CIV-421 · Lesson 10 of 12 · 19 min

Masonry shear walls

Masonry walls are very stiff in their own plane, and in many buildings they are the whole lateral system. At the operations center, the wash bay is a concrete block addition on the east end of the steel-framed garage, and its two end walls, each 24 ft long and 20 ft high, will replace the steel X-bracing in the end bay of the garage that CIV-331 designed. The roof diaphragm of the garage delivers a strength-level wind shear of 34 kips to each wall. This lesson designs those walls for in-plane shear and flexure by TMS 402 strength design.

How shear walls behave

A masonry shear wall loaded in its plane is a deep cantilever. Depending on its aspect ratio, reinforcement and axial load, it fails in one of three main ways. A squat wall with little vertical load tends to fail in shear, with diagonal cracks running from corner to corner through the units and joints. A taller, slender wall tends to fail in flexure, with the vertical bars at the heel yielding and the toe crushing. A lightly loaded wall without enough dowels can slide on a bed joint or at its base. The flexural mode is ductile and is the mode designers want; the shear mode is brittle.

Elevation of the wash bay end wall, 7.2 m (24 ft) long and 6.0 m (20 ft) high, with the roof shear V at the top, the axial dead load P, vertical bars in grouted cores at spacing s with extra bars at each end, horizontal bond beams, and three failure modes sketched beside it: diagonal shear cracking, flexural yielding with toe crushing, and sliding at the base.
The wash bay shear wall and its possible failure modes.

Shear resistance

TMS 402 gives the nominal in-plane shear strength as the sum of a masonry contribution, including a share of the axial compression, and a steel contribution from horizontal reinforcement, multiplied by a grouting factor:

\[V_n = (V_{nm} + V_{ns})\gamma_g \qquad V_{nm} = \Big[4.0 - 1.75\Big(\frac{M_u}{V_ud_v}\Big)\Big]A_n\sqrt{f'_m} + 0.25P_u \qquad V_{ns} = 0.5\Big(\frac{A_v}{s}\Big)f_yd_v\]

Here \(d_v\) is the actual depth of the masonry in the direction of shear, the wall length; \(A_n\) the net area; \(P_u\) the axial compression, using 0.9 of the dead load; and \(\gamma_g\) a factor of 1.0 for fully grouted walls and 0.75 for partially grouted walls. The ratio \(M_u/(V_ud_v)\) need not exceed 1.0: squat walls get more masonry shear strength than slender ones because more of the shear goes directly to the base in a diagonal strut. The steel term counts half the yield force of the horizontal bars crossing a diagonal crack. An upper limit, between \(6A_n\sqrt{f'_m}\) for squat walls and \(4A_n\sqrt{f'_m}\) for slender ones, times \(\gamma_g\), protects against crushing of the diagonal strut. The strength reduction factor for shear is \(\phi = 0.80\).

Flexure in the plane of the wall

The overturning moment at the base of the wall is resisted by the vertical bars on the tension side and the masonry compression at the toe, with the axial load helping. With bars distributed along the length, an exact calculation uses strain compatibility, like a reinforced concrete column with many layers of steel. For walls with uniformly distributed bars and moderate axial load, a simple and widely used approximation is

\[M_r \approx \frac{1}{2}A_sf_yl_w\Big(1 + \frac{P}{A_sf_y}\Big)\Big(1 - \frac{c}{l_w}\Big) \qquad \frac{c}{l_w} = \frac{\omega + \alpha}{2\omega + 0.85\beta_1}\]

where \(A_s\) is the total vertical steel, \(\omega = A_sf_y/(l_wtf'_m)\) and \(\alpha = P/(l_wtf'_m)\), with resistance factors applied to the materials as the standard requires. It assumes that all the bars in tension yield. A designer who wants more precision runs a strain compatibility calculation, which Sofia's script could do with a few more lines.

The wash bay end wall

The wall is 8 in. CMU, partially grouted, with #5 vertical bars at 32 in. and two extra #5 bars in the end cores at each end, and #5 horizontal bars in bond beams every 48 in. (six courses). \(f'_m = 2000\) psi and \(f_y = 60\) ksi. The roof and the wall itself give a dead load on the wall of 250 plf from the roof and 54 psf of wall area.

In-plane shear strength of the wash bay end wall and its flexural strength by the approximate equation.
# Lesson 10: in-plane shear wall, wash-bay end wall, 8 in. CMU, lw 24 ft, h 20 ft (TMS 402 strength design)
import math
lw, h, t = 288.0, 240.0, 7.625               # in.
fm, fy = 2000.0, 60000.0
Vu = 34.0                                    # kips, strength-level wind at roof level
Mu = Vu * h                                  # kip-in. at the base
Pu = 0.9 * (0.25 * 24 + 0.054 * 20 * 24)     # 0.9D on the wall (kips)
An = 43.8 * 24                               # net area, cells grouted at 32 in. (in2)
dv = lw
r = min(Mu / (Vu * dv), 1.0)
Vnm = ((4.0 - 1.75 * r) * An * math.sqrt(fm) + 0.25 * Pu * 1000) / 1000
Av, s = 0.31, 48.0                           # #5 in a bond beam every 48 in.
Vns = 0.5 * (Av / s) * fy * dv / 1000
gamma_g = 0.75                               # partially grouted wall
lim = 6.0 - 2.0 * (max(r, 0.25) - 0.25) / 0.75 if r < 1.0 else 4.0
Vn_max = lim * An * math.sqrt(fm) * gamma_g / 1000
Vn = min((Vnm + Vns) * gamma_g, Vn_max)
print(f"Mu/(Vu dv) = {r:.3f}, Pu = {Pu:.1f} kips, An = {An:.0f} in2")
print(f"Vnm = {Vnm:.1f} kips, Vns = {Vns:.1f} kips, upper limit {Vn_max:.1f} kips")
print(f"phi Vn = 0.8 x {Vn:.1f} = {0.8*Vn:.1f} kips vs Vu = {Vu:.0f} kips, ratio {Vu/(0.8*Vn):.2f}")

# In-plane flexure: #5 @ 32 in. plus 2 #5 at each end (approximate design equation)
As_total = 0.31 * (lw / 32 - 1) + 4 * 0.31
w_ = As_total * fy / (lw * t * fm)
alpha = Pu * 1000 / (lw * t * fm)
c_lw = (w_ + alpha) / (2 * w_ + 0.80 * 0.80)
Mn = 0.5 * As_total * fy * lw * (1 + Pu * 1000 / (As_total * fy)) * (1 - c_lw) / 1000
print(f"As = {As_total:.2f} in2, c/lw = {c_lw:.3f}, phi Mn = {0.9*Mn/12:.0f} kip-ft vs Mu = {Mu/12:.0f} kip-ft, ratio {Mu/(0.9*Mn):.2f}")
Output
Mu/(Vu dv) = 0.833, Pu = 28.7 kips, An = 1051 in2
Vnm = 126.7 kips, Vns = 55.8 kips, upper limit 156.7 kips
phi Vn = 0.8 x 136.9 = 109.5 kips vs Vu = 34 kips, ratio 0.31
As = 3.72 in2, c/lw = 0.077, phi Mn = 2510 kip-ft vs Mu = 680 kip-ft, ratio 0.27

Worked problem

Shear design of the wash bay wall

Check the in-plane shear of the 24 ft by 20 ft wall for \(V_u = 34\) kips at the top.

Show solution
  1. \[M_u = 34(20) = 680\ \text{kip·ft} \qquad d_v = 288\ \text{in.} \qquad \frac{M_u}{V_ud_v} = 0.833\]
  2. \[P_u = 0.9[0.25(24) + 0.054(20)(24)] = 28.7\ \text{kips} \qquad A_n = 43.8(24) = 1051\ \text{in}^2\]
  3. \[V_{nm} = [4.0 - 1.75(0.833)](1051)\sqrt{2000} + 0.25(28{,}700) = 126.7\ \text{kips}\]
  4. \[V_{ns} = 0.5\Big(\frac{0.31}{48}\Big)(60{,}000)(288) = 55.8\ \text{kips}\]
  5. \[V_n = (126.7 + 55.8)(0.75) = 136.9\ \text{kips} \le 156.7\ \text{kips} \qquad \phi V_n = 0.80(136.9) = 109.5\ \text{kips}\]

Answer: The wall has ample shear strength, at a ratio of 0.31, and in-plane flexure is at 0.27. With a strength-level wind shear this low compared with the capacity, minimum reinforcement and detailing, not strength, set the design.

Minimum reinforcement and detailing

Because the wind shear is small, the reinforcement in the wash bay walls is set mostly by minimum requirements and good practice. Vertical bars go at every corner, at each side of every opening, at the ends of each wall and at regular spacing between, so that every part of the wall can span to a bar. Horizontal reinforcement goes in a bond beam at the roof, at each floor or lintel level and at regular spacing, or as joint reinforcement in the bed joints. TMS 402 attaches prescriptive minimums to each shear wall type, such as bars of at least 0.2 in² at corners, openings and wall ends and maximum spacings of 120 in. for ordinary reinforced walls and less for higher categories; even in low-hazard regions, a minimum amount of reinforcement controls cracking and gives the wall some ability to redistribute load if it is damaged.

The same thinking applies to the pump house. Its walls are short and stiff in plane, and the wind on the building is small, so their in-plane shear capacity is many times the demand. The team checks it quickly, confirms that the bars already chosen for out-of-plane bending satisfy the minimums, and adds two bars at each door jamb and corner. Most of the design effort in a small masonry building goes into openings, lintels, corners, bearings and the connection to the roof, not into the shear calculation.

Sliding and the base connection

The shear must pass from the wall into the foundation. Friction under the dead load carries some of it, but for a lightly loaded wall the dowels from the footing into the grouted cores do most of the work in shear friction. The dowels lap with the vertical bars of the wall, so they also anchor the tension side for overturning. At the top, the roof diaphragm delivers the shear through anchor bolts or welded plates into the bond beam, and the bond beam acts as a collector along the wall. The wash bay walls share the steel garage's end bay, so the connection between the steel roof and the masonry wall must allow the steel frame to move with temperature while transferring shear, which slotted connections in the direction of the wall achieve.

Stiffness and load sharing

When several walls resist lateral load together through a rigid diaphragm, they share the load in proportion to their stiffness. A masonry wall's stiffness depends on both flexure and shear deformation and on whether it is cracked; squat walls are much stiffer than slender piers, and a wall with a large opening behaves as two piers with a coupling beam. Long walls attract most of the load. Where a stiff masonry wall and a flexible steel frame share the load, as in the garage, the masonry takes almost all of it until it cracks, which is why the design gives the wash bay walls the full end-bay shear and does not count on the steel frame at all.

For a cantilever wall of height \(h\), the lateral stiffness combines bending and shear deformation:

\[k = \frac{1}{\dfrac{h^3}{3E_mI} + \dfrac{1.2h}{G_mA}} \qquad G_m \approx 0.4E_m\]

For the wash bay wall, whose height is less than its length, the shear term is about as large as the bending term, and the uncracked wall is roughly ten times stiffer than the X-braced steel bay it replaces. Cracking reduces the stiffness of masonry walls considerably, so designers who need to distribute load between walls of different types often use cracked or effective stiffness values, typically a half or less of the uncracked value.

Earthquake design and ductility

Duluth and Thunder Bay have low seismic hazard, and wind governs. In regions of higher hazard, TMS 402 and ASCE 7 classify masonry shear walls as ordinary plain, detailed plain, ordinary reinforced, intermediate reinforced and special reinforced, with response modification coefficients from 1.5 to 5.0 and increasing requirements for prescriptive reinforcement, bar spacing, grouting and capacity design. Special reinforced shear walls must be designed so that their shear strength exceeds the shear corresponding to their flexural strength, so that the ductile flexural mode governs. Plain (unreinforced) masonry is not permitted as the lateral system in higher seismic design categories.

Unreinforced masonry buildings are among the most vulnerable structures in earthquakes, and many older ones in North American cities have been strengthened with steel frames, reinforced cores or fiber-reinforced overlays. The lesson for new design is that reinforcement, grouting and good connections to floors and roofs make masonry a reliable lateral system; their absence makes it a hazard.

Common mistakes

  • MistakeUsing the gross wall thickness and full grouting factor for a partially grouted wall.

    InsteadUse the net area and \(\gamma_g = 0.75\) for partially grouted walls.

  • MistakeUsing the full dead load to help shear and overturning resistance.

    InsteadUse 0.9D (or 0.6D in some combinations) so that dead load is not overestimated.

  • MistakeDesigning the wall but not the base and top connections.

    InsteadDowels, anchor bolts and the bond beam collector must carry the full shear and overturning.

  • MistakeAllowing a brittle shear failure before flexural yielding in seismic design.

    InsteadProvide shear strength greater than the shear associated with flexural strength in ductile walls.

Summary and key terms

  • In-plane shear walls fail in shear, flexure or sliding; flexural yielding is the ductile mode.
  • \(V_n = (V_{nm} + V_{ns})\gamma_g\), with \(V_{nm} = [4.0 - 1.75(M_u/V_ud_v)]A_n\sqrt{f'_m} + 0.25P_u\), \(V_{ns} = 0.5(A_v/s)f_yd_v\), \(\gamma_g = 0.75\) for partial grouting and \(\phi = 0.80\).
  • In-plane flexure with distributed bars can be estimated with \(M \approx \frac12A_sf_yl_w(1 + P/A_sf_y)(1 - c/l_w)\).
  • The wash bay wall: \(\phi V_n = 109.5\) kips against 34 kips (0.31) and in-plane flexure at 0.27.
  • Seismic detailing and capacity design make masonry shear walls ductile.
In-plane shear
Shear acting parallel to the face of a wall.
Shear friction
Transfer of shear across a crack or joint by friction and dowel clamping.
Grouting factor
A factor that reduces the shear strength of partially grouted walls.
Diagonal tension cracking
Cracking across a wall at an angle, caused by in-plane shear.
Toe crushing
Crushing of the masonry at the compression end of a wall in flexure.
Capacity design
Design that ensures a ductile mode forms before brittle modes.
Pier
A narrow segment of wall between openings.
Collector
An element that gathers diaphragm shear and delivers it to a wall.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. Why do squat walls have higher masonry shear strength per unit area than slender walls?

    Show answer

    More of the shear goes directly to the base in a diagonal compression strut.

  2. If the wall were fully grouted, what would the strength become?

    Show answer

    With \(\gamma_g = 1.0\) and \(A_n = 91.5(24) = 2196\) in², \(V_{nm}\) rises to about 257 kips and the upper limit to about 436 kips; \(\phi V_n\) would be about 250 kips.

  3. Why is 0.9D used rather than the full dead load in the shear and overturning checks?

    Show answer

    Dead load helps resist shear and overturning, so a lower value is conservative.

  4. A wall has a large door opening near one end. How does that change its behavior?

    Show answer

    It becomes a pier and a short wall segment linked by a lintel, with much lower stiffness and capacity than the solid wall.

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