The design of Elements of Lifts and Escalator from Civil Engineering point of view. Mainly Raft foundation, Machine Foundation, and Shear walls are discussed.

1. (Civil Engineering Aspect)
By- Shubham Arora

2. Introduction-LIFT/ELEVATOR
 The work of a Civil Engineer in the design of Lifts is
the design of two main components, namely-
1. Lift Well.
2. Footings.
The design of Lift well is composed of design of Shear
Walls mainly for Wind Loads and Seismic Load.
The design of footings is basically decided on the basis
of loads and types of buildings/structure.

3. What is Shear Wall & why Needed?
• Shear Wall is a Structural Element used to resist lateral
or horizontal forces parallel to the plane of the wall.
Wind and seismic loads are most common for which
shear walls are designed.
• Plywood is the basic material of which shear walls are
designed but with improvement in technology, there
are many other injectors designed which can carry
lateral loads easily.
These may also be called as reinforced walls serving the
purpose of lateral loads.

4. Design of Shear Wall(IS456:2000)
 General Requirements (As per 13920:1993, Cl.9)
The thickness of any part of shear wall should not be
less than 150mm(preferably).
Reinforcement should be provided in both
longitudinal and transverse direction, distributed
uniformly. Minimum reinforcement ratio is .0025 of
gross area.
The diameter of bars should not be more than 1/10th of
thickness of cross-section.

5. Design Steps(Cl.32.4)
1.Critical Section for Shear: The critical section for maximum
shear shall be at .5Lw or .5Hw from the base, whichever is
less.
2.Calculate Nominal Shear Stress, Tvw
Tvw=Vuw/twdw, where
Vuw=factored shear force
tw=wall thickness
dw=.8*Lw where Lw is length of wall.
Check: Tvw<=.17fck for limit state method,
<=.12fck for working stress method.
(As per Cl.32.4.2.1)

6. Design Steps
3.Calculate Tc:
(a) For Hw/Lw<=1;Squat walls;
Tcw=(3-Hw/Lw)*k1*fck^.5
Where k1=.2 for limit state method; and
k1=.13 for working stress method;
(b) For Hw/Lw>1; Slender walls;
Lesser of calculated above and
Tcw=k2*fck^.5*((Hw/Lw )+1)/((Hw/Lw) -1);
k2=.045 for limit state and .03 for working stress;

7. Design Steps
4. Shear Reinforcement:
Shear force to be carried=Vu-Tcw*tw*.8*Lw
Area(Aav)=Pw*.8*Lw*tw; where Pw=
(a) Hw/Lw<=1,Pw shall be lesser of ratio of vertical
reinforcement or horizontal reinforcement to the
respective cross sectional area of wall in respective
direction.
(b) Hw/Lw>1; it is ratio of horizontal reinforcement area
to the cross section of wall per vertical meter.
Provide reinforcement conforming Cl.32.5.

8. Design Steps
5. Calculate spacing of reinforcement;
Vusw=.87fyAhdw/Svw; Ah is area of horizontal shear
reinforcement.
6. The vertical reinforcement should not be less than
horizontal reinforcement in any case.
7. In walls if there is no boundary element, then vertical
reinforcement should be concentrated at ends.
 Each concentration shall consist of a minimum of 4 bars of
12 mm diameter arranged in at least 2 layers.
 Provide development length if required as per Cl.9.9 in
IS13920:1993.

9. Boundary Elements(Cl.9.4;IS13920)
 Boundary elements are portions along the wall edges
that are strengthened by longitudinal and transverse
reinforcement. Though they may have the same
thickness as that of the wall web it is advantageous to
provide them with greater thickness.
 If Seismic Load + Gravity Loads>.2fck, boundary
elements are necessary; if <.15fck, they can be ignored.

10. Design of Boundary elements
 A boundary element shall have adequate axial load
carrying capacity, assuming short column action, so as to
enable it to carry an axial compression equal to the sum of
factored gravity load on it and the additional compressive
load induced by the seismic force.
 Load Carrying capacity may be calculated as:
Mu – Muv/Cw; where
 Mu=factored design moment on the entire wall section,
 Muv=moment of resistance provided by distributed
vertical reinforcement across a wall section, (Annexure A)
 Cw=center to center distance between the boundary
elements along the two vertical edges of the wall.

11. Footings
 Depending upon the type of building, its usage, height
and considering many other factors;
There are two types of footing that can be provided
under the shear wall and boundary elements; basically
the lift well-
1. Raft Foundation.
2. Wall Footing/Strip Footing.
On the basis of different site conditions(type ,BC of soil),
space available, Economy and type of structure, the type
of footing can be designed.

12. Design of Strip/Wall Footing
 Calculate the Serviceable Load on the and using the
bearing the capacity, find the area of footing to be
provided.
 Provide area more than required, then calculate the
corresponding bearing capacity.
 Provide the depth for one-way shear at the critical section
and then verify it for two-way shear.
 Design the flexural reinforcement, verify the depth and
decide the spacing of main and distribution steel.
 Provide development length and transfer of force at base if
required.

13. Design of Raft Foundation
 Calculate the total load that is to be taken up by
foundation.
1. Loads on Columns.
2. Load due to wind.
3. Self weight, etc.
Note-When wind load is considered the stresses can be
increased by 33.3%. However, if wind load reactions are
less than one-third of other loads, wind is not
considered in the design. Take wind loads on leeward
and windward side both.

14. Design Steps
 Calculate the total load and max. and min.
eccentricities and hence the upward soil pressure.
 Design of Slab- It is designed for two moments
namely,
1. Bending Moment in Cantilever Portion.
2. Bending Moment in Continuous Slab.
 Provide Area of Steel Corresponding to Max. bending
moment and apply the checks for shear and
development length.

15. Design Steps
 Design of Intermediate Secondary Beams- Pressure
Intensity here does not vary, as slab provides reaction
to beams, rest is same, calculate bending moment,
check depth and provide required area of steel.
 Design of Central Beams- Since connected at two
different ends, calculate the additional moments along
with moments transferred from slab below, Rest is
same.
 Design of End Secondary Beam- These may get less
load width of slab offering reaction gets reduced, but
provide same depth for the sake of symmetry.

16. Design Steps
 Design of Main (Longitudinal) Beam- Three Cases-
1. Ignoring Wind Effect- Design as simple continuous
beam, calculate point of max. bending moment,
check depth, provide steel and apply checks.
2. Taking wind effect: leeward Beam- Net Reactions
will change in this case as you need to take the effect
of wind also, rest is same.
3. Taking wind effect: Windward Beam, same goes for
this case as mentioned above.
 Design for the most critical condition of the 3 above.

17. Footings for Escalators
 Escalators need special kind of foundation and this
foundation being safe in shear and settlement needs to
comprehend with the vibrations from machine
installed, which unanimously cause dynamic loads.
These special foundations are called machine
foundations.
 A rotary machine is installed for movement of belt and
the frequency and power of machine to be used is
chosen on basis of Load{live load and dead
load(steps+ machine)}, speed and height of
transportation required.

18. Design of Machine Foundation
 There are various parameters that need to be
considered for design of machine foundation namely,
Damping factor
Frequency Ratio
Coefficient of Elastic Uniform Compression.
Engine mass or Machine Weight.

19. Design Criteria{IS 2974(part 3)}
 It should be safe against shear, settlement should be in
limits, and there should not be any possibility of
resonance.
 The amplitude, displacement should be within limits
and C.G of machine and foundation must lie on same
line to avoid additional moments.
 Vibrations produced should not be annoying to
persons, adjacent footings of buildings.
 The depth of ground water table should be 1/4th to
width of foundation below base plane.

20. Design Steps
 Calculate the total live load, dead load and accordingly decide
the motor to be employed for conveyor belt to operate, keeping
in mind the velocity and height.
 Decide frequency Ratio(r); It should be >2 or <1/2.
 Calculate natural frequency;
W/Wn =2;
• Calculate mass;
Wn= (Cu*A/m)^1/2 or (k/m)^1/2
Cu= Coefficient of Elastic Uniform compression;
Cu=1.13E/(A)^1/2*(1-u^2); u= dynamic viscosity;
• Mass = Density* A*h
• Calculate height of foundation to be provided.

21. Design Steps
 Apply Checks-
 Calculate displacement-
Z= F/ k*(1-r^2)^2+4(Dr)^2;
 Calculate Magnification Factor-
M= Z/Zst; Zst=F/k;
 Calculate Force Transmitted-
T= Ft/Fo;
=M{1+(2Dr)^2}^1/2
D=C/Cc=C/(mk)^1/2; D= damping factor.
0.01<D<0.1

22. Construction Details
 The reinforcement in concrete should not be less than
25kg/m^3 and it should run in all 3 directions.
 Min. Cover should be 75mm at bottom, 50mm on top
and sides and min. 12mm diameter bars should be
used @200-250mm spacing.
 Foundation should be casted in single operation with
at least M-15 preferably.
 For foundation block of height above 5m, construction
joint can be provided.