06 07 Solutions

7/28/2019 06 07 Solutions

http://slidepdf.com/reader/full/06-07-solutions 1/16

1. (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, 2600mmhigh 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)

f y = 460N/mm2.

Strength of concrete infill = 30N/mm2

Block thickness (width of wall) = 215mm.

Block shell thickness = 35mm.

Percentage of block which is solid = 60%.

SOLUTION.

1. (a) Partial safety factors for materials are recommended in BS ; 5628 : Part

2 which depend on the control of manufacturing of the units. The codeassumes 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

7/28/2019 06 07 Solutions


- 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 required

Insurance 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 = 0.6

7/28/2019 06 07 Solutions


Effective 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 ρ = 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 f k x bd2)/ γmu

Therefore f k = (30.05 x 2.3 x 106)/(0.4 x 1000 x 150

2) = 7.68N/mm

2

For steel MD = (As x f y x Z)/ γms

So z = [MDγms]/[Asf y]

And z = d[(1 – (0.5Asf yγmu)/ (bdf k γms)]

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 = 693mm2

Use 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 height

Provide 2 No. R5 bars in alternative courses = 87mm2

Units strength

For f k = 7.68N/mm2,

Compressive strength of units = 15N/mm2

(net)

Compressive strength units = 9.0N/mm

2

(gross – assuming gross/net ratio =60%)

Use blocks of 7.0N/mm2

Shear.

Design shear strength = 6 x 1.4 x 2.675 = 22.5kN/m run

Shear strength = f v 0.35 + 17.5ρ= 0.35 + 17.5 x As /bd

7/28/2019 06 07 Solutions


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

= 0.35 + 0.104

= 0.454MPa not > 0.7 so OK

Check on enhancement.

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

No enhancement

Shear stress = (22.5 x 1000)/(1000 x 150) = 0.15 MPa

f y / γmv = 0.454/2.3 = 0.20 > Applied shear = 0.15MPa so OK

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 µ 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]

(b) 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.

[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 characteristicwind load on the panel is 0.8kN/m2, design the wall.

i. With an allowance for a modified µ 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

7/28/2019 06 07 Solutions


concrete columns either end. Take the density of brickwork as

20kN/m3. Comment on the two different answers.

Limiting panel dimensions.

Clause 32.3 Height x length ≤ 1350tef 3000 x 7500 ≤ 1350 tef

So tef = 129mm

Try wall thickness = 340mm

Clause 18 Design 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 32 Applied 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%

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

[340 x 1000]

γf = 0.9 γf = 0.9

0.427 0.442

f kp 1.1 1.1 1.1

Clause 32 µ 0.36 0.39 0.40

h/L = 3.0/7.5 = 0.4 0.4 0.4

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

α Using linear interpolation.

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

0.52 0.055 0.052 0.055

0.055 – 1/5(0.055-0.052) 0.055-4/5(0.003)

0.0544 0.0526 0.052

Applied mt αwk γf L2

7/28/2019 06 07 Solutions


0.0544 x 0.8 x 1.4 x 7.52 0.0526 x 0.8 x 1.4 x 7.5

2 0.052x 0.8 x 1.4 x 7.5

2

3.43 3.31 3.28

Moment of resistance = f kxz/ γm = 1.1 x 1000 x 3402 /6 x 3.0 = 7.06kNm

Note. A thinner wall is possible but since this example was undertaken, the partialfactors have reduced.

2.

(b) 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 wintertemperatures 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 / DPC’s / 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 themortar 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.

7/28/2019 06 07 Solutions


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 andsuggest 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 limitedmoment 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.

free

Return

Built in

DPM – simple support

Continuous

– built in

7/28/2019 06 07 Solutions


Limiting dimensions. Clause 32.3.

Area = 3 x 4 x 106

= 12 x 106

mm2

tef = 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/m2

Design bending moment = αwk γf L2

(Clause 32.4.2)

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

µ = 0.5/1.5 = 0.33 (Table 3)

So α = 0.045 (Table 8c)

Partial safety factor γf = 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 = f kxz/ γm (Clause 32.4.3)

z = (1 x 1000 x 102.52)/6 = 1.75 x 10

6mm

3 /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 = (f kx x 3.5 )/3.0 = (1.5 x 3.5)/3.0 = 1.75kNm >> 0.81kNm

Shear 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 50.0mlength 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 peroC x 10

-6

Coefficient of expansion of clay brickwork 5 – 8 peroC x 10

-6

7/28/2019 06 07 Solutions


The 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.

7/28/2019 06 07 Solutions


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]

2 9 0 0 m m

FIGURE 4

Wall x

7/28/2019 06 07 Solutions


Solution

4a). 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, β = 0.845 (linear interpolation)

Design equation

N = βtf k / γm

(127.4 + 33.6) x 103

= 0.845f k x 140 x 103 /3.5

f k = 4.76MPa

Unit selection.

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

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

f k = 3.2 + [(6.4 – 3.2) x (1.36- 0.6)]/1.4 = 4.94MPa so OK.

t

t/6

33.6kN127.4kN

7/28/2019 06 07 Solutions


Use unit as selected.

4b). From clause 32.4.4,

qlat = 4tn/ha2

= 4 x 0.14 x 45 /(2.92)

= 2.99kN/m2

7/28/2019 06 07 Solutions


7/28/2019 06 07 Solutions


5.

(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 areawhich caters for thousands of children. Wind speeds are high.

• A masonry cavity infill panel for a single storey factorybuilding 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 f kp strength is as for

a reinforced section and then it is designed in accordance with Part 1. No limits are

assigned to f kp which is of concern.

7/28/2019 06 07 Solutions


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 accountedfor 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 somust be considered.

Movement. Provide expansion joints every 10 – 12mm and ensure the joints can

accommodate up to 15mm of movement otherwise reduce the spacing.


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

7/28/2019 06 07 Solutions


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 tiesconnecting 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.


Documents

06 07 Solutions