1. CEE 626 MASONRY DESIGN Slide Set 6 Reinforced Masonry Spring 2015

2. 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.

3. Shear Walls Definition: 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

4. MDG TMS Shopping Center TMS Shopping Center Plan C B 40’-6” rib metal deck W14 W16 control joint for Option A A 1 2 3 102’-4” 41’-0” K6 steel joists N Shear Wall Wind or Seismic loads Shear Wall

5. MDG TMS Shopping Center TMS Shopping Center Plan C B 40’-6” rib metal deck W14 W16 control joint for Option A A 1 2 3 102’-4” 41’-0” K6 steel joists N Shear Wall Wind or Seismic loads Shear Wall

6. Reinforced Masonry Shear Walls IN-PLANE LOADING h V L P V= base shear M = over turning moment May or may not have vertical loads

7. 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

8. 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 M = over turning moment Multiple rebar locations

9. Reinforced Masonry (ASD) In Plane Loading P = 0 L P- self weight only ignore V= base shear M = over turning moment fm k*d* F sc /N TiTiTiTi F s1 /N f si /N F sn /N<= Fs/N di – location to centroid of each bar Cm T sn = Fs As TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi T1T1T1T1 Tension Compression f si /N M only!!! d* As*

10. 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  *=A s */bd* Get k* = ((  *n) 2 + 2  *n) 1/2 – n  * Unless tied ignore compression steel Check k* d* to ensure assume tension bars correct – iterate if not Determine f si from similar triangles and T i M capacity (Σabout C)= Σ (T i x (d i – k*d*/3))

11. 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

12. Flexure Only P = 0 H= 16’ V L P- self weight only ignore V= base shear M = over turning moment #5 rebar @2’ OC 13 bars – 8” from ends L = 25’-4”

13. Reinforced Masonry (ASD) In Plane Loading P = 0 L V= base shear M(cl) = over turning moment fmfm K*d* di – location to centroid of each bar Cm T sn = Fy As TiTi Tension Compression 32 f si /n f sn /n<= Fs/n `` `` f si /n TiTi TiTi TiTi TiTi TiTi TiTi TiTi TiTi TiTi

15. 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) 10.3124.673.719.92232.46 20.3122.673.718.97192.33 30.3120.673.718.03155.99 40.3118.673.717.08123.43 50.3116.673.716.1394.66 60.3114.673.715.1969.68 70.3112.673.714.2448.48 80.3110.673.713.2931.07 90.318.673.712.3517.45 100.316.673.711.407.61 110.314.673.710.451.56 Sum=57.06 Sum M =974.74Kip.ft Check compression Stresses and total Force

16. FOR COMBINED LOADING ON SHEAR WALLS (ASD) Net vertical Force is not zero That is  F  0 That is  F  0 Must assume a stress distribution then back calculate P and M Must assume a stress distribution then back calculate P and M

17. 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.

18. Reinforced Masonry (ASD) In Plane Loading P ≠ 0 L P- is to be found V= base shear M(cl) = over turning moment fmfmfmfm Kd=  d di – location to centroid of each bar Cm T sn = Fy As TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi Tension Compression TiTiTiTi 3 f si /N f sn /N<= Fs/N

19. P-M ASD - In Plane Unless tied ignore compression steel To locate Neutral Axis (c) – guess at  d Cm = ½ (  d x b x Fb ) Get f si for each bar (use similar triangles) also check to see is f si max is ≤ Fs, then T i = As x f si P  Cm -  Ti (note stress in Steel in tension is negative) M capacity (Σabout center of wall length)= Cm x(L/2 –  d/3) + (Σ (T i x (L/2- d i )) 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.

20. Reinforced Masonry (ASD) In Plane Loading P ≠ 0 L P- is to be found V= base shear M(cl) = over turning moment fmfmfmfm Kd=  d di – location to centroid of each bar Cm T sn = Fy As TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi Tension Compression TiTiTiTi 3 f si /N f sn /N<= Fs/N

21. MUST also check Shear- f v =V/A nv ≤F v

22. MUST ALSO CHECK SHEAR (SEE SLIDE SET 5) ASD Masonry subjected to net flexural tension (reinforced); f v = V/A nv (this is an average stress) f vmax ≤ F v F v = (F vm +F vs )  g

23. P-M ASD In Plane Example by ASD (will do this same example by Strength Design later) – See MDG

24. NOW for Strength Design

25. Reinforced Masonry Shear Walls - SD h VuVuVuVu L PuPuPuPu V u = base shear M = over turning moment

26. 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) P u applied to ≤  P n - Eq 9-19 and Eq 9-20 In plane load produces moment and thus moment capacity is dealt with slightly differently

27. Also have Maximum Reinforcement Limits STRENGTH DESIGN ONLY  = 3 for intermediate shear walls shear walls  = 4 for special shear walls For walls designed with Seismic R ≤ 1.5 & M/Vd≤1 There is no limit on reinforcing all other  = 1.5

28. 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 M = over turning moment mmmm c  sc TiTiTiTi  si  s max di – location to centroid of each bar Cm T sn = Fy As TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi T1T1T1T1 Tension Compression cccc  si  s1  m =.0025 or.0035

29. SD For In Plane Max Reinf. Get c from similar triangles Get C m = 0.8c x 0.8 f’ m x b Assume all bars are in tension have yielded (F y ) then T i = A s F y Is C m – P + A scomp F y (in compression) ≥  T i =  A sten F y ? If not you have too much reinforcing - reduce number or size of bars or increase f’ m, or size of units.

30. 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  y, if not get f s and then force in the bars not exceeding  y All other bars are at F y then T i = A s F y C m = 0.8c x 0.8 f’ m x b where “b” is the wall thickness (for solid) P n = C m -  Ti – Note this must note exceed 9-19 or 9-20 reduced using out-of-plane R factor M n capacity (Σabout wall center)= Σ (T i x (L/2 -d i ) + C m 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=   c P cut off is for out of plane (and b eff is length of wall)

31. Reinforced Masonry (SD) In Plane Loading P ≠ 0 L PnPnPnPn V= base shear Mn =  about CL P n =  of forces mmmm c  sc TiTiTiTi  si  s max di – location to centroid of each bar Cm T sn = Fy As TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi T1T1T1T1 Tension Compression cccc  si  s1  m =.0025 or.0035  =0.8

32. MUST ALSO CHECK SHEAR (SEE SLIDE SET 5) SD Reinforced masonry :  V n ≥ V umax & V n = (V m +V s )  g

33. 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 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 no axial forces

34. Flexure Only P = 0 H= 16’ V L P- self weight only ignore V= base shear M = over turning moment #5 rebar @2’ OC 13 bars – 8” from ends L = 25’-4”

35. Reinforced Masonry (SD) In Plane Loading (shear Walls) Flexure Only P = 0 on diagram h VuVuVuVu L P- self weight only ignore V u = base shear M = over turning moment Multiple rebar locations

36. Moment only SD In Plane Unless tied ignore compression steel To locate Neutral Axis (c) – use Equilibrium C m = 0.8c x 0.8 f’ m x b =  T i (assume all steel yields then check using similar triangles  s >  y ) Get C m = 0.8c x 0.8 f’ m x b M capacity (Σabout C)= Σ (T i x (d i – 0.8c/2)) Check maximum reinforcing not exceeded

37. SD Example In Plane M only on Board Lets check maximum reinforcing is not exceeded first no - point in going on if this not so. Get c from similar triangles

38. Get C m = 0.8c x 0.8 f’ m x b Assumed all bars are in tension have yielded (F y ) then T i = A s F y Is C m – P + A scomp F y ≥  T i =  A sten F y ?

39. L V= base shear mmmm c  sc TiTiTiTi  si  s max di – location to centroid of each bar Cm T sn = Fy As TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi TiTiTiTi T1T1T1T1 Tension Compression cccc  si  s1  m =.0025 or.0035  =0.8 TiTiTiTi TiTiTiTi  si  bars  5 bars To locate Neutral Axis (c) – get C m, M n

41. Strength Design Example Inpane Moment example Mabout Cm As (in2)di (ft)a (ft)Ti (kips)(kip.ft) 0.3124.671.2418.60447.33 0.3122.671.2418.60410.13 0.3120.671.2418.60372.93 0.3118.671.2418.60335.73 0.3116.671.2418.60298.53 0.3114.671.2418.60261.33 0.3112.671.2418.60224.13 0.3110.671.2418.60186.93 0.318.671.2418.60149.73 0.316.671.2418.60112.53 0.314.671.2418.6075.33 0.312.671.2418.6038.13 Sum=223.20 Sum M =2912.76Kip.ft Phi Mn2621.484