher).
A force of magnitude 5 N moves through a distance of 4mm in a direction, inclined at 60° to the direction of force. The magnitude of the work done by the force is ______.
- 10√3 N.mm
- 0 N.mm
- 10 N.mm
- 20 N.mm
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Answer: (3) 10 N.mm
Work = F × d × cos θ = 5 × 4 × cos 60° = 5 × 4 × 0.5 = 10 N.mm. Only the component of displacement along the force does work.
The shear centre for an angle section is located ______.
- at the tip of the flange
- at the intersection of flanges
- at the centroid of the angle
- at the rigidity centre of the angle
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Answer: (2) at the intersection of flanges
In an angle section the shear flows in both legs act along the leg centre-lines, which meet at the corner, so their resultant passes through that point. Hence the shear centre lies at the intersection of the two flanges (legs), not at the centroid.
A beam has a circular cross section. If the plane of loading on the beam does not coincide with the centroidal axis of the beam, the member is subjected to ______.
- axial force, shear force and bending moment
- torque, shear force
- torque, shear force and bending moment
- axial force, shear force and torque
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Answer: (3) torque, shear force and bending moment
A transverse load always produces shear force and bending moment; when its plane does not pass through the axis (shear centre) of the section, the eccentricity also produces a twisting moment (torque). There is no axial force because the load is transverse.
The stiffness of a close coiled spring is more when ______.
- wire diameter, rigidity modulus are more and mean radius, number of turns are lesser
- number of turns, rigidity modulus are more and mean radius, wire diameter are lesser
- mean radius, rigidity modulus are more and wire diameter, number of turns are lesser
- rigidity modulus is more and wire diameter, mean radius, number of turns are lesser
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Answer: (1) wire diameter, rigidity modulus are more and mean radius, number of turns are lesser
Stiffness of a close-coiled helical spring k = Gd⁴/(64R³n), so it increases with larger wire diameter d and rigidity modulus G, and with smaller mean coil radius R and fewer turns n.
A pressure vessel in the form of a thin cylinder of 1m diameter and 1mm plate thickness is subjected to an internal fluid pressure of 0.2 MPa. The maximum shear stress in the material is ______.
- 50 MPa
- 0 MPa
- 25 MPa
- 37.5 MPa
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Answer: (3) 25 MPa
Hoop stress = pd/2t = 0.2 × 1000/(2 × 1) = 100 MPa and longitudinal stress = pd/4t = 50 MPa, so the maximum (in-plane) shear stress = (100 − 50)/2 = 25 MPa. The key takes this in-plane value; 50 MPa would be the absolute maximum if the near-zero radial stress were also considered.
Two circular shafts of same length, weight and material are compared for strength. The first one is a solid shaft and the other is a hollow shaft of outer to inner diameter ratio as 2. The ratio of the torque carrying capacity of the hollow shaft to solid shaft considering the shear stress criterion alone is ______.
- 2.5/√3
- 5√3/6
- √3
- √3/2
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Answer: (2) 5√3/6
Equal weight and length means equal areas: D² − (D/2)² = d², so D = 2d/√3. T_hollow/T_solid = [(D⁴ − (D/2)⁴)/D]/d³ = (15/16)(2/√3)³ = 5/(2√3) = 5√3/6 ≈ 1.44. Note that option (1), 2.5/√3, is the same number written differently; the key gives (2).
For a circular cross section subjected to shear force, the ratio of maximum shear stress to average shear stress is ______.
- 1.0
- 2.0
- 1.5
- 4/3
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Answer: (4) 4/3
For a solid circular section the maximum shear stress, at the neutral axis, is (4/3) times the average shear stress V/(πd²/4). The ratio 1.5 belongs to a rectangular section.
A stepped bar A-B-C of total length 2L carries an axial load P at B as shown in Fig.5. Axial rigidity of segment AB is 2AE and that of BC is AE. The displacement at B is ______.
- PL/3AE
- 0
- PL/2AE
- PL/AE
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Answer: (1) PL/3AE
Both ends are fixed, so AB and BC resist the load at B like two springs in parallel with stiffnesses 2AE/L and AE/L. Displacement at B = P/(3AE/L) = PL/3AE; AB then carries 2P/3 (tension) and BC carries P/3 (compression).
A cantilever beam AB is fixed at left end A and free at B. The corresponding conjugate beam will supports ______.
- fixed at A and free at B
- free at A and simple support at B
- free at A and fixed at B
- simple supports at both A and B
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Answer: (3) free at A and fixed at B
In the conjugate beam a fixed end (zero slope and deflection) becomes a free end, and a free end (with slope and deflection) becomes a fixed end. So a cantilever fixed at A and free at B has a conjugate beam free at A and fixed at B.
Structural Analysis
The plane truss shown in Fig.2 carries a point load W and a moment M at the location B. Force carried by member AB is ______.
- 5W/6 (compressive)
- W/2 (compressive)
- 5W/6 + M/L (compressive)
- W/2 - M/2 (compressive)
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Answer: (1) 5W/6 (compressive)
Truss members are pin-connected and carry only axial force, so the moment at the pin B does not enter the member forces; both members slope 3 vertical : 4 horizontal (sin θ = 3/5), so by symmetry 2F(3/5) = W gives F_AB = 5W/6, compressive.
Analysis of trusses is based on the following assumptions. The correct combination of assumptions is ______. (i) All loads act at joints (ii) All joints are smooth friction less pin joints. (iii) Truss is made with material which is homogeneous, elastic and isotropic. (iv) The axis of all members is straight and their self weights are ignored.
- i, ii and iv
- i, ii, iii
- i, iv
- i, iii, iv
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Answer: (1) i, ii and iv
The ideal-truss assumptions are that loads act only at the joints, joints are frictionless pins, and members are straight with self-weight neglected, so every member carries only axial force; the material being homogeneous and isotropic is not one of the basic truss assumptions.
The correct combination of conditions that defines a rigid joint is ______ i) All members meeting at the joint are rigid. ii) All deflections and rotations at the joint are zero. iii) All members meeting at the joint undergo same deflections at that joint. iv) All members meeting at the joint undergo same rotations at that joint.
- i, ii
- i, iii
- ii, iv
- iii, iv
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Answer: (4) iii, iv
A rigid joint keeps the angles between the connected members unchanged, so all member ends at the joint have the same displacement and the same rotation; the joint itself may still deflect and rotate (ii is a fixed support), and the members need not be rigid (i).
The kinematic indeterminacy of the 2-D portal frame shown in Fig.3 including axial deformations is ______.
- 6
- 11
- 8
- 9
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Answer: (2) 11
The two fixed bases have no freedom; the hinged corner at the top of the left column has 2 translations + 2 independent rotations = 4, the internal hinge in the beam also has 4, and the rigid right corner has 3, so D_k = 4 + 4 + 3 = 11.
In the moment distribution method of plane frame analysis, the distribution factor for a member at a joint depends on ______ combination of conditions. i) The slenderness ratio of all the members meeting at the joint. ii) The flexural rigidity of all the members meeting at the joint. iii) The I/L values of each member meeting at the joint. iv) Support conditions at the farther ends of members meeting at the joint.
- ii, iii
- i, iii
- iii, iv
- i, iv
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Answer: (3) iii, iv
The distribution factor of a member is its stiffness divided by the sum of the stiffnesses of all members at the joint, where stiffness is 4EI/L for a far end fixed and 3EI/L for a far end hinged. So it depends on the relative I/L values and on the far-end support conditions; slenderness ratio plays no part.
A cantilever beam of span L and flexural rigidity EI carries a point load W, vertically downwards at its free end. The free end of the beam is resting at the centre of another simply supported beam of span L and flexural rigidity EI. The support reactions for the simply supported beam are ______ and ______.
- W/4, W/4
- W/2, W/2
- 8W/17, 8W/17
- 8W/3, 8W/3
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Answer: (3) 8W/17, 8W/17
Let R be the force between the beams. Equal deflection at the contact gives (W − R)L³/(3EI) = RL³/(48EI), so 16(W − R) = R and R = 16W/17. This acts at mid-span of the simply supported beam, so each reaction = R/2 = 8W/17.
A simply supported beam of span L and flexural rigidity EI carries a UDL of intensity w/unit run all along its span. The beam is supported at its centre by a linear spring of stiffness k. The force carried by the spring is ______.
- 5kwL⁴/(kL³+8EI)
- 5kwL⁴/8(kL³+48EI)
- 5wL⁴/8(kL³+8EI)
- 5wL⁴/8(kL³+48EI)
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Answer: (2) 5kwL⁴/8(kL³+48EI)
The UDL alone deflects mid-span by 5wL⁴/384EI and the spring force R lifts it back by RL³/48EI; the spring compresses by the net deflection, so R = k(5wL⁴/384EI − RL³/48EI). Multiplying by 48EI gives R(48EI + kL³) = 5kwL⁴/8, so R = 5kwL⁴/8(kL³ + 48EI).
The three member plane truss A-B-C-D, shown in Fig.4 supports a vertical load W at B. The magnitude of the force carried by member BD is ______.
- W (compressive)
- 0.5 W (compressive)
- (√3/2) W (compressive)
- 0.5W (tensile)
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Answer: (3) (√3/2) W (compressive)
AB and BC are both inclined at 30° to the horizontal, so A, B and C lie on one straight line, and BD (at 60° to the horizontal) is perpendicular to it. Resolving forces at B perpendicular to line ABC, only W and BD have components, so F_BD = W cos 30° = (√3/2)W, and since it pushes back on joint B it is compressive.
The symmetry of flexibility matrix is due to ______.
- Betty's theorem
- Maxwell's reciprocal theorem
- Eddy's theorem
- Castigliano's theorem
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Answer: (2) Maxwell's reciprocal theorem
Maxwell's reciprocal theorem states that the deflection at i due to a unit load at j equals the deflection at j due to a unit load at i, i.e. f_ij = f_ji, which is exactly the symmetry of the flexibility matrix. Eddy's theorem relates to arches and Castigliano's theorem to strain energy.
A fixed beam of span L and uniform flexural rigidity EI carries a vertical downward load W at its mid span. If a hinge is introduced in the beam at the location of the load, the deflection under the load is ______.
- WL³/12EI
- WL³/24EI
- WL³/16EI
- WL³/48EI
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Answer: (4) WL³/48EI
With a hinge at mid-span the fixed beam becomes two cantilevers of length L/2 meeting at the hinge, each carrying W/2 by symmetry. Deflection under the load = (W/2)(L/2)³/(3EI) = WL³/48EI.
The static indeterminacy for the continuous beam shown in Fig.6 is ______.
- 6
- 3
- 2
- 4
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Answer: (3) 2
The fixed support at the left gives 3 reactions, the roller at the middle 1 and the inclined roller at the right end 1, a total of 5. With 3 equations of static equilibrium, the static indeterminacy = 5 − 3 = 2.
A simply supported girder of span 10m is traversed by a 5m long UDL segment of intensity 10kN/m. The magnitude of absolute maximum moment in the beam is ______.
- 93.75kN.m
- 98.75kN.m
- 90.25kN.m
- 88.75kN.m
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Answer: (1) 93.75kN.m
For a UDL shorter than the span, the absolute maximum moment occurs at mid-span with the load placed centrally: M = (W/4)(L − a/2), where W = 10 × 5 = 50 kN and a = 5 m. M = (50/4)(10 − 2.5) = 12.5 × 7.5 = 93.75 kN.m.
For a simply supported girder, EUDLL is defined as the intensity of the UDL acting on full span whose ______.
- intensity is larger than all other loads acting on the span.
- moment diagram is above the bending moment ordinates caused by all loads on the span.
- bending moment diagram just envelops bending moment caused by all loads on the span.
- intensity is larger than the sum of intensities of all loads on the span.
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Answer: (3) bending moment diagram just envelops bending moment caused by all loads on the span.
The equivalent uniformly distributed live load (EUDLL) is the uniform load over the whole span whose bending moment diagram just envelops the maximum bending moments produced at every section by the actual moving loads. It lets girders and bridge decks be designed with one simple uniform load.
For a member subjected to loads on its span, equivalent joint loads are determined in such a way that ______.
- The member end actions caused by equivalent joint loads are same as those caused by loads on the member.
- The equivalent joint loads cause same set of nodal displacements as caused by the member loads.
- The equivalent joint loads together with member loads ensure the equilibrium of the structure.
- The equivalent joint loads together with member loads ensure the compatibility of displacements for the structure.
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Answer: (1) The member end actions caused by equivalent joint loads are same as those caused by loads on the member.
Equivalent joint loads are obtained by reversing the fixed-end actions that the span loads produce at the member ends, so they stand in at the joints for the end actions of the member loads, which is the reasoning of the keyed option. Standard matrix-analysis texts, however, define them as joint loads that produce the same joint (nodal) displacements as the actual member loads, with final end actions = fixed-end actions + actions due to the equivalent joint loads. (Official final key; some sources differ.)
An orthogrid structure is one which satisfies following combinations of conditions ______. i) A two dimensional framed structure consisting of members oriented in different directions. ii) Loading plane perpendicular to the plane of structure. iii) Member end actions are the axial forces, shear forces and bending moments. iv) Angle between members is always 90°. v) Member end actions are the shear forces, torques and bending moments.
- i, ii, iii, iv, v
- i, ii, iii, iv
- ii, iii, iv, v
- i, ii, iv, v
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Answer: (4) i, ii, iv, v
A grid is a plane framework loaded perpendicular to its own plane, so its member end actions are shear force, twisting moment (torque) and bending moment, with no axial force; in an orthogrid the members meet at 90°. Statement iii (axial force, shear force and bending moment) describes a plane frame, so it is excluded.
In the stiffness matrix method, the boundary conditions are needed to avert ______.
- divergent solution
- singularity
- irrational solution
- zero displacement vector
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Answer: (2) singularity
Without support conditions the assembled structure stiffness matrix permits rigid-body movement and is singular (its determinant is zero), so it cannot be inverted. Applying the boundary conditions removes these rigid-body modes and makes the matrix non-singular.
In the direct element method of structural analysis, the transformation matrices are needed for following combinations of actions ______. i) They facilitate the transformation of element stiffness and element load matrices to global stiffness and global load matrices respectively. ii) They are useful for determining the displacement matrices. iii) They facilitate the conversion of resultant displacement matrices into member oriented displacement matrices. iv) They are useful for determining member end actions.
- i, iii
- ii, iii
- iii, iv
- i, ii
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Answer: (1) i, iii
Transformation matrices rotate each element's stiffness and load matrices from local (member) axes to global axes before assembly, and after the solution they convert the global (resultant) displacements back into member-oriented displacements. The displacements and end actions themselves come from solving the equations and from member stiffness, so i and iii describe the role of the transformation matrices.
For a fixed beam of span L and uniform flexural rigidity EI, with both supports at same level, the support at right rotate by α radians anti clockwise. The support reactions at left and right supports (assuming upward displacement and anti clockwise rotations as positive) respectively are ______.
- 6EI α/L², 2EI α/L, 6EI α/L², 2EI α/L
- 6EI α/L², 4EI α/L, - 6EI α/L², 2EI α/L
- 6EI α/L², 2EI α/L, - 6EI α/L², 4EI α/L
- 6EI α/L², 2EI α/L, - 6EI α/L², 2EI α/L
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Answer: (3) 6EI α/L², 2EI α/L, - 6EI α/L², 4EI α/L
Rotating the right end by α induces a moment 4EIα/L at that end and a carry-over moment 2EIα/L at the left end, both anticlockwise. These end moments (total 6EIα/L) are balanced by a couple of vertical reactions 6EIα/L², upward at the left and downward at the right, giving 6EIα/L², 2EIα/L, −6EIα/L², 4EIα/L.
Three members AB, BC and BD 3m, 3m and 4m long each, with uniform flexural rigidity are rigidly connected at B. If the supports at A and C are fixed while at D it is hinged the distribution factors in directions of BA, BC and BD at the joint B are ______ respectively.
- 3/11, 4/11, 4/11
- 4/11, 3/11, 4/11
- 4/11, 4/11, 3/11
- 1/3, 1/3, 1/3
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Answer: (4) 1/3, 1/3, 1/3
The key takes the three members as sharing the moment at B equally (1/3 each). Working it out with stiffnesses 4EI/3 (BA), 4EI/3 (BC) and 3EI/4 (BD, far end hinged) gives 16/41, 16/41 and 9/41, which is not among the options, while option (3) would follow if D were treated as fixed (4EI/4). (Official final key; some sources differ.)
In a three hinged parabolic arch, the third hinge is located at the crown. The beam moment coinciding with H-moment indicates ______.
- Arch all along carries only radial shear.
- Arch all along carries only beam moment
- Arch all along carries only normal thrust.
- Arch all along carries only H- moment.
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Answer: (3) Arch all along carries only normal thrust.
In a three-hinged arch the net bending moment at any section = beam moment − H·y (the H-moment). If the beam moment diagram coincides with the H-moment diagram, the net moment is zero everywhere and the arch carries only normal (axial) thrust, as a parabolic arch does under a uniformly distributed load.
A cable carrying a load of w/unit run is stretched between supports L apart. The supports are at the same level and the central dip is h. The greatest and least tensions in the cable are ______.
- wL/2, wL²/8h
- √((wL/2)²+(wL²/8h)), wL²/16h
- √((wL/2)²+(wL²/8h)), wL²/8h
- √((wL/2)²+(wL²/2h)), wL²/8h
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Answer: (3) √((wL/2)²+(wL²/8h)), wL²/8h
The horizontal tension H = (wL²/8)/h = wL²/8h is the least tension, at the lowest point, and each support also has a vertical reaction wL/2. The greatest tension, at the supports, is √((wL/2)² + (wL²/8h)²); the booklet's options omit the square on the second term.
A fixed beam of span L carries a UDL of intensity w/unit run. If the plastic moment of the beam section is M_p, collapse occurs when ______ number of plastic hinges are formed and wL = ______.
- 2 and 8M_p/L
- 3 and 8M_p/L
- 3 and 16M_p/L
- 2 and 16M_p/L
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Answer: (3) 3 and 16M_p/L
A fixed beam under UDL collapses when plastic hinges form at both supports and at mid-span, i.e. 3 hinges. The free moment wL²/8 must equal M_p (support) + M_p (mid-span) = 2M_p, so w = 16M_p/L² and wL = 16M_p/L.
A cantilever beam of span L carries a point load W at its free end. Plastic moment of the section from support to middle of span is 1.5 M_p and from there to free end is M_p. Collapse occurs when W = ______.
- 1.5M_p/L
- M_p/L
- 2M_p/L
- 0.75M_p/L
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Answer: (1) 1.5M_p/L
The moment is WL at the support (capacity 1.5M_p) and WL/2 at mid-span (capacity M_p). A hinge forms at the support when W = 1.5M_p/L, before mid-span reaches its capacity (which would need W = 2M_p/L), so the collapse load is W = 1.5M_p/L.