This set of Irrigation Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Design of Gravity Dams”.
1. In high dams, the safety against sliding should be checked only for friction.
Explanation: The safety against sliding should be checked only for friction in case of low dams and in high dams, the shear strength of the joint (i.e. an additional shear resistance) is also considered for economical design. The dam section is given an extra slope or batter on the U/s or D/s side as per requirements for achieving stability.
2. For full reservoir condition in a gravity dam, the critical combination of vertical and horizontal earthquake accelerations to be considered for checking the stability is ________________________
a) vertically upward and horizontally downstream
b) vertically downward and horizontally downstream
c) vertically upward and horizontally upstream
d) vertically downward and horizontally upstream
Explanation: Horizontal inertia force should be considered to be acting at the center of the gravity of the mass regardless of the shape of cross-section and it acts horizontally downstream in worst cases under full reservoir case. This force would produce the worst results is it is additive to the hydrostatic water pressure (acting towards the downstream).
3. The base width of a solid gravity dam is 35 m and the specific gravity of dam material is 2.45. What is the approximate allowable height of the dam having an elementary profile without considering the uplift?
a) 64.68 m
b) 54.80 m
c) 164 m
d) 80 m
Explanation: The base width at bottom is given by B = H/Sc1/2 (c = 0 since uplift is not considered).
B = 35 m and Sc = 2.45
Allowable height of the dam H = 35 x 2.451/2 = 54.8 m.
4. A low gravity dam of elementary profile made up of concrete of relative density 2.57 and safe allowable stress of foundation material 4.2 MPa. What is the maximum height of the dam without considering the uplift force?
a) 120 m
b) 217 m
c) 279 m
d) 325 m
Explanation: The maximum possible height of low gravity dam is H = f / ϒw (Sc + 1) where f = allowable stress of dam material = 4.2 MPa, Sc = 2.57, and ϒw = 9.81 KN/m2.
H = [4.2 / (9.81 x 3.57)] x 1000 = 119.92 m.
5. The vertical stress at the toe was found to be 3.44 MPa at the base of the gravity dam section. If the downstream face of the dam has a slope of 0.617 horizontal: 1 vertical, the maximum principal stress at the toe of the dam when there is no tailwater is _______________
a) 1.7 MPa
b) 2.4 MPa
c) 3.6 MPa
d) 4.8 MPa
Explanation: The principal stress at the toe is given by Pat toe = Pv. secΦ2 (without considering the tailwater) where Pv = 3.44 MPa and tan Φ = 0.617/1 i.e.Φ = 31.67°
Pat toe = 3.44 x sec(31.67°)2 = 4.74 MPa.
6. What is the recommended value of shear friction factor against sliding?
a) More than unity
b) Less than unity
c) More than 3 to 5
d) Less than 3
Explanation: Shear Friction Factor is given by –
SFF = sliding factor (SF) + B.q / ∑Ph where B = width of joint or section area = B x 1, q is the shear strength of the joint, and Ph is the sum of horizontal force causing sliding. SF must be greater than 1 and SFF must be greater than 3 to 5. This analysis is carried out for a full reservoir case as well as an empty case.
7. The small openings made in the huge body of a concrete gravity dam such as sluices and inspection galleries can be assumed to be causing only local effects without any appreciable effect on the distribution of stresses as per the principle of_____________________
b) St. Venant
d) St. Francis
Explanation: Small openings made in the dam only produce local effects as per St. Venant’s principle. They do not affect the general distribution of stresses. This is one of the most important assumptions made in the two-dimensional analysis of gravity dams.
8. A concrete gravity dam having a maximum reservoir level at 200 m and the RL of the bottom of the dam 100 m. The maximum allowable compressive stress in concrete is 3000 KN/m2 and the specific gravity of concrete is 2.4. Calculate the height of the dam and check whether it is a high dam or low dam.
a) H = 90 m High gravity dam
b) H = 90 m Low gravity dam
c) H = 214.2 m High gravity dam
d) H = 214.2 m Low gravity dam
Explanation: The limiting height of the dam is given by-
H = f / ϒw (Sc + 1) where f = allowable stress of dam material = 3000 KN/m2, Sc = 2.4, and ϒw = 9.81 KN/m2.
H = 3000 / 9.81 x 3.4 = 89.9 m.
This value is more than the height of the dam so it is a high gravity dam.
9. The axis of a gravity dam is the ______________________
a) line of the crown of the dam on the downstream side
b) line of the crown of the dam on the upstream side
c) centre-line of the top width of the dam
d) line joining mid-points of the base
Explanation: The axis of the dam is taken as the reference line which is defined separately in the plan and in the cross-section of the dam. In plan, it is the horizontal trace of the U/s edge of the top of the dam. In the cross-section, the vertical line passing through the U/s edge of the top of the dam is considered as the axis of the dam.
10. Presence of tail-water in a gravity dam ____________________
a) increases the principal stress and decreases the shear stress
b) increases both the principal stress and the shear stress
c) decreases the principal stress and increases the shear stress
d) decreases both the principal stress and the shear stress
Explanation: The principal stress is given by the formula –
P = Pv sec(Φ)2 – p’ tan(Φ)2 where Pv is the intensity of vertical pressure and p’ is the tail-water pressure
The shear stress on the horizontal plane near the toe is given by –
S = (Pv – p’) tan(Φ)
From both the equations, it is clear that the tail-water pressure is opposite in nature and it reduces the value of principal stress and shear stress.
Sanfoundry Global Education & Learning Series – Irrigation Engineering.
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