Structural Analysis Questions and Answers – Arches

«
»

This set of Structural Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Arches”.

1. An arch is a beam except for ____
a) It does not resist inclined load
b) It does not resist transverse forces
c) It does not allow rotation at any point
d) It does not allow horizontal movement
View Answer

Answer: d
Explanation: An arch is a curved member in which horizontal displacements are prevented at the supports/springings/abutments.

2. An arch is more economical than a beam for a shorter span length.
a) True
b) False
View Answer

Answer: a
Explanation: Bending Moment for an arch is given by the bending moment produced in simply supported for same loading minus bending moment produced due to horizontal thrust. Since the bending moment produced is lower for the same loading, it is more economical than the beam.

3. Two hinged arches is a determinate structure.
a) True
b) False
View Answer

Answer: b
Explanation: Two hinged arches is an indeterminate structure. We can calculate vertical reactions by using ∑M = 0 and ∑V = 0 but the horizontal reaction cannot be computed by any of equilibrium equations. Thus, two hinged arches is an indeterminate structure.
Note: Join free Sanfoundry classes at Telegram or Youtube
advertisement
advertisement

4. Calculate the horizontal thrust for the two hinged parabolic arch loaded uniformly throughout with distributed load.
Find horizontal thrust for hinged parabolic arch loaded uniformly with distributed load
a) \(\frac{WL^2}{32H} \)
b) \(\frac{WL^2}{16H} \)
c) \(\frac{WL^2}{8H} \)
d) \(\frac{WL^2}{2H} \)
View Answer

Answer: c
Explanation: ∑H = 0
H = \(\frac{∫M.y dy}{∫y^2 dy}\)
Where, y=\(\frac{4 H x ( L-x )}{L^2}\)
Hence, H = \(\frac{WL^2}{8H} \)

5. Calculate the horizontal thrust for the two hinged parabolic arch loaded uniformly for the left half span of the arch with distributed load.
Find horizontal thrust for hinged parabolic arch loaded for left half span of arch
a) \(\frac{WL^2}{32H} \)
b) \(\frac{WL^2}{16H} \)
c) \(\frac{WL^2}{8H} \)
d) \(\frac{WL^2}{2H} \)
View Answer

Answer: b
Explanation: ∑H = 0
H = \(\frac{∫M.y dy}{∫y^2 dy}\)
Where, y=\(\frac{4 H x ( L-x )}{L^2}\)
Hence, H = \(\frac{WL^2}{16H} \)

6. Calculate the horizontal thrust for the two hinged semicircular arch loaded uniformly throughout with distributed load.
Find horizontal thrust for hinged semicircular arch loaded uniformly with distributed load
a) \(\frac{W}{\pi}\)
b) \(\frac{W}{\pi}\) sin2
c) \(\frac{4RW}{3\pi}\)
d) \(\frac{W}{2\pi}\)
View Answer

Answer: c
Explanation: ∑H = 0
H = \(\frac{∫M.y dy}{∫y^2 dy}\)
Y = \(\sqrt{R^2- x^2} – \sqrt{R^2-(\frac{L^2}{2})} \)
Hence, H = \(\frac{4RW}{3\pi}.\)

7. Calculate the horizontal thrust for the two hinged semicircular arch loaded with point load at its crown.
Find horizontal thrust for hinged semicircular arch loaded with point load at crown
a) \(\frac{W}{\pi}\)
b) \(\frac{W}{\pi}\) 2
c) \(\frac{4RW}{3\pi}\)
d) \(\frac{W}{2\pi}\)
View Answer

Answer: a
Explanation: ∑H = 0
H = \(\frac{∫M.y dy}{∫y^2 dy}\)
Y = \(\sqrt{R^2- x^2} – \sqrt{R^2-(\frac{L^2}{2})} \)
Hence, H = \(\frac{W}{\pi}\)
advertisement

8. Calculate the horizontal thrust for the two hinged parabolic arch loaded with point load at its crown.
Find horizontal thrust for two hinged parabolic arch loaded with point load at its crow
a) \(\frac{W}{\pi}\)
b) \(\frac{W}{\pi}\)sin2
c) \(\frac{4RW}{3\pi}\)
d) \(\frac{25WL}{128H}\)
View Answer

Answer: d
Explanation: ∑H = 0
H = \(\frac{∫M.y dy}{∫y^2 dy}\)
Where, y=\(\frac{4 H x (L-x)}{L^2}\)
Hence, H = \(\frac{25WL}{128H}.\)

9. Calculate the horizontal thrust for the two hinged semicircular arch loaded with point load at inclination of α with horizontal axis on the left span.
Find horizontal thrust for hinged semicircular arch loaded with point load at inclination
a) \(\frac{W}{\pi}\)
b) \(\frac{W}{\pi}\)sin2
c) \(\frac{4RW}{3\pi}\)
d) \(\frac{25WL}{128H}\)
View Answer

Answer: b
Explanation: ∑H = 0
H = \(\frac{∫M.y dy}{∫y^2 dy}\)
Y = \(\sqrt{R^2- x^2} – \sqrt{R^2-(\frac{L}{2^2})} \)
Hence, H = \(\frac{W}{\pi}\)sin2∞.
advertisement

10. Identify the incorrect statement according to the hinged arches.
a) Three hinged arch is a statically determinate structure
b) To analyze three hinged arch, equlibrium equations are sufficient
c) For three hinged parabolic arch subjected to u.d.l over the entire span, the bending moment is constant throughout the span
d) For two hinged parabolic arch subjected to u.d.l over the entire span, the bending moment is zero throughout the span
View Answer

Answer: c
Explanation: For three hinged parabolic arch subjected to u.d.l over the entire span, the bending moment and radial shear at any section is zero throughout the span.

Sanfoundry Global Education & Learning Series – Structural Analysis.

To practice all areas of Structural Analysis, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & technical discussions at Telegram SanfoundryClasses.