1 2009 by R.C. Hibbeler. Published by Prentice Hall, Pearson Education South Asia Pte Ltd. All rights reserved. This material is protected under copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1-1. The floor of a light storage warehouse is made of 150-mm-thick cinder concrete. If the floor is a slab having a length of 3 m and width of 2.4 m, determine the resultant force caused by the dead load and that caused by the live load.

From Table 13,

DL (150 mm)(0.017 kN/m2/mm(2.4 m)(3.0 m) 18.4 kN Ans

From Table 14,

LL (6 kN/m2)(2.4 m)(3.0 m) 43.2 kN Ans

From Table 13.

DL (3.78 kN/m2)(3 m) + (0.57 kN/m2)(3m) 13.1 kN/m Ans

3 m

1-2. The building wall consists of 200-mm clay brick. In the interior, the wall is made from 50 mm 100 mm wood studs, plastered on one side. If the wall is 3 m high, determine the load in kN per meter of length of wall that the wall exerts on the flloor.

13. The second floor of a light manufacturing building is constructed from a 100-mm thick reinforced-stone concrete slab with an added 75-mm cinder concrete fill as shown. If the suspended ceiling of the first floor consists of metal lath and gypsum plaster, determine the dead load for design in Newtons per square meter of floor area.

100 mm concrete slab

ceiling

75 mm cinder fill

From Table 1 3.

100 mm reinforced stone slab 100(0.023) 2.30 kN/m275 mm cinder concrete 75(0.017) 1.28 kN/m2Plaster and lath 0.48 kN/m2

Total p 4.06 kN/m2 Ans

2 2009 by R.C. Hibbeler. Published by Prentice Hall, Pearson Education South Asia Pte Ltd. All rights reserved. This material is protected under copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

14. The hollow core panel is made from plain stone concrete. Determine the dead weight of the panel. The holes each have a diameter of 100 mm.

175 mm

300 m 300 m 300 m 300 m300 m 300 m

3.6 m

W = (22.6 kN/m2)[(3.6 m)(1.8 m)(0.175 m) 5(3.6 m)(p)(0.05 m)2]

= 22.4 kN Ans

200 mm

200 mm

150 mm150 mm150 mm

375 mm

500 mm

375 mm

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17. The second floor of a light manufacturing building is constructed from a 125-mm-thick stone concrete slab with an added 100-mm cinder concrete fill as shown. If the suspended ceiling of the first floor consists of metal lath and gypsum plaster, determine the dead load for design in kN per square meter of floor area.

100 mm cinder fill

125 mm concrete slab

ceiling

From Tables 1 2 and 1 3,

125-mm concrete slab (22.6)(0.125) 2.825100-mm cinder fill (17.0)(0.100) 1.700metal lath & plaster 0.48Total dead load 5.0 kN/m2 Ans

*18. The T-beam used in a heavy storage warehouse is made of concrete having a specific weight of 19.6 kN/m3. Determine the dead load per meter length of beam, and the live load on the top of the beam per meter length of beam. Neglect the weight of the steel reinforcement.

A = 900(150) + 200(500) + 300(250) = 310 000 mm2DL = (310 000 106 m2)(19.6 kN/m3) = 6.08 kN/m AnsFrom Table 1 4,

LL = (11.97 kN/m2) 900 mm1000 mm/m

= 10.77 kN/m Ans

900 mm

900 mm

200 mm

250 mm

150 mm

300 mm

19. The beam supports the roof made from asphalt shingles and wood sheathing boards. If the boards have a thickness of 38 mm and a specific weight of 7.86 kN/m3, and the roofs angle of slope is 30, determine the dead load of the roofing per square meter that is supported in the x and y directions by the purlins.

r = 0.40 kN/m2

rx = (0.40) sin 30o = 0.20 kN/m2 Ansry = (0.40) cos 30o = 0.35 kN/m2 Ans

Weight per square meter = (7.86 kN/m3) 38 mm1000 mm/m

= 0.30 kN/m2From Table 1 3

Shingles = 0.10 kN/m2Total p = 0.40 kN/m2

4 2009 by R.C. Hibbeler. Published by Prentice Hall, Pearson Education South Asia Pte Ltd. All rights reserved. This material is protected under copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

a) For ground floor column:L = 1.45 > 0.5 LO = 0.96FF = 1.45(20.25) = 29.46 kNFg = FF + FR = 29.46 + 20.25 = 49.71 kN Ans

b) For second floor column:F = FR = 20.25 kN Ans

Tributary area AT = (4.5)(4.5) = 20.25 m2FR = 1(20.25) = 20.25 kNSince KLLAT = 4(20.25) > 37.2Live load for second floor can be reduced.

L = LO(0.25 + 4 57.K ALL t )

L = 1.92(0.25 +4 57

4 20 25.

( )( . ) = 1.45 kN/m2

110. A two-storey school has interior columns thatare spaced 4.5 m apart in two perpendiculardirections. If the loading on the flat roof is estimatedto be 1 kN/m2, determine the reduced live loadsupported by a typical interior column at (a) theground-floor level, and (b) the second-floor level.

*112. A four-storey office building has interiorcolumns spaced 9 m apart in two perpendiculardirections. If the flat-roof loading is estimated to be1.5 kN/m2, determine the reduced live load supportedby a typical interior column located at ground level.

Floor load:LO = 2.40 kN/m2At = (9)(9) = 81 m2

L = LO(0.25 + 4 57.K ALL t )

= 2.40(0.25 + 4 574 81.( ) ) = 1.21 kN/m2

% reduction = 1 212 40.. = 50% > 40% (OK)

Fg = 3[1.21 kN/m2)(9 m)(9 m)] + 1.5 kN/m2(9 m)(9 m) = 415.5 kN Ans

1.11.

a) Fg = 2[(36 m2)(1.21 kN/m2)] + (36 m2)(1.5 kN/m2)= 141.2 kN Ans

b) F2F = (36 m2)(1.21 kN/m2) + (36 m2)(1.5 kN/m2)= 97.6 kN Ans

111. A three-storey hotel has interior columns thatare spaced 6 m apart in two perpendicular directions.If the loading on the flat roof is estimated to be 1.5kN/m2, determine the live load supported by a typicalinterior column at (a) the ground-floor level, and (b)the second-floor level.At = (6)(6) = 36 m2LO = 1.92 kN/m2

L = LO (0.25 + 4 57.K ALL t )

=1.92(0.25 + 4 574 36.( ) ) = 1.21 kN/m2

*1.12.

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y

x

z

3 m

1.35 m2.4 m

1.8 m

6 2009 by R.C. Hibbeler. Published by Prentice Hall, Pearson Education South Asia Pte Ltd. All rights reserved. This material is protected under copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

q1 0.613KzKz1KdV 2I 0.613Kz(1)(1)(40)2(0.87)q15 0.613(0.85)(1)(1)(40)2(0.87) 725 N/m2q20 0.613(0.90)(1)(1)(40)2(0.87) 768 N/m2

h 15 + 12

(7.5 tan 10) 5.16 m

qh

=

7255 16 4 6

768 7256 1 4 6. . . .

qh 741 N/m2

External pressure on windward side of roofp qhGCphL = =

=

5 1615 0 344

0 9 0 70 5 0 25

. .[( . ) ( . )]

( . . )( 00 90 5 0 344

. )( . . )

Cp

Cp 0.7752p 741(0.85)(0.7752) 488 N/m2 Ans

External pressure on leeward side of roof[ . ( . )]

( . . )( . )

( . .

=

0 5 0 30 5 0 25

0 50 5 0 344

Cp))

Cp 0.3752p qhGCp 741(0.85)(0.3752) = 236 N/m2 Ans

Internal pressurep qh(GCp1) 741(0.18) 134 N/m2 Ans

30 m

15 mwind

A

B

C

D

4.5 m

7 2009 by R.C. Hibbeler. Published by Prentice Hall, Pearson Education South Asia Pte Ltd. All rights reserved. This material is protected under copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30 m

15 mwind

A

B

C

D

4.5 m

q1 0.613KzKz1KdV 2I 0.613Kz(1)(1)(40)2(0.87)q15 0.613(0.85)(1)(1)(40)2(0.87) 725 N/m2q20 0.613(0.90)(1)(1)(40)2(0.87) 768 N/m2

q 4.5 + 12

(7.5 tan 10) 5.16 m

qh

=

7255 16 4 6

768 7256 1 4 6. . . .

qh 741 N/m2

External pressure on windward wallp qzG Cp 725(0.85)(0.8) 493 N/m2 Ans

External pressure on leeward wall LB = =1530 0 5.

p qhGCp 741(0.85)(0.5) 315 N/m2

External pressure on side wallsp qhGCp 741(0.85)(0.7) 441 N/m2 Ans

Internal pressurep qh(GCp1) 741(0.18) 134 N/m2 Ans

8 2009 by R.C. Hibbeler. Published by Prentice Hall, Pearson Education South Asia Pte Ltd. All rights reserved. This material is protected under copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9 2009 by R.C. Hibbeler. Published by Prentice Hall, Pearson Education South Asia Pte Ltd. All rights reserved. This material is protected under copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

119. A hospital located in Chicago, Illinois, has a flat roof, where the ground snow load is 1.2 kN/m2. Determine the design snow load on the roof of the hospital.

CI 1.3

CI 1.0

I 1.2

pf 0.7CICIIpgpf 0.7(1.3)(1.0)(1.2)(1.2) 1.31 kN/m2

Since pg > 0.96 kN/m2, then use

pf I(0.96 kN/m2) 1.2(0.96 kN/m2) 1.15 kN/m2 Ans

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