Soil mechanics solutions for Assignment 1

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1
UNIVERSITY OF ZIMBABWE
NAME VOTE MARANGE
REG NUMBER R1811272
PROGRAM HBAE
LEVEL 3.2
COURSE CODE AGEN 314
LECTURER’S NAME PROF W. GWENZI
ASSIGNMENT NUMBER ONE

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AGEN 314 Assignment 1
SOLUTION
1.(a) How shrubs may contribute to slope failure
 Roots of the trees growing on an earth dam embankment widen the fractures on
slopes and consequently failure of the body occurs.
 On the slopes, which are having vegetation, uprooting during storms is common.
The pit on the surface caused by this uprooting many times become the nucleus for
further slope failure.
 Slope failure occurs when driving force is greater than resisting force
 The occurrence of a major storm and mass erosion is closely correlated. Excess soil
water is generally accepted to be the principal factor causing slope failures.
 Pore water pressure produced by the head of water in a saturated soil can reduce
shear strength.
 Rising pore water pressures can reduce the effective weight of the soil mass by
producing an uplift force.
 From modified soil strength equation ,𝑆 = (𝑅 + 𝑟) + (𝑊 cos 𝛼 − 𝜌) tan ∅
considering pore water pressure.active pore water pressures can reduce soil shear
strength
 Increased soil water may also decrease cohesion (C) of some soils through leaching
and eluviation.
 Soil failure is caused since shrubs have short roots and cannot take all the water in
the soil.
 The canopy of the shrubs act as a barrier for sunlight to penetrate,when there is
moisture accumulation, increased moisture decreases cohesion of soil hence soil
failure occurs
1.(b) How shrubs may contribute to slope stability
Hydromechanical effects of vegetation on soil stability
Hydrological mechanisms
 Foliage intercepts rainfall, causing absorptive and evaporative losses that reduce
precipitation to infiltrate.
 Roots and stems increased the roughness to the ground surface and soil permeability
thereby expanding the infiltration capacity.
 The roots absorb water from the soil and it is released into the atmosphere through a
transpiration mechanism that causes the pore water to decrease.
 Depletion of soil moisture by the root absorption may accentuate the soil to crack,
thus increasing the infiltration capacity.
Mechanical mechanisms
 Roots reinforce the soil, increasing soil shear strength.
 Vegetation roots anchor to the deep soil layer, providing support to the upslope soil
mantle through buttressing and arching.

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AGEN 314 Assignment 1
 Weight of vegetation surcharges the slope and increases normal and downhill force
components.
 Plant exposed to the wind transmits dynamic forces into the hill.
 Roots bind soil particles and reduce their susceptibility to erosion
Solution
2.(a) 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑚𝑒𝑛, 𝐴 =
50
1000
×
50
1000
= 0.0025𝑚2
𝑁𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜎
́ (𝑘𝑁 𝑚2
⁄ ) =
𝑁𝑜𝑟𝑚𝑎𝑙 𝑓𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑚𝑒𝑛
=
𝑁𝑜𝑟𝑚𝑎𝑙 𝑓𝑜𝑟𝑐𝑒
1000 × 𝐴
𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜏𝑓(𝑘𝑁 𝑚2
⁄ ) =
𝑆ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑚𝑒𝑛
=
𝑆ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒
1000×𝐴
For test 1
𝑁𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜎
́ (𝑘𝑁 𝑚2
⁄ ) =
150
1000 × 0.0025
3.(a) =
150
2.5
= 60 𝑘𝑁 𝑚2
⁄
𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜏𝑓(𝑘𝑁 𝑚2
⁄ ) =
157.5
1000×0.0025
=
157.5
2.5
= 63 𝑘𝑁 𝑚2
⁄
For test 2
𝑁𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜎
́ (𝑘𝑁 𝑚2
⁄ ) =
250
1000×0.0025
=
250
2.5
= 100𝑘𝑁/𝑚2
𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜏𝑓(𝑘𝑁 𝑚2
⁄ ) =
199.9
1000×0.0025
=
199.9
2.5
= 79.96 𝑘𝑁 𝑚2
⁄

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AGEN 314 Assignment 1
For Test 3
𝑁𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜎
́ (𝑘𝑁 𝑚2
⁄ ) =
350
1000×0.0025
=
350
2.5
= 140𝑘𝑁/𝑚2
𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜏𝑓(𝑘𝑁 𝑚2
⁄ ) =
257.6
1000×0.0025
=
257.6
2.5
= 103.04 𝑘𝑁 𝑚2
⁄
For test 4
𝑁𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜎
́ (𝑘𝑁 𝑚2
⁄ ) =
550
1000 × 0.0025
=
550
2.5
= 220𝑘𝑁/𝑚2
𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠, 𝜏𝑓(𝑘𝑁 𝑚2
⁄ ) =
363.4
1000 × 0.0025
=
363.4
2.5
= 145.36 𝑘𝑁 𝑚2
⁄
Table of results
Test Number Normal
stress 𝝈 = 𝝈
́ (𝒌𝑵 𝒎𝟐
⁄ )
Shear stress at failure
𝝉𝒇 (𝒌𝑵 𝒎𝟐
)
⁄
1 60 63
2 100 79.96
3 140 103.04
4 220 145.36

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AGEN 314 Assignment 1
2. From the above graph we deduce the value of c and ∅
2. Equation of a line 𝜏𝑓 = 0.5217𝜎́ + 30.017 = 𝑐 + 𝜎́𝑡𝑎𝑛∅
This yields 𝑐 = 30.0𝑘𝑁/𝑚2
and 𝑡𝑎𝑛∅ = 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
∆𝜏𝑓
∆𝜎
́
= 0.5217
𝑡𝑎𝑛∅ = 0.5217
∅ = tan−1(0.5217)
∅ = 27.55° ≈ 27.6°
2.(a) Cohesion 𝑐 = 𝑡ℎ𝑒 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ = 30.0𝑘𝑁/𝑚2
2.(b) Angle of internal friction, ∅ = 27.6°
2.(c) Angle of inclination of failure plane ,𝜃𝑓 = 45 +
∅
2
=
∅+90
2
=
90+27.6
2
= 58.8°
Figure 1 shows a graph of normal stress against shear stress