cee 626 masonry design slide set 6 reinforced masonry spring 2015

CEE 626 MASONRY DESIGN

Slide Set 6 Reinforced Masonry

Spring 2015

Learning Objectives

Students will be able to determine the capacity of Shear walls using Allowable Stress (ASD) design for Flexure only, then with axial forces and flexure combined.

Students will be able to determine the capacity of Shear walls using Strength Design for Flexure only, then with axial forces and flexure combined.

Shear WallsDefinition: vertical elements that resist lateral forces

parallel with the plane of the wall may or may not be bearing walls may be exterior or interior walls resist story shears in proportion to

lateral stiffness ( ~ plan length ) for rigid diaphragms

tributary area for flexible diaphragms

MDG TMS Shopping Center

TMS Shopping Center PlanC

B

40’-6” rib metal deck

W14 W14 W14 W16W16

control joint for Option A

A

1 2 3102’-4” 102’-4”

41’-0” K6 steel joists

N

Shear Wall

Wind or Seismic loads

Shear Wall

Reinforced Masonry Shear WallsIN-PLANE LOADING

h

V

L

P

V= base shear

M = over turning moment

May or may not have vertical loads

Design of Reinforced Masonry (ASD&SD)

In Plane Loading (shear Walls) Still use interaction diagrams Axial Load is still dealt with out of

plane slenderness (M=0) In plane horizontal load produces

moment and thus moment capacity is dealt with slightly differently.

LETS LOOK FIRST AT MOMENT CAPACITY ONLY

Reinforced Masonry (ASD) In Plane Loading (shear Walls)

Flexure Only P = 0 on diagram

h

V

L

P- self weight only ignore

V= base shear


Multiple rebar locations

Reinforced Masonry (ASD) In Plane Loading P = 0

L


V= base shear


fm

k*d*

Fsc/NTi

Fs1/Nfsi/N

Fsn/N<= Fs/N

di – location to centroid of each bar

CmTsn= Fs As Ti Ti Ti Ti Ti Ti Ti T1

Tension Compression

fsi/N fsi/N fsi/N fsi/N fsi/N fsi/N fsi/N fsi/N

M only!!!d*

As*

Moment only P= 0 ASD In Plane

To locate Neutral Axis – Guess how many bars on tension side – As*

Find d* (centroid of tension bars) and r*=As*/bd*

Get k* = ((r*n)2 + 2r*n)1/2 – nr* Unless tied ignore compression steel Check k* d* to ensure assume tension bars

correct – iterate if not Determine fsi from similar triangles and Ti M capacity (Σabout C)= Σ (Ti x (di – k*d*/3))

Design of Reinforced Masonry (ASD) In Plane Loading (shear Walls)

MOMENT ONLY EXAMPLE

Assume a Hollow Clay unit 7.5 in thick

Axial Load is only self wt – ignore – moment only

In plane load produces moment and thus moment capacity is dealt with slightly differently

Assume f’m = 2500 psi and #5 rebar grade 60 similar to Example 9.4-5 MDG

but axial forces only

Flexure Only P = 0

H= 16’

V

L


V= base shear


#5 rebar @2’ OC 13 bars – 8” from ends

L = 25’-4”

Reinforced Masonry (ASD) In Plane Loading P = 0

L V= base shear

M(cl) = over turning moment

fm

K*d*


CmTsn= Fy As Ti

Tension Compression

32

fsi /n

fsn/n<= Fs/n

````fsi /nfsi /n

fsi /n

fsi /nfsi /nfsi /nfsi /n fsi /n fsi /n

Ti Ti TiTi Ti Ti Ti Ti

fsi /n

Ti

Allowable Stress Design Examplen = 29/1.750 =16.571

Inpane Moment example Mabout Cm

As (in2) di (ft) k*d* (ft) Ti (kips (kip.ft)1 0.31 24.67 3.71 9.92 232.462 0.31 22.67 3.71 8.97 192.333 0.31 20.67 3.71 8.03 155.994 0.31 18.67 3.71 7.08 123.435 0.31 16.67 3.71 6.13 94.666 0.31 14.67 3.71 5.19 69.687 0.31 12.67 3.71 4.24 48.488 0.31 10.67 3.71 3.29 31.079 0.31 8.67 3.71 2.35 17.45

10 0.31 6.67 3.71 1.40 7.6111 0.31 4.67 3.71 0.45 1.56

Sum= 57.06Sum M = 974.74 Kip.ft

Check compression Stresses and total Force

FOR COMBINED LOADING ON SHEAR WALLS (ASD)

Net vertical Force is not zero

That is SF 0 Must assume a stress distribution then

back calculate P and M

Design of Reinforced Masonry (ASD&SD)

In Plane Loading (shear Walls) Still use interaction

diagrams Axial Load is still dealt with

out of plane slenderness (M=0)

(b= L and r is based on OOP) In plane load produces

moment and thus moment capacity is dealt with slightly differently.

Reinforced Masonry (ASD) In Plane Loading P ≠ 0

L

P- is to be found

V= base shear


fm

Kd=ad


CmTsn= Fy As Ti

Ti Ti Ti

Tension Compression

Ti

3

fsi/Nfsi/N fsi/N fsi/Nfsi/N

fsn/N<= Fs/N

P-M ASD - In Plane

Unless tied ignore compression steel To locate Neutral Axis (c) – guess at ad Cm = ½ (ad x b x Fb ) Get fsi for each bar (use similar triangles) also check

to see is fsi max is ≤ Fs, then Ti = As x fsi P = Cm -S Ti (note stress in Steel in tension is

negative) M capacity (Σabout center of wall length)= Cm x(L/2 –

ad/3) + (Σ (Ti x (L/2- di )) Note this assumes steel in tension is negative)

Note P (cutoff for M = 0) is out of plane with whole length of wall as beff.

Reinforced Masonry (ASD) In Plane Loading P ≠ 0

L

P- is to be found

V= base shear


fm

Kd=ad


CmTsn= Fy As Ti

Ti Ti Ti

Tension Compression

Ti

3

fsi/Nfsi/N fsi/N fsi/Nfsi/N

fsn/N<= Fs/N

MUST also check Shear- fv=V/Anv≤Fv

MUST ALSO CHECK SHEAR(SEE SLIDE SET 5)

ASDMasonry subjected to net flexural tension (reinforced); fv = V/Anv (this is an average stress) fvmax ≤ Fv

Fv = (Fvm +Fvs)g

P-M ASD In PlaneExample by ASD (will do this same example by Strength Design later) – See MDG

NOW for Strength Design

Reinforced Masonry Shear Walls - SD

h

Vu

L

Pu

Vu= base shear


Design of Reinforced Masonry (SD) In Plane Loading (shear Walls)

Still use interaction diagrams

Axial Load is still dealt with as out of plane (M=0)

Pu applied to ≤ Pn - Eq 9-19 and Eq 9-20


99for 70))(80(.8.0

99 for 140

1))(80(.8.0

2'

2'

=

=

rh

hrfAAAfP

rh

rhfAAAfP

ystsnmn

ystsnmn

Also have Maximum Reinforcement Limits

STRENGTH DESIGN ONLYa = 3 for intermediate shear wallsa = 4 for special shear wallsFor walls designed with Seismic R ≤ 1.5 & M/Vd≤1There is no limit on reinforcing all other a = 1.5

Reinforced Masonry (SD) In Plane Loading P at LC for maximum Reinforcing limits – Stress and Strains

L

P- due to D + .75 L + .525Qe

V= base shear


em

c

esc

Ti

esi

aesmax


CmTsn= Fy As Ti Ti Ti Ti Ti Ti Ti T1

Tension Compression

bcesi esi esi esi esi esi esi

es1em=.0025 or .0035

SD For In Plane Max Reinf.

Get c from similar triangles Get Cm = 0.8c x 0.8 f’m x b Assume all bars are in tension have yielded

(Fy) then Ti = As Fy Is Cm – P + Ascomp Fy (in compression) ≥ S Ti = S Asten

Fy?

If not you have too much reinforcing - reduce number or size of bars or increase f’m, or size of units.

P-M Diagrams SD In Plane Check maximum reinforcing is not exceeded. Initially assume c (Neutral Axis) is some small value get

strains in steel – check they are greater than ey , if not get fs and then force in the bars not exceeding ey

All other bars are at Fy then Ti = As Fy Cm = 0.8c x 0.8 f’m x b where “b” is the wall thickness

(for solid) Pn = Cm -S Ti – Note this must note exceed 9-19 or 9-20

reduced using out-of-plane R factor Mn capacity (Σabout wall center)= Σ (Ti x (L/2 -di) + Cm

x(L/2 – a/2)) -be careful some bars are producing negative moments about Center line (CL) should take care of itself if tension in bars is negative. Note a=b1c

P cut off is for out of plane (and beff is length of wall)

Reinforced Masonry (SD) In Plane Loading P ≠ 0

LPn

V= base shear

Mn = S about CLPn = S of forces

em

c

esc

Ti

esi

esmax



Tension Compression


es1

em=.0025 or .0035

b =0.8

MUST ALSO CHECK SHEAR(SEE SLIDE SET 5)

SDReinforced masonry : Vn ≥ Vumax & Vn = (Vm +Vs)g

Design of Reinforced Masonry (SD) In Plane Loading (shear Walls)

MOMENT ONLY EXAMPLE

Assume a Hollow Clay unit 7.5 in thick

Axial Load is only self wt – ignore – moment only


Assume f’m = 2500 psi and #5 rebar grade 60 similar to Example 9.4-5 MDG

but no axial forces

Flexure Only P = 0

H= 16’

V

L


V= base shear


#5 rebar @2’ OC 13 bars – 8” from ends

L = 25’-4”

Reinforced Masonry (SD) In Plane Loading (shear Walls)

Flexure Only P = 0 on diagram

h

Vu

L


Vu= base shear


Multiple rebar locations

Moment only SD In Plane

Unless tied ignore compression steel To locate Neutral Axis (c) – use Equilibrium Cm = 0.8c x 0.8 f’m x b = S Ti (assume all

steel yields then check using similar triangles es>ey)

Get Cm = 0.8c x 0.8 f’m x b M capacity (Σabout C)= Σ (Ti x (di – 0.8c/2)) Check maximum reinforcing not exceeded

SD Example In Plane M only on BoardLets check maximum reinforcing is not exceeded first no - point in going on if this not so.

Get c from similar triangles

Get Cm = 0.8c x 0.8 f’m x b

Assumed all bars are in tension have yielded (Fy) then Ti = As Fy

Is Cm – P + Ascomp Fy ≥ S Ti = S Asten Fy?

L V= base shear

em

c

esc

Ti

esi

esmax



Tension Compression


es1

em=.0025 or .0035

b =0.8

TiTi

esiesi

13 bars #5 bars

To locate Neutral Axis (c) – get Cm , Mn

Strength Design ExampleInpane Moment example

Mabout CmAs (in2) di (ft) a (ft) Ti (kips) (kip.ft)

0.31 24.67 1.24 18.60 447.330.31 22.67 1.24 18.60 410.130.31 20.67 1.24 18.60 372.930.31 18.67 1.24 18.60 335.730.31 16.67 1.24 18.60 298.530.31 14.67 1.24 18.60 261.330.31 12.67 1.24 18.60 224.130.31 10.67 1.24 18.60 186.930.31 8.67 1.24 18.60 149.730.31 6.67 1.24 18.60 112.530.31 4.67 1.24 18.60 75.330.31 2.67 1.24 18.60 38.13

Sum= 223.20Sum M = 2912.76 Kip.ftPhi Mn 2621.484

cee 626 masonry design slide set 6 reinforced masonry spring 2015

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