(a) Partial safety factors for materials are recommended in BS ; 5628 : Part 2 which depend on the control of manufacturing of the units. The code assumes a high quality of control of construction will be followed. Discuss the method and implication of controls in practice when building reinforced hollow blockwork.[6 marks](b) The durability of reinforced masonry structures is dependant on a number of factors. Discuss the effects of the following and how to maximise durability where appropriate :-

i. Masonry units and mortar.

ii. Concrete infill.

iii. Cover to reinforcement.

iv. Exposure conditions.[9 marks](c) A cantilevered reinforced concrete blockwork masonry wall, 2600mm high is required to store grain which exerts a uniform nominal (characteristic) horizontal load of 6.0kN/m2 against the wall. The wall is to be constructed on a reinforced concrete base of thickness 300mm. Assuming normal manufacturing control of units, that the wall is constructed and functions in e3xposure conditions E1 and neglecting the self weight of the wall, design a suitable construction.[10 marks]Design information.

Mortar designation (ii)

fy = 460N/mm2.

Strength of concrete infill = 30N/mm2Block thickness (width of wall) = 215mm.

Block shell thickness = 35mm.

Percentage of block which is solid = 60%.SOLUTION.(a) Partial safety factors for materials are recommended in BS ; 5628 : Part 2 which depend on the control of manufacturing of the units. The code assumes a high quality of control of construction will be followed. Discuss the method and implication of controls in practice when building reinforced hollow blockwork.Methods : Correct plumbness and alignment

Protection available in cold weather

Lifts not than 1.5m per day

Infill concrete- Clause 10.1.2.5 BS 5628 : Part 2

- workability

- compaction / aggregate size relationship

- cover to reinforcement

- stainless steel reinforcement

Reinforcement- spacers / laps

- cleanliness

Masonry- good bed joints / clean cores

- Lifts generally not 900mm.

(50mm clearance to top of block, clean outside of blocks, compact grout after pouring etc).

Implications :Cost {Contractor will try to reduce, client wants quality}

Balance requiredInsurance costs may be reduced with quality schemes.

Safety and Quality need to be balanced.

(b) The durability of reinforced masonry structures is dependant on a number of factors. Discuss the effects of the following and how to maximise durability where appropriate :-

i. Masonry units and mortar.Greater durability results from high strength units with low water absorption formed into masonry using a strong mortar. Problems arise if the water absorption exceeds about 10% and / or the concrete density is below 1500kg/m3. Durability can be maximised through correct specification of materials.ii. Concrete infill.Greater durability is achieved with a low permeability product. A strong well compacted concrete achieves this. Durability can be maximised by good site control and well specified materials.

iii. Cover to reinforcement.Greater durability can be obtained with greater cover. The use of spacers and good detailing together with good quality control will improve the durability.iv. Exposure conditions.

Harsher weather conditions will reduce the durability of reinforced masonry. Sensible and practical designs will achieve the design life.

(c)A cantilevered reinforced concrete blockwork masonry wall, 2600mm high is required to store grain which exerts a uniform nominal (characteristic) horizontal load of 6.0kN/m2 against the wall. The wall is to be constructed on a reinforced concrete base of thickness 300mm. Assuming normal manufacturing control of units, that the wall is constructed and functions in exposure conditions E1 and neglecting the self weight of the wall, design a suitable construction. Assume the ratio of net/gross area = 0.6Effective depth d = 215 35 20 10 = 150mm

Effective span :

Least of (i). Wall ht + support depth/2 = 2600 + 150 = 2750mm

(ii). Wall ht + effective depth/2 = 2600 + 75 = 2675mm

Check on limiting dimensions :

Providing = As/bd not > 0.005, then

Span/effective depth = 2675/150 = 17.83 not > 18 so OK.

Design moment.

MD = (6 x 1.4 x 2.6752)/2 = 30.05kNm/m width

For the masonry, MD not > (0.4 x fk x bd2)/muTherefore fk = (30.05 x 2.3 x 106)/(0.4 x 1000 x 1502) = 7.68N/mm2For steel MD = (As x fy x Z)/msSo z = [MDms]/[Asfy]

And z = d[(1 (0.5Asfymu)/ (bdfkms)]Combining the above two equations gives an equation in As.

[30.05 x 106 x 1.15]/[As/460]

= 150[1-[0.5As x 460 x 2.3]/[103 x 150 x 7.68 x 1.15]

Solving gives As = 693mm2Use 1 No T16 in each hole. This provides 893mm2 per metre length of wall.

Min area of secondary reinforcement = 0.0005bd = 0.0005 x 1000 x 150

= 75mm2 per m heightProvide 2 No. R5 bars in alternative courses = 87mm2

Units strength

For fk = 7.68N/mm2,

Compressive strength of units = 15N/mm2 (net)

Compressive strength units = 9.0N/mm2 (gross assuming gross/net ratio = 60%)Use blocks of 7.0N/mm2Shear. Design shear strength = 6 x 1.4 x 2.675 = 22.5kN/m runShear strength = fv 0.35 + 17.5= 0.35 + 17.5 x As/bd

= 0.35 + 17.5 x 893/(1000 x 150)

= 0.35 + 0.104

= 0.454MPa not > 0.7 so OKCheck on enhancement.

Shear span a = Max design moment/Maximum design shear force = 30.05/22.5 = 1.34 not > 6

No enhancementShear stress = (22.5 x 1000)/(1000 x 150) = 0.15 MPa

fy/mv = 0.454/2.3 = 0.20 > Applied shear = 0.15MPa so OK2.

(a) En external brickwork boundary wall is 3.0m in height from the top of the foundation.. The horizontal span is 7.5m. If the characteristic wind load on the panel is 0.8kN/m2, design the wall.

i. With an allowance for a modified due to self weight.

ii. With no allowance for self weight.

Normal construction control and unit manufacturing procedures are being used. Wall ties every third course connect the wall to reinforced concrete columns either end. Take the density of brickwork as 20kN/m3. Comment on the two different answers.

[16 marks]

iii. Write notes on the problems of sulphate attack in brickwork. Suggest methods of minimising sulphate attack in the 3.0m boundary wall if the wall was constructed in a highly industrialised area with high rainfall and the ground water was known to contain sulphates.

iv. Write notes on the problem of frost attack on brickwork and suggest how this could be minimised. In a 3.0m high boundary wall constructed in an area of high rainfall and winter temperatures which often move below freezing.[9 marks]2.

(a) En external brickwork boundary wall is 3.0m in height from the top of the foundation.. The horizontal span is 7.5m. If the characteristic wind load on the panel is 0.8kN/m2, design the wall.

i. With an allowance for a modified due to self weight.

ii. With no allowance for self weight.

Normal construction control and unit manufacturing procedures are being used. Wall ties every third course connect the wall to reinforced concrete columns either end. Take the density of brickwork as 20kN/m3. Comment on the two different answers.

Limiting panel dimensions.

Clause 32.3Height x length 1350tef

3000 x 7500 1350 tef

So tef = 129mm

Try wall thickness = 340mm

Clause 18Design wind load = 1.4 x 0.8 = 1.12kN/m2 [This may be reduced to 1.2 x 0.8 but has not been undertaken in this example]Clause 32Applied design bending moment

No dead load

Dead load at mid height included

Clause 20 Table 3

Use M4 mortar (designation (iii) with clay units WA between 7 12%

fkb0.4

0.4 + [1.5 x 20 x 0.34 x 103 x 0.9 or 1.4]

[340 x 1000]

f = 0.9

f = 0.9

0.427 0.442

fkp

1.1

1.1

1.1Clause 32 0.36

0.39

0.40

h/L =

3.0/7.5 = 0.4

0.4

0.4Table 8 Assume simple support all round i.e. Panel type A

Using linear interpolation.

(0.043+0.061)/2 (0.045+0.064)/2

0.052 0.055

0.052

0.055

0.055 1/5(0.055-0.052)

0.055-4/5(0.003)

0.0544

0.0526

0.052Applied mtwkfL2

0.0544 x 0.8 x 1.4 x 7.52

0.0526 x 0.8 x 1.4 x 7.52

0.052x 0.8 x 1.4 x 7.52

3.43

3.31

3.28Moment of resistance = fkxz/m = 1.1 x 1000 x 3402/6 x 3.0 = 7.06kNmNote. A thinner wall is possible but since this example was undertaken, the partial factors have reduced.

2.i. Write notes on the problems of sulphate attack in brickwork. Suggest methods of minimising sulphate attack in the 3.0m boundary wall if the wall was constructed in a highly industrialised area with high rainfall and the ground water was known to contain sulphates.

ii. Write notes on the problem of frost attack on brickwork and suggest how this could be minimised. In a 3.0m high boundary wall constructed in an area of high rainfall and winter temperatures which often move below freezing.

bi). Sulphate attack. The C3A part of the mortar reacts with sulphates in solution and expansion results.

C3A is present in OPC (CEM class II 42.5) so use a sulphate resisting cement which has the C3A removed.

Sulphates come from ground water so masonry in these environments should be protected.

Some clay bricks enhance the possibility of sulphate attack. Use low soluble salt content clay bricks.

To reduce sulphate attack use dense mortar.

Improved detailing should reduce water ingress.

Copings / cappings / DPCs / Surface treatments etc.

Sulphate attack requires a flow of water.

bii).

Frost action.

Saturated brickwork is vulnerable to frost attack if temperatures fall below freezing and bricks are incorrectly specified.

Repeated freezing and thawing increases damage.

Valve action reduces the vulnerability of mortar and concrete to frost action. i.e include air entrainers.

Winter construction should be avoided if there is the likelihood of freezing before the mortar has hardened.

Frost resistant clay bricks can be specified.

Calcium silicate bricks are vulnerable to freeze thaw especially when in the vicinity of sea water.

3.

(a) Design a cavity wall, 3.0m high and 4.0m long with both skins comprising low absorption clay brickwork, 102.5mm thick and using M4 (designation (iii) mortar. The left hand edge of the panel has a 2.0m return wall whilst the right hand edge is part of a continuous wall extending beyond the vertical wall support and into an adjacent panel. A damp proof membrane along the base of the wall results in limited moment resistance and the top of the wall is free. A characteristic wind load of 0.8kN/m2 acts laterally over the face of the wall. Assume that the weight of masonry is 20.0kN/m3 and that normal construction practice and unit manufacture apply.

[16 marks](b) A load bearing cavity wall with a clay brickwork outer and concrete blockwork inner skin is used to construct a 50.0m long four storey building. Discuss the problems which may result along this 50.0m length due to the relative movements between the two materials and suggest and illustrate appropriate details to surmount these difficulties .

[9 marks]3.

(a) Design a cavity wall, 3.0m high and 4.0m long with both skins comprising low absorption clay brickwork, 102.5mm thick and using M4 (designation (iii) mortar. The left hand edge of the panel has a 2.0m return wall whilst the right hand edge is part of a continuous wall extending beyond the vertical wall support and into an adjacent panel. A damp proof membrane along the base of the wall results in limited moment resistance and the top of the wall is free. A characteristic wind load of 0.8kN/m2 acts laterally over the face of the wall. Assume that the weight of masonry is 20.0kN/m3 and that normal construction practice and unit manufacture apply.

Sketch of wall.

Limiting dimensions. Clause 32.3.

Area = 3 x 4 x 106 = 12 x 106 mm2tef = 2/3(102.5 + 102.5) = 137 (Clause 24.4.1 Figure 2)

1500tef = 1500 x 137 = 28 x 106 > 12 x 106 so OK

Panel satisfies limiting dimensions.

Characteristic wind load = 0.8kN/m2Design bending moment = wkfL2

(Clause 32.4.2)

Aspect ratio of panel = h/L = 3.0/4.0 = 0.75

= 0.5/1.5 = 0.33

(Table 3)So = 0.045

(Table 8c)

Partial safety factor f = 1.4

(Clause 18)

Moment about vertical axis = 0.045 x 1.4 x 0.8 x 4.02 = 0.81kNm/m

Design moment of resistance MD = fkxz/m

(Clause 32.4.3)

z = (1 x 1000 x 102.52)/6 = 1.75 x 106 mm3/m width per leaf

But cavity wall equals sum of strength of both walls (Clause 32.4.5)

z(cavity) = 3.5 x 106 mm3/m width

MD = (fkx x 3.5 )/3.0 = (1.5 x 3.5)/3.0 = 1.75kNm >> 0.81kNmShear aspects will not be critical and need not be considered.

If the design had failed then the effects of including self weight to enhance flexural strength would be considered.3.(b) A load bearing cavity wall with a clay brickwork outer and concrete blockwork inner skin is used to construct a 50.0m long four storey building. Discuss the problems which may result along this 50.0m length due to the relative movements between the two materials and suggest and illustrate appropriate details to surmount these difficulties .

Temperature effects.

Coefficient of expansion of concrete is 7 14 per oC x 10-6Coefficient of expansion of clay brickwork 5 8 per oC x 10-6The expansive effects of concrete with temperature differences will be a problem. However, the fact that the blockwork skin is on the inside will assist in reducing the differential.

Reversible moisture movements.

Concrete blockwork ; +/- (0.02 0.1)%

Clay brickwork+/- (0.02)%Clearly the range of moisture movement in concreteis much greater than in brickwork. Using concrete in the inner skin wil again be an advantage.

Irreversible moisture movements.

Concrere blockwork : -(0.03 0.08)%

Clay brickwork+(0.02 0.07)%

Clearly these irreversible movements may cause serious distress to a structure.

It is possible to envisage a situation where

Irreversible Concreret blockwork shrinkage = 0.5%

- Clay brickwork expansion

= 0.4%

And additionally the clay brick may be temporarily wetted and expand further

= 0.02%

Therefore, relative movement

= 0.5 + 0.4 + 0.2 = 1.1%

Over 1.0m this will represent 11.0mm, a very significant movement.

Vertically, the shrinkage of concrete may cause bulging of the outer skin if no provision is made fro this.

Expansion joints are required. In concrete these allow for shrinkage and should be at 6.0m intervals whereas in clay brickwork, the joints may be at 9.0m intervals but allow for expansion. In addition, consideration of how the relative movement between the walls affects wall ties is necessary.

4.

(a) A load bearing masonry wall X carries four stories above and supports a first floor concrete slab as indicated in Figure 4. The load from the roof and four storeys above is 127.4kN/m. The load from the concrete slab is 33.6kN/m. Design the wall assuming normal manufacture and construction.

[16 marks]

(b) Check what lateral load the wall in a) could sustain if the axial load available to resist lateral load is 45kN/m length.

[9 marks]Solution4a). Assume the wall uses 140 wide units of height 190mm.

Eccentricity of load.

Position of resultant load.

(127.4 + 33.6) e = 33.6 x t/6

Therefore e = 0.035t

Slenderness ratio.

= 0.75 x 2900/140 = 15.53.

From Table 7, = 0.845 (linear interpolation)

Design equation

N = tfk/m(127.4 + 33.6) x 103 = 0.845fk x 140 x 103/3.5 fk = 4.76MPa

Unit selection.

Ht/least horiz distance = 190/140 = 1.36.

Using M4 (designation (iii) mortar, try 7.3MPa units.

fk = 3.2 + [(6.4 3.2) x (1.36- 0.6)]/1.4 = 4.94MPa so OK. Use unit as selected.4b). From clause 32.4.4,

qlat = 4tn/ha2 = 4 x 0.14 x 45 /(2.92)

= 2.99kN/m25.

(a) You are required to design a diaphragm wall which is 6.4m long and continuous either side. The height of the wall is 5.75m. A reinforced concrete beam is placed on the wall to carry the roof and is of such weight that it counteracts uplift forces acting on the roof. The dead load of the roof which spans 10.5m onto the diaphragm wall is 3.5kN/m2 and a live load of 2.0kN/m2 is to be anticipated. A characteristic wind load of 0.9kN/m2 is likely and normal control of unit manufacture and construction are expected. M4 (designation (iii) mortar is specified.[15 marks](b) Sketch three arrangements of blocks which could be used to construct a diaphragm wall. How would you ensure adequate bond if a vertical damp proof membrane was required between the ribs and leaves.?[6 marks](c) Why are deflections not likely to be a problem in diaphragm walls. Describe with the aid of sketches two forms of cracking which could develop in poorly designed unreinforced diaphragm walls.[10 marks]Q5 Solution

5a). Length of wall = 6.4m, ht of wall = 5.7m.

DL of roof = 10.95 x (3.5/2) = 19.16kN/m

LL of roof = 10.95 x 2.0/2 = 10.95kN/mWind load = 0.8kN/m2

Load case 0.9Gk + 1.4Wk

For reinforcement ms = 1.15. Normal control m = 3.5

From ACBM (diaphragm walls Fig 3.2) D = 550mm

Table 3 BS 5628 : part 1 fkb = 0.25. (NOTE 3 Table 3)

Equation 3.4 ACBM B = {(0.25 x 1002/3.5)/(0.6 x 1.4 x 0.8 x 103)}0.5

= 1.03m

Try 550 x 1180 bonded.

Area of block Am = 0.23m2/m

(Table 3.1 ACBM)

Ku = 2fdAmD/h = 2 x 0.9 x 18 x 0.23 x 0.55/5.7 = 0.72kN/m2 (eq 3.7 ACBM)

The design ultimate load Wu = 0.72/(1.4 (1 0.375)) = 0.82kN/m2Applied load = 1.4 Wk = 1.4 x 0.8 = 1.12kN/m2 > 0.82 so not OK

Try 780 x 1180 bonded.

Area of block Am = 0.23m2/m

(Table 3.1 ACBM)

Ku = 2fdAmD/h = 2 x 0.9 x 18 x 0.23 x 0.78/5.7 = 1.05kN/m2 (eq 3.7 ACBM)

The design ultimate load Wu = 1.05/(1.4 (1 0.375)) = 1.2kN/m2Applied load = 1.4 Wk = 1.4 x 0.8 = 1.12kN/m2 not > 1.2 so t OK

Now for 7.3N/mm2 concrete blocks, 440 x 215 x 100mm,

fk = 6.4MPa

(BS 5628 : Part 2)

From clause 28.2.1 of BS 5628 : Part 1,

fk = Nm/t

But N = 1.4 x 19.6 + 1.6 x 10.5 = 44.34kN/m run.

Effective height hef = 1.0 x 5.7 = 5.7m (Bs 5628 :Part 1 clause 24.3.2)Slenderness ratio = 5700/780 = 7.3

Assume eccentricity at top of wall = 780/6 = 130mm

Therefore = 0.748

(Table 7 : BS 5628 : Part 1)

Required fk = (44.34 x 103 x 3.5)/(0.793 x 780 x 103)= 0.252MPa not > 6.4MPa OKConsider shear :

V at base = 0.625 x 5.7 x 1.4 x 0.8 = 4.0kN(propped cantilever)

Shear stress at base v = (4.0 x 103 x 0.9)/(780 x 100) = 0.05MPa

fv/mv = 0.35/2.5 = 0.14 so OK

5b).

5c). With diaphragm walls the second moment of area is large which implies flexural cracking is unlikely before ULS reached. Deflection is usually +/- span/750

Cracking

1. Flexural tensile cracking

2. Tensile cracking in webs.

6.(a) Discuss the merits of enhancing the lateral load capacity of wall panels using reinforcement. Examine the four methods given in Appendix A of BS 5628-Part 2 and comment on the appropriateness of each.

(b) Consider the following three situations.

A concrete blockwork infill panel on the sixteenth story of a reinforced concrete framed residential high rise building in London. The building is heated by gas.

A solid brickwork garden wall in an entertainment area which caters for thousands of children. Wind speeds are high.

A masonry cavity infill panel for a single storey factory building in an industrial park which is surrounded by tall trees and buildings. The frame is steel.

For each wall :-

i. Recommend if bed joint reinforcement should be included and give reasons.

[6 marks]

ii. Make recommendations on how you would support the wall.

[3 marks]

iii. Describe any movement joist to be included.

[3 marks]

iv. Indicate which design method you would use.

[3 marks]

Solution Q6a).

Bed joint reinforcement distributes cracking and maintains slender sections.

Method 1. Most conservative. Treats the wall as a horizontal beam. Walls will usually span in two directions in reality. As this is conservative, enhancements are limited to 50% of unreinforced load. With cavity walls, add the strengths of both walls.

Method 2. The method requires the unreinforced load capacity to be determined and for this to be enhanced by 30%. The enhancement of 30% is carried by the reinforcement which is then designed in accordance with BS 5628 : Part 2 of the code. The technique has no theoretical basis as the unreinforced and hence uncracked wall is assumed to contribute to wall strength together with the reinforced and hence cracked section.

Method 3. The lateral load capacity of unreinforced masonry walls as described in Clause 32 of BS 5628 : Part 1 uses the concept of orthogonal strength ratios. In Method 3 the orthogonal strength ratio is altered by assuming the fkp strength is as for a reinforced section and then it is designed in accordance with Part 1. No limits are assigned to fkp which is of concern.

Method 4. Experimental evidence indicates that walls with bed joint reinforcement first crack at the ultimate load of an equivalent unreinforced wall. The ultimate load capacity of the unreinforced wall is determined with partial safety factors set to 1. The characteristic wind load of the panel is then determined by dividing this load by the partial safety factor for materials appropriate to serviceability. Using one of the other methods, a check at ultimate load is made.

Q6bi).An infill panel high up in a building should have two main functions. Firstly it must keep the occupants safely protected from the elements, and secondly it must not contain a blast which in very rare circumstances may arise. The former situation may certainly warrant including bed joint reinforcement, but correct lateral load design may abrogate this need. Further, arching effects, whether intentional or not will enhance the strength. The second aspect of the wall is to enable blasts to escape from the interior of the building without damaging additional internal structural elements. Again it is debatable whether a lightly reinforced infill panel will have significant effects on the progress of a blast.

Q6bi). A boundary garden wall in an entertainment centre may have serious consequences if it falls. In this instance the wall should have bed joint reinforcement included. The wall will also have a damp proof course which needs to be accounted for in the flexural strength calculations by assuming the base is simply supported. Adequate piers to ensure stability should also be considered.Q6bi). A single storey masonry in-fill panel wall in a factory which is well protected from wind should be designed for lateral loading without any bed joint reinforcement. These walls usually comprise an inner blockwork skin which spans between supports and needs to be adequately tied into the uprights which is connected to a continuous outer skin usually of brickwork. The outer skin will need expansion joints.

Q6bii, iii, and iv). Wall support, movement and design method.

A concrete blockwork infill panel on the sixteenth story of a reinforced concrete framed residential high rise building in London. The building is heated by gas.Supports. Wall ties attached to the frame and built into the bed joints of the wall. Angle or other bracket attached to floor and ceiling to fix base and top.Movement. Providing movement joints is not critical. Both materials will be shrinking

Design method. Clause 32 BS 5628:Part 1

A solid brickwork garden wall in an entertainment area which caters for thousands of children. Wind speeds are high.Supports. Piers of brickwork integral with the wall. A dpc will be placed below the wall to minimise the upward movement of water and will affect this boundary so must be considered.

Movement. Provide expansion joints every 10 12mm and ensure the joints can accommodate up to 15mm of movement otherwise reduce the spacing.

Design method. Clause 32 BS 5628:Part 1

A masonry cavity infill panel for a single storey factory building in an industrial park which is surrounded by tall trees and buildings. The frame is steel

Supports. The inner block skin will span between the steel uprights and as such needs to be attached to the uprights using appropriate wall ties. The upper edge of this wall is unlikely to have any lateral support . The base of the wall will have a damp proof membrane which will affect the support conditions. The facing brickwork will be continuous in most cases and needs wall ties to enable it to be connected to the concrete leaf but there need to be additional ties at the vertical uprights.

Movement. The inner skin is unlikely to need movement joints but the ties connecting the wall to the columns should not be rigid. The brick wall will need expansion joints every 10 12mm and ensure the joints can accommodate up to 15mm of movement otherwise reduce the spacing. These should be positioned at the uprights.Design method. Clause 32 BS 5628:Part 1

Continuous built in

DPM simple support

Return

Built in

free

FIGURE 4

127.4kN

33.6kN

t/6

t

2900mm

Bonded wall

Quoin bond

Tied wall

Adequate bond using ties

Tensile flexural cracks (unusual)

Webs in wall

Web crack due to shear

Wall x