Theory Of Structures — Part 1 MCQs with Explanations

$$\eqalign{ & {\text{Deflection max }}\left( {\text{A}} \right) = \frac{{{\text{P}}{{\text{L}}^3}}}{{48{\text{E}}I}} \cr & {\text{Deflection max }}\left( {\text{B}} \right) = \frac{{5{\text{w}}{{\text{L}}^4}}}{{384{\text{E}}I}};\,\,\,\left( {{\text{w}} = \frac{{\text{P}}}{{\text{L}}}} \right) \cr & = \frac{{5{\text{P}}{{\text{L}}^3}}}{{384{\text{E}}I}} \cr & {\text{Ratio}} = \frac{{{\text{P}}{{\text{L}}^3}}}{{48{\text{E}}I}} \times \frac{{384{\text{E}}I}}{{5{\text{P}}{{\text{L}}^3}}} = \frac{8}{5} \cr} $$

Principal planes are special planes within a stressed body. They are unique because at these specific planes, something very important happens: There are NO shear stresses acting . This means that the only stresses present on a principal plane are normal stresses (also called direct stresses), which act perpendicular to the plane. Therefore, the correct answer is that principal planes are subjected to normal stresses only . The other options are incorrect because shear stress is absent on principal planes.

We have a circular bar with diameter d and length l . It extends by an amount e under a tensile pull W . We want to find the deflection when the same load W is applied at the mid-span of the bar, making it act like a beam. 1. Relate Elongation (e) to Tensile Stress and Strain: When the bar is under a tensile pull W , we know: Stress \( \sigma \) = Force (W) / Area (A), where \( A = \frac{\pi d^2}{4} \). Strain \( \varepsilon \) = Change in Length (e) / Original Length (l). Also, Stress \( \sigma \) = Young's Modulus \( E \) * Strain \( \varepsilon \). Therefore, \( E = \frac{\sigma}{\varepsilon} \). 2. Find Young's Modulus (E) in terms of e, l, d, and W: \( E = \frac{W}{\left( \frac{\pi d^2}{4} \right)} \div \frac{e}{l} \) \( E = \frac{4Wl}{\pi d^2 e} \). 3. Deflection Formula for a Simply Supported Beam with a Point Load at Mid-Span: The standard formula for the deflection (\( \delta \)) of a simply supported beam with a point load (W) at the mid-span is: \( \delta = \frac{Wl^3}{48EI} \), where \( I \) is the area moment of inertia. 4. Area Moment of Inertia (I) for a Circular Cross-Section: \( I = \frac{\pi d^4}{64} \). 5. Substitute I and E into the Deflection Formula: \( \delta = \frac{Wl^3}{48 \times E \times \left( \frac{\pi d^4}{64} \right)} \) \( \delta = \frac{Wl^3}{48 \times \left( \frac{4Wl}{\pi d^2 e} \right) \times \left( \frac{\pi d^4}{64} \right)} \) \( \delta = \frac{Wl^3 \times \pi d^2 e \times 64}{48 \times 4Wl \times \pi d^4} \) \( \delta = \frac{64Wl^3 \pi d^2 e}{192Wl \pi d^4} \) \( \delta = \frac{e l^2}{3 d^2} \). Therefore, the deflection is: \( \frac{el^2}{3d^2} \). Correct answer: Option C \( \frac{el^2}{3d^2} \).

A three-hinged arch is a structural form used in civil engineering to efficiently distribute loads and resist bending moments. It consists of three hinges, which allow rotation at the supports to accommodate the arch's deformation under load. The three hinges are located at the supports and at the crown of the arch. The crown of an arch refers to the highest point along the arch's curve, where the arch is subjected to compressive forces. Placing a hinge at the crown allows the arch to rotate and adapt to changes in loading and deformation. This rotational flexibility is crucial for the arch's stability and ability to carry loads without excessive stress. Note: Options A, C, and D are not accurate descriptions of the typical hinge locations in a three-hinged arch. The correct configuration involves hinges at the supports and at the crown to ensure proper load distribution and structural performance.

$$\left(\frac{2}{3}\right)^3$$

A close coil helical spring is designed primarily to resist axial loads, but when a moment is applied along the axis of the helix, it behaves differently. In this case, the axial moment causes each coil to bend rather than twist or compress. This results in pure bending of the spring wire, without torsion or axial compression/stretching. Option B: Its mean diameter will decrease – Not true in this scenario; there is no significant radial compression due to moment. Option C: Its number of coils will increase – This is incorrect; the number of coils is fixed and does not change with loading. Therefore, only Option A is correct, as the spring is subjected to pure bending under the applied moment.

1. Locus of Reactions: The reaction locus refers to the path traced by the reaction forces at the hinge supports as the load moves along the arch. 2. Two-Hinged Semi-Circular Arch Behavior: A two-hinged arch is statically indeterminate to one degree , meaning it requires additional equations (beyond equilibrium equations) to determine its reactions. The horizontal thrust at the hinges plays a crucial role in defining the locus of reactions. 3. Mathematical Justification: When deriving the equation of the reaction locus , it is found to follow the equation of a hyperbola rather than a circle. This happens because the reactions depend on the horizontal thrust, which varies with the load position in a hyperbolic manner. 4. Why Not a Circle? If the locus were a circle, it would imply a uniform reaction distribution independent of the load position. However, in a two-hinged arch, the reactions change non-linearly, leading to a hyperbolic relationship rather than a circular one. Conclusion: The correct answer is Hyperbola (Option D) , as it accurately represents the mathematical and structural behavior of the reaction locus in a two-hinged semi-circular arch.

Shear stress is the stress induced when a force is applied parallel or tangential to the surface of a material. Average shear stress is calculated as the total shear force divided by the cross-sectional area. Maximum shear stress is the highest shear stress value in a beam's cross-section, and in the case of a circular cross-section, it occurs at the neutral axis. For a circular beam , the relationship between maximum shear stress and average shear stress is given by: τ max = (4/3) × τ avg Therefore, the ratio of maximum shear stress to average shear stress is: τ max / τ avg = 4⁄3 This means the maximum shear stress is 1.33 times the average shear stress in a circular beam Hence, the correct answer is Option D: 4⁄3

Both ends hinged $$l = {\text{L}}$$ Both ends fixed $$l = \frac{{\text{L}}}{2}$$ One end hinged and other end fixed $$l = \frac{{\text{L}}}{{\sqrt 2 }}$$ One end hinged another end free $$l = 2{\text{L}}$$

The correct answer is A: Circle . Let's break down why: Understanding the Concepts: * Stress on an Inclined Plane: When a material is subjected to stress, we can analyze the stress components acting on any inclined plane within that material. * Normal Component: This is the stress acting perpendicular (or normal) to the inclined plane. * Tangential Component: This is the stress acting parallel (or tangential) to the inclined plane. It's also known as shear stress. * Resultant Stress: This is the combination of the normal and tangential stress components. Think of it as a single stress that represents the combined effect of both. * Locus: In this context, locus refers to the path traced by the end point of the resultant stress vector as the inclination of the plane changes. Why a Circle? * Mohr's Circle: The relationship between normal and tangential stresses on an inclined plane is elegantly represented by Mohr's Circle . * Mohr's Circle Visualization: Mohr's Circle is a graphical representation of the state of stress at a point. The coordinates of any point on the circle represent the normal and shear stresses on a particular plane through that point. * Resultant as a Radius: The radius of Mohr's Circle represents the magnitude of the resultant stress. * Tracing the End Point: As the angle of the inclined plane changes, the point representing the stress state moves around the circumference of Mohr's Circle. Therefore, the end point of the resultant stress (which is related to the radius and center of the circle) traces a circular path . Why not the other options? * Parabola, Ellipse, Straight Line: These shapes don't accurately depict the relationship between normal and tangential stresses on an inclined plane as defined by Mohr's Circle and the fundamental principles of stress transformation. In summary: The locus (path) of the end point of the resultant stress on an inclined plane is a circle , which is directly related to the visual representation of stresses on Mohr's Circle.

The point of contraflexure is the location on a beam where the bending moment (B.M.) changes its sign. This means that the bending moment transitions from positive to negative or vice versa at this point. Key Points: B.M. changes sign : The point of contraflexure is characterized by a change in the sign of the bending moment, indicating a shift in the direction of bending. B.M. is maximum or minimum : Maximum or minimum bending moments occur at different points along the beam and are not necessarily at the point of contraflexure. S.F. is zero : While shear force (S.F.) being zero might be a condition at some points, it is not a defining characteristic of the point of contraflexure. The point of contraflexure specifically involves a change in the bending moment direction.

Shear stress in a shaft is not directly proportional to the length of the shaft. Key Points: Shear stress in a shaft is primarily dependent on the torque applied, the radius of the shaft, and the modulus of rigidity. The angle of twist is related to the length of the shaft, but the shear stress itself is independent of the length. Shear stress varies linearly with the radius of the shaft and is directly proportional to the modulus of rigidity. Therefore: The shear stress in a shaft is not directly proportional to the length of the shaft .

$$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{F}}}{{\text{Y}}}$$ is the correct answer to “Y are the bending moment, moment of inertia, radius of curvature, modulus of If M, I, R, E, F and elasticity stress and the depth of the neutral axis at section, then”. This question is standard in competitive examination syllabi for Theory Of Structures.

$$\frac{{2{\text{M}}}}{{\text{T}}}$$ is the correct answer to “A shaft is subjected to bending moment M and a torque T simultaneously. The ratio of the maximum bending stress to maximum shear stress developed in the shaft, is”. This question is standard in competitive examination syllabi for Theory Of Structures.

7,500 N/m2 is the correct answer to “A masonry dam (density = 20,000 N/m 3 ) 6 m high, one metre wide at the top and 4 m wide at the base, has vertical water face. The minimum stress at the base of the dam when the reservoir is full, will be”. This question is standard in competitive examination syllabi for Theory Of Structures.

$$\frac{1}{{16}}$$ is the correct answer to “The ratio of the maximum deflections of a simply supported beam with a central load W and of a cantilever of same length and with a load W at its free end, is”. This question is standard in competitive examination syllabi for Theory Of Structures.

$$\frac{2}{3}$$ is the correct answer to “The ratio of the length and depth of a simply supported rectangular beam which experiences maximum bending stress equal to tensile stress, due to same load at its mid span, is”. This question is standard in competitive examination syllabi for Theory Of Structures.

$$\frac{1}{2}$$ is the correct answer to “The ratio of the length and diameter of a simply supported uniform circular beam which experiences maximum bending stress equal to tensile stress due to same load at its mid span, is”. This question is standard in competitive examination syllabi for Theory Of Structures.

Von Mises is the correct answer to “Shear strain energy theory for the failure of a material at elastic limit, is due to”. This question is standard in competitive examination syllabi for Theory Of Structures.

$$\frac{{3{\text{F}}}}{{2{\text{A}}}}$$ is the correct answer to “The maximum magnitude of shear stress due to shear force F on a rectangular section of area A at the neutral axis, is”. This question is standard in competitive examination syllabi for Theory Of Structures.

400 mm is the correct answer to “A simply supported rolled steel joist 8 m long carries a uniformly distributed load over it span so that the maximum bending stress is 75 N/mm 2 . If the slope at the ends is 0.005 radian and the value of E = 0.2 × 10 6 N/mm 2 , the depth of the joist, is”. This question is standard in competitive examination syllabi for Theory Of Structures.

$$\frac{8}{5}$$ is the correct answer to “A short column (30 cm × 20 cm) carries a load P 1 at 4 cm on one side and another load P 2 at 8 cm on the other side along a principal section parallel to longer dimension. If the extreme intensity on either side is same, the ratio of P 1 to P 2 will be”. This question is standard in competitive examination syllabi for Theory Of Structures.

Three bars is the correct answer to “A compound truss may be formed by connecting two simple rigid frames, by”. This question is standard in competitive examination syllabi for Theory Of Structures.

All the above is the correct answer to “If a three hinged parabolic arch, (span $$l$$, rise h) is carrying a uniformly distributed load w/unit length over the entire span,”. This question is standard in competitive examination syllabi for Theory Of Structures.

$$\sqrt {\frac{{3{\text{w}}}}{{\sigma {\text{b}}}} \times l} {\text{ at the fixed end}}$$ is the correct answer to “Keeping breadth constant, depth of a cantilever of length $$l$$ of uniform strength loaded with uniformly distributed load w varies from zero at the free end and”. This question is standard in competitive examination syllabi for Theory Of Structures.

5 mm is the correct answer to “A cantilever of length 2 cm and depth 10 cm tapers in plan from a width 24 cm to zero at its free end. If the modulus of elasticity of the material is 0.2 × 10 6 N/mm 2 , the deflection of the free end, is”. This question is standard in competitive examination syllabi for Theory Of Structures.

t compressive is the correct answer to “In the truss, the force in the member AC is”. This question is standard in competitive examination syllabi for Theory Of Structures.

284 MN is the correct answer to “If a concrete column 200 × 200 mm in cross-section is reinforced with four steel bars of 1200 mm 2 total cross-sectional area. Calculate the safe load for the column if permissible stress in concrete is 5 N/mm 2 and E s is 15 E c”. This question is standard in competitive examination syllabi for Theory Of Structures.

Length of column divided by least radius of gyration is the correct answer to “Slenderness ratio of a long column, is”. This question is standard in competitive examination syllabi for Theory Of Structures.

16 is the correct answer to “The ratio of the area of cross-section of a circular section to the area of its core, is”. This question is standard in competitive examination syllabi for Theory Of Structures.

m = 2j - 3 is the correct answer to “A truss containing j joints and m members, will be a simple truss if”. This question is standard in competitive examination syllabi for Theory Of Structures.

Depth $${\text{d}} \propto \sqrt {\text{M}} $$ is the correct answer to “For beams breadth is constant,”. This question is standard in competitive examination syllabi for Theory Of Structures.

All the above is the correct answer to “Flat spiral springs”. This question is standard in competitive examination syllabi for Theory Of Structures.

Poisson’s ratio is the correct answer to “The ratio of lateral strain to axial strain of a homogeneous material, is known”. This question is standard in competitive examination syllabi for Theory Of Structures.

120 N/mm2 is the correct answer to “A steel rod of sectional area 250 sq. mm connects two parallel walls 5 m apart. The nuts at the ends were tightened when the rod was heated to 100°C. If α steel = 0.000012/C°, E steel = 0.2 MN/mm 2 , the tensile force developed at a temperature of 50°C, is”. This question is standard in competitive examination syllabi for Theory Of Structures.

Product of moment of inertia is zero is the correct answer to “In case of principal axes of a section”. This question is standard in competitive examination syllabi for Theory Of Structures.

All the above is the correct answer to “The forces in the members of simple trusses, may be analysed by”. This question is standard in competitive examination syllabi for Theory Of Structures.

$$\frac{{{\text{b}}{{\text{h}}^3}}}{{12}}$$ is the correct answer to “The moment of inertia of a triangular section (height h, base b) about its base, is”. This question is standard in competitive examination syllabi for Theory Of Structures.

Guest's or Trecas' theory is the correct answer to “Maximum shear stress theory for the failure of a material at the elastic limit, is known”. This question is standard in competitive examination syllabi for Theory Of Structures.

Haig’s theory is the correct answer to “Total strain energy theory for the failure of a material at elastic limit, is known”. This question is standard in competitive examination syllabi for Theory Of Structures.