Strength Of Materials Miscellaneous
- Consider a cantilever beam, having negligible mass and uniform flexural rigidity, with length 0.01 m. The frequency of vibration of the beam, with a 0.5 kg mass attached at the free tip, is 100 Hz. The flexural rigidity (in Nm2) of the beam is ______.
-
View Hint View Answer Discuss in Forum
S = FL3 3EI k = F 3EI S L3 k = 3EI (0.01)3
k = 3, 000, 000 EICorrect Option: A
S = FL3 3EI k = F 3EI S L3 k = 3EI (0.01)3
k = 3, 000, 000 EI
- A cantilever beam carries the anti symmetric load shown, where W0 is the peak intensity of the distributed load. Qualitatively, the correct bending moment diagram for this beam is
-
View Hint View Answer Discuss in Forum
Here at both the ends, slope is zero and SF is zero. As moving form left to right, rate of shear force decreases due to varying uniform distributed load.Correct Option: C
Here at both the ends, slope is zero and SF is zero. As moving form left to right, rate of shear force decreases due to varying uniform distributed load.
Direction: A steel beam of breadth 120 mm and height 750 mm is loaded as shown in the figure. Assume Esteel = 200 GPa.
- The value of maximum deflection of the beam is
-
View Hint View Answer Discuss in Forum
Moment of inertia
I = bh3 = 120 × (750)3 = 42.1875 × 108 mm4 12 12
Maximum deflection in the beam is given byδ = 5 wL4 384 EI = 5 × 120 × 103 × (15 × 103)4 384 103 200 × 103 × 42.1875 × 108
= 93.75 mmCorrect Option: A
Moment of inertia
I = bh3 = 120 × (750)3 = 42.1875 × 108 mm4 12 12
Maximum deflection in the beam is given byδ = 5 wL4 384 EI = 5 × 120 × 103 × (15 × 103)4 384 103 200 × 103 × 42.1875 × 108
= 93.75 mm
- The beam is subjected to a maximum bending moment of
-
View Hint View Answer Discuss in Forum
Maximum bending moment occurs at the centre
= wL2 = 20 × (15)2 = 3375 kNm 8 8 Correct Option: A
Maximum bending moment occurs at the centre
= wL2 = 20 × (15)2 = 3375 kNm 8 8
- A concentrated load P acts on a simply supported beam of span L at a distance L/3 from the left support. The bending moment at the point of application of the load is given by
-
View Hint View Answer Discuss in Forum
For horizontal equilibrium
R1 + R2 = P
Taking moment about R1, we have
R2 × L = P × L/3R2 = P 3 ∴ R1 = 2P 3
The bending moment at the point of application of load = Area under OABC= 2P × L = 2PL 3 3 9 Correct Option: D
For horizontal equilibrium
R1 + R2 = P
Taking moment about R1, we have
R2 × L = P × L/3R2 = P 3 ∴ R1 = 2P 3
The bending moment at the point of application of load = Area under OABC= 2P × L = 2PL 3 3 9