1. SEISMIC LOADS
LATERAL LOAD FLOW
FRAMES and SHEAR WALLS

2. SEISMIC LOAD

3. Sec. 11.4.1 Mapped Acceleration Parameters Determine SS and S1.
These are determined from two maps
SS is the 0.2 second spectral response, and S1 is the 1.0 second spectral
response shown on figures 22-3 and 22-4 respectively.
You can find these maps by going to the USGS website and try your luck with the
java applet. Note these maps show whole number percentages, so 1.75 means 175%g which
translates to 1.75 [SS is 1.5 and S1 is 0.647] These
values are from the centroid of the zip-code range of values and adequate for our purposes.
Sec Assume Site Class D, because we do not have a geotechnical report for our site.
Sec Site Coefficients and Adjusted Maximum Considered Earthquake (MCE) Spectral
Response Parameters.
Refer to tables and to determine the Site coefficients Fa &
Fv respectively. You have to match the SS and S1 that you mapped with the values in the
charts. [Fa = 1.0 Fv = 1.5]
Sec Design Spectral Acceleration Parameters: Take the values that you have
determined in Sec to determine SDS and SD1 (SDS turns out to be important) [SDS =
1.005 SD1 = ]
Sec Importance Factor & Occupancy Factor. Choose your importance factor from table
Assume that you are occupancy category II (I is agricultural buildings, II is most
buildings, III includes buildings having assembly spaces for 300+ persons, Elementary &
Secondary Schools, etc., IV is Hospitals, Emergency Facilities, etc.)
Sec Seismic Design Category, read the notes and use the chart Table to
determine your seismic design category, it is based on the value of SDS that you found in Sec.
above. The seismic design category in many ways replaces the older Uniform Building
Code method of determining seismic forces that separated the US into seismic zones graded
from 1 to 4 (four being the worst). [SDC = D]

4. Determine Spectral Response Parameters at design location
SS is the 0.2 second spectral response, and S1 is the 1.0 second spectral
response
At N , W :
Ss = 1.50
S1 = 0.60

5. Determine Site Coefficients
Site Class : D
Ss > 1.25
Fa = 1.0
S1 > 0.5
Fv = 1.5
Determine Design Spectral Acceleration Parameters
Assume Site Class D, because we do not have a geotechnical report for our site.
Sec Site Coefficients and Adjusted Maximum Considered Earthquake (MCE) Spectral
Response Parameters.
Refer to tables and to determine the Site coefficients Fa &
Fv respectively. You have to match the SS and S1 that you mapped with the values in the
charts. [Fa = 1.0 Fv = 1.5]
Sec Design Spectral Acceleration Parameters: Take the values that you have
determined in Sec to determine SDS and SD1 (SDS turns out to be important) [SDS =
1.005 SD1 = ]
SMS = (1.0)(1.5) = 1.5
SDS = (2/3)(1.5) = 1.0

6. Cs = SDS /(R/I)
=1.0/(R/I)
Class II : I = 1.0
Ordinary Moment Resisting Frame :
R = 3.5
V = 1.0/3.5 W
0.3 W

7. Seismic Load is generated by the inertia of the mass of the structure : VBASE
Redistributed (based on relative height and weight) to each level as a ‘Point Load’ at the center of mass of the structure or element in question : FX
VBASE Wx hx
S(w h)
VBASE =
(Cs)(W)
( VBASE )
Fx =

8. Total Seismic Loading : VBASE = 0.3 W
W = Wroof + Wsecond

9. Wroof

10. Wsecond flr

11. W = Wroof + Wsecond flr

12. VBASE

13. Redistribute Total Seismic Load to each level based on relative height and weight
Froof
Fsecond flr
VBASE (wx)(hx)
S (w h)
Fx =

14. VBASE (wx)(hx) S (w h) Fx =
In order to solve the equivalent lateral force distribution equation,
we suggest you break it up into a spreadsheet layout
Floor w h (w)(h) (w)(h)/S(w)(h) Vbase Fx
Roof k 30ft 5000k-ft k 68.75k
2nd 200k 15ft 3000k-ft k 41.25k
S (366.67k) S(8000k-ft) S (110k)
Vbase = 0.3W = 0.3(166.67k+200k) = 0.3(366.67k) = 110k

15. Load Flow to Lateral Resisting System :
Distribution based on Relative Rigidity
Assume Relative Rigidity :
Single Bay MF :
Rel Rigidity = 1
2 - Bay MF :
Rel Rigidity = 2
3 - Bay MF :
Rel Rigidity = 3

16. Distribution based on Relative Rigidity :
SR = = 4
Px = ( Rx / SR ) (Ptotal)
PMF1 = 1/4 Ptotal

17. Lateral Load Flow
diaphragm > collectors/drags > frames

18. STRUCTURAL DIAPHRAGM
A structural diaphragm is a horizontal structural system used to transfer lateral loads to shear walls or frames primarily through in-plane shear stress
Basically, combined with vertical shear walls or frames IT ACTS LIKE A LARGE I-BEAM
a diaphragm is a structural system used to transfer lateral loads to shear walls or frames primarily through in-plane shear stress
The diaphragm of a structure often does double duty as the floor system or roof system in a building, or the deck of a bridge, which simultaneously supports gravity loads.
Diaphragms are usually constructed of plywood or oriented strand board in timber construction; metal deck or composite metal deck in steel construction; or a concrete slab in concrete construction.
The two primary types of diaphragm are flexible and rigid:
Flexible diaphragms resist lateral forces depending on the tributary area, irrespective of the flexibility of the members that they are transferring force to.
Rigid diaphragms transfer load to frames or shear walls depending on their flexibility and their location in the structure. The flexibility of a diaphragm affects the distribution of lateral forces to the vertical components of the lateral force resisting elements in a structure. [1]
Parts of a diaphragm include:
the membrane, used as a shear panel to carry in-plane shear
the drag strut member, used to transfer the load to the shear walls or frames
the chord, used to resist the tension and compression forces that develop in the diaphragm, since the membrane is usually incapable of handling these loads alone.

19. STRUCTURAL DIAPHRAGM Flexible or Semi-flexible Type: Plywood
Metal Decking
a diaphragm is a structural system used to transfer lateral loads to shear walls or frames primarily through in-plane shear stress
The diaphragm of a structure often does double duty as the floor system or roof system in a building, or the deck of a bridge, which simultaneously supports gravity loads.
Diaphragms are usually constructed of plywood or oriented strand board in timber construction; metal deck or composite metal deck in steel construction; or a concrete slab in concrete construction.
The two primary types of diaphragm are flexible and rigid:
Flexible diaphragms resist lateral forces depending on the tributary area, irrespective of the flexibility of the members that they are transferring force to.
Rigid diaphragms transfer load to frames or shear walls depending on their flexibility and their location in the structure. The flexibility of a diaphragm affects the distribution of lateral forces to the vertical components of the lateral force resisting elements in a structure. [1]
Parts of a diaphragm include:
the membrane, used as a shear panel to carry in-plane shear
the drag strut member, used to transfer the load to the shear walls or frames
the chord, used to resist the tension and compression forces that develop in the diaphragm, since the membrane is usually incapable of handling these loads alone.

20. STRUCTURAL DIAPHRAGM Rigid Diaphragm Type: Reinforced Concrete Slab
Concrete-filled Metal Deck composite Slab
Braced/horizontal truss
a diaphragm is a structural system used to transfer lateral loads to shear walls or frames primarily through in-plane shear stress
The diaphragm of a structure often does double duty as the floor system or roof system in a building, or the deck of a bridge, which simultaneously supports gravity loads.
Diaphragms are usually constructed of plywood or oriented strand board in timber construction; metal deck or composite metal deck in steel construction; or a concrete slab in concrete construction.
The two primary types of diaphragm are flexible and rigid:
Flexible diaphragms resist lateral forces depending on the tributary area, irrespective of the flexibility of the members that they are transferring force to.
Rigid diaphragms transfer load to frames or shear walls depending on their flexibility and their location in the structure. The flexibility of a diaphragm affects the distribution of lateral forces to the vertical components of the lateral force resisting elements in a structure. [1]
Parts of a diaphragm include:
the membrane, used as a shear panel to carry in-plane shear
the drag strut member, used to transfer the load to the shear walls or frames
the chord, used to resist the tension and compression forces that develop in the diaphragm, since the membrane is usually incapable of handling these loads alone.

21. STRUCTURAL DIAPHRAGM Rigid Diaphragm: Almost no deflection
Can transmit loads through torsion
Flexible Diaphragm:
Deflects horizontally
Cannot transmit loads through torsion
a diaphragm is a structural system used to transfer lateral loads to shear walls or frames primarily through in-plane shear stress
The diaphragm of a structure often does double duty as the floor system or roof system in a building, or the deck of a bridge, which simultaneously supports gravity loads.
Diaphragms are usually constructed of plywood or oriented strand board in timber construction; metal deck or composite metal deck in steel construction; or a concrete slab in concrete construction.
The two primary types of diaphragm are flexible and rigid:
Flexible diaphragms resist lateral forces depending on the tributary area, irrespective of the flexibility of the members that they are transferring force to.
Rigid diaphragms transfer load to frames or shear walls depending on their flexibility and their location in the structure. The flexibility of a diaphragm affects the distribution of lateral forces to the vertical components of the lateral force resisting elements in a structure. [1]
Parts of a diaphragm include:
the membrane, used as a shear panel to carry in-plane shear
the drag strut member, used to transfer the load to the shear walls or frames
the chord, used to resist the tension and compression forces that develop in the diaphragm, since the membrane is usually incapable of handling these loads alone.

22. COLLECTORS and DRAGS

23. COLLECTORS and DRAG STRUTS
A beam element or line of reinforcement that carries or “collects” loads from a diaphragm and carries them axially to shear walls or frames.
A drag strut or collector behaves like a column.

24. diaphragm > collectors/drags > frames
Lateral Load Flow
diaphragm > collectors/drags > frames

25. diaphragm > collectors/drags > frames
LATERAL
LOAD
DIAPHRAGM
COLLECTOR
FRAME
Lateral Load Flow
diaphragm > collectors/drags > frames

26. diaphragm > collectors/drags > frames
LATERAL
LOAD
DIAPHRAGM
COLLECTOR
FRAME
Lateral Load Flow
diaphragm > collectors/drags > frames

27. LATERAL
LOAD
COLLECTOR
FRAME
COLLECTOR
FRAME
DIAPHRAGM
COLLECTOR
COLLECTOR
FRAME

28. LATERAL FORCE RESISTING SYSTEMS:
MOMENT Resisting frames
Diagonally BRACED frames
SHEAR walls

29. INSTABILITY OF THE FRAME
Pinned connections cannot resist rotation. This is not a structure but rather a mechanism.

30. STABILIZE THE FRAME
FIX ONE OR MORE OF THE BASES

31. STABILIZE THE FRAME
FIX ONE OR MORE OF THE CORNERS

32. STABILIZE THE FRAME
ADD A DIAGONAL BRACE

33. RELATIVE STIFFNESS OF FRAMES AND WALLS
LOW DEFLECTION
HIGH STIFFNESS
ATTRACTS MORE LOAD
HIGH DEFLECTION
LOW STIFFNESS
ATTRACTS LESS LOAD

34. BRACED FRAMES

35. BRACED FRAMES

36. SHEAR WALLS

37. SHEAR WALLS

38. SHEAR WALLS

39. SHEAR WALLS

40. SHEAR WALLS

41. MOMENT FRAMES

42. MOMENT FRAMES

43. MOMENT FRAMES
INDETERMINATE STRUCTURES
SOLVE BY “PORTAL FRAME METHOD”

44. MOMENT FRAMES
PINNED BASE =4 UNKNOWNS, 3 EQUATIONS, STATICALLY INDETERMINATE TO FIRST DEGREE
SOLVE BY “PORTAL FRAME METHOD”
The portal method
This method is satisfactory for buildings up to 25 stories, hence is the
most commonly used approximate method for analyzing tall buildings. The following are the simplifying assumptions made in the portal method:
1. A point of contraflexure occurs at the center of each beam.
2. A point of contraflexure occurs at the center of each column.
3. The total horizontal shear at each story is distributed between the columns of that story in such a way that each interior column carries twice the shear carried by each exterior column.
The above assumptions convert the indeterminate multi-storY frame to a determinate structure. The steps involved in the analysis of the frame are detailed below:
1. The horizontal shears on each level are distributed between the columns of that floor according to assumption (3).
2. The moment in each column is equal to the column shear multiplied by half the column height according to assumption (2).
3. The girder moments are determined by applying moment equilibrium equation to the joints: by noting that the sum of the girder moments at any joint equals the sum of the column moments at that joint. These calculations are easily made by starting at the upper left joint and working joint by joint across to the right end.
4. The shear in each girder is equal to its moment divided by half the girder length. This is according to assumption (1).
5. Finally, the column axial forces are determined by summing up the beam shears and other axial forces at each joint. These calculations again are easily made by working from left to right and from the top floor down.

45. MOMENT FRAMES
FIXED BASE =6 UNKNOWNS, 3 AVAILABLE EQUATIONS OF EQUILIBRIUM
STATICALLY INDETERMINATE TO THE 3RD DEGREE
SOLVE BY “PORTAL FRAME METHOD”
The portal method
This method is satisfactory for buildings up to 25 stories, hence is the
most commonly used approximate method for analyzing tall buildings. The following are the simplifying assumptions made in the portal method:
1. A point of contraflexure occurs at the centre of each beam.
2. A point of contraflexure occurs at the centre of each column.
3. The total horizontal shear at each storey is distributed between the columns of that storey in such a way that each interior column carries twice the shear carried by each exterior column.
The above assumptions convert the indeterminate multi-storey frame to a determinate structure. The steps involved in the analysis of the frame are detailed below:
1. The horizontal shears on each level are distributed between the columns of that floor according to assumption (3).
2. The moment in each column is equal to the column shear multiplied by half the column height according to assumption (2).
3. The girder moments are determined by applying moment equilibrium equation to the joints: by noting that the sum of the girder moments at any joint equals the sum of the column moments at that joint. These calculations are easily made by starting at the upper left joint and working joint by joint across to the right end.
4. The shear in each girder is equal to its moment divided by half the girder length. This is according to assumption (1).
5. Finally, the column axial forces are determined by summing up the beam shears and other axial forces at each joint. These calculations again are easily made by working from left to right and from the top floor down.