# Strength of Materials Questions and Answers – Dams Stability Analysis

This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Dams Stability Analysis”.

1. To ensure economy in dam sections, the ______ should be minimum.
a) Base width
b) Top width
c) Spillway length
d) Toe of the wall

Explanation: In dams, the economy prevails if the cross section of a dam is provided with minimum base width. Base width “b” can be determined from the equation: b2 +ab+ a2 = H2/ S.

2. Calculate the maximum stress at the base section is the self weight is 4400 kN. The top and bottom width of them are 3 and 8 m respectively. Take (e) = 2.97.
a) 1658.15 kN/m2
b) 1775.12 kN/m2
c) 1897.45 kN/m2
d) 2336.67 kN/m2

Explanation: The maximum stress at the base section = W/b (1+6e/b)
P= 4400/8 (1+6×2.97/8)
= 1775.125 kN/m2.

3. If the maximum stress is positive, then the nature of stress is ____
a) Tensile
b) Shearing
c) Compressive
d) Bending

Explanation: If the stress developed in the dam section at the base is positive there it indicates the nature of stress to be compressive.

4. Determine the eccentricity of the dam section, if the base width of the dam be 6m. Take Z = 5.5m.
a) 2.5
b) 1.5
c) 3.5
d) 4.5

Explanation: e = Z – b/2. [The value of Z = 5.5m] = 5.5 – 6/2
= 5.5 – 3
= 2.5 m.

5. Calculate the minimum stress developed at the heel of the dam, if the self weight of the dam is 924 kN and the base with is 6 metres [Take e = 0.0945m].
a) 145 kN/m2
b) 139 kN/m2
c) 167 kN/m2
d) 183 kN/m2

Explanation: The minimum stress developed at the heel of the dam is W/b ×(1-6e/b).
= 920/6 ×(1-6×0.0945/6).
= 139 kN/m2.
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6. The side slopes depend on ____________ conditions of a proposed dam.
a) Toe width
b) Height of foundation
c) Character of material
d) Free board allowance

Explanation: The upstream and downstream slopes of the Dam should be stable in all situations the side slopes depend upon
1.Height of the dam
2.Character of material
3.Nature of foundation.

7. Molitor’s formula can be used for calculation of ___________
a) Freeboard
b) Toe width
c) Wave height
d) Base drop

Explanation: The minimum height of the free board for wave action is generally taken to be equal to 1.5 hw. Where hw = height of wave action.
It can be calculated using 0.032(VF)1/2.

8. The height of the dam = free board + ___________
a) FTL
b) MWL
c) FRL
d) HFL

Explanation: The dam’s height mainly depends on HFL of the reservoir. The height of the dam above surface is given by the HFL plus the free board.
Height of the dam = HFL + free board.

9. _____________ sections allow the surplus discharge to flow in dams.
a) Mulching
b) Over reinforced
c) Breaching
d) Balanced

Explanation: Sometimes during construction, the surplus works are avoided in many tanks. In such circumstances, breaching sections are provided in the dams to allow the surplus flood discharge. With this the scouring of section can be avoided.

10. If the minimum stress developed is negative, then the nature of stress is ___________
a) Shearing
b) Tensile
c) Bending
d) Compressive

Explanation: If the minimum stress developed at the base of the dam section is negative, then the respective nature of the stress will be in tensile condition.

11. ____________ creates concentrated seepage in dams section.
a) Longitudinal cracks
b) Transverse cracks
c) Construction cracks
d) Contraction cracks

Explanation: Due to differential settlement cracks may be developed in the bund. These are of two types 1) longitudinal cracks 2) transverse cracks. The transverse cracks are more dangerous to the dam section because they can create concentrated seepage.

12. The upstream slope recommended for sand and gravel with RCC core wall is __________
a) 1:2
b) 3:1
c) 2.5:1
d) 1.5:1

Explanation: According to Terazghi, the following upstream slopes should be recommended.
For:
1. Homogeneous well graded – 2.5 : 1
2. Homogeneous coarse silt – 3 : 1
3. Sand and gravel with RCC core wall – 2.5 : 1.

13. Major distributaries discharge varies from ____________
a) 0.25 to 5 cumecs
b) 2 to 4 cumecs
c) 1.5 to 5 cumecs
d) 1.2 to 5 cumecs

Explanation: The major distributary takes off from a branch canal to distribute the water to various parts of the field. The supply of the water varies from 0.25 to 5 cumecs.

14. Field channels are also known as ______
a) Branch canals
b) Slope channels
c) Water courses
d) Contour canals

Explanation: The water courses are the channel that carries irrigation water to the fields. The water courses derive their supply from distributaries through outlets. They are also called as field channels.

15. The structures constructed along are distributaries are called as _______
a) Inlets
b) Outlets
c) Distributaries
d) Channels

Explanation: An outlet is a simple and small irrigation structure which is constructed along the distributaries. The amount of water that is withdrawn through the outlet is in proportion to the area that is irrigated below respective point.

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