Geomechanics & Ground Control

Slope Stability (Planar Failure)

Factor of safety for sliding on a discontinuity — resisting friction and cohesion versus the driving weight component.

PART 1

Topic Breakdown & Traps

The Engineering Principle

An open-pit bench or hillside fails when a block slides along a discontinuity dipping out of the slope. The driving force is the down-dip component of the block's weight,

W sin β

; the resisting force is the friction

W cos β tan ϕ

plus the cohesion acting over the failure-plane area,

c A

. Their ratio is the factor of safety — below 1 the slope moves. When cohesion is zero the result collapses to the elegant

F S = tan ϕ / tan β

The Core Formula Matrix

Factor of safety, planar (dry) failure:

F S = \frac{c A + W cos β tan ϕ}{W sin β}

β

= dip of failure plane,

ϕ

= friction angle,

c

= cohesion,

A

= plane area,

W

= block weight.

Cohesionless special case (

c = 0

F S = \frac{tan ϕ}{tan β}

The ‘IIT Traps’

⚠
**Driving uses $sin β$ , resisting friction uses $cos β$ .** Swapping the trig functions inverts the safety factor.
⚠
**Cohesion acts over the plane area $A$ **, giving a force $c A$ — don't add $c$ (a stress) directly to forces.
⚠
** $c = 0 \Rightarrow F S = tan ϕ / tan β$ **: weight cancels, so a cohesionless slope's stability is independent of block size.

PART 2

Progressive 3-Tier Question Suite

Q1BASIC1 Mark · MCQ

A block weighing

1000 kN

rests on a plane dipping at

3 0^{\circ}

. The driving (down-dip) force is:

Q2MEDIUM2 Marks · NAT

A cohesionless (

c = 0

) discontinuity has friction angle

ϕ = 3 5^{\circ}

and dips at

β = 3 0^{\circ}

out of the slope. The factor of safety against planar sliding is ______. (Round off to two decimal places.)

Q3HARD2 Marks · NAT

A block of weight

W = 2000 kN

rests on a plane of area

A = 20 m^{2}

dipping at

β = 3 5^{\circ}

. The discontinuity has

c = 25 kPa

and

ϕ = 3 0^{\circ}

. The factor of safety is ______. (Round off to two decimal places.)