© 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 1–3, DL (150 mm)(0.017 kN/m2/mm(2.4 m)(3.0 m) 18.4 kN Ans From Table 1–4, LL (6 kN/m2)(2.4 m)(3.0 m) 43.2 kN
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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.
3m
From Table 1–3. DL (3.78 kN/m2)(3 m) + (0.57 kN/m2)(3m) 13.1 kN/m
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1–3. 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.
75 mm cinder fill 100 mm concrete slab ceiling
From Table 1 – 3. 100 mm – reinforced – stone slab 100(0.023) 2.30 kN/m2 75 mm – cinder concrete 75(0.017) 1.28 kN/m2 Plaster and lath 0.48 kN/m2 Total
p 4.06 kN/m2
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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.
1–4. 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
3.6 m 300 m
300 m
300 m 300 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
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200 mm 200 mm 500 mm
375 mm
375 mm 150 mm 150 mm 150 mm
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© 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.
1–7. 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.
From Tables 1 – 2 and 1 – 3,
100 mm cinder fill 125 mm concrete slab
ceiling
125-mm concrete slab (22.6)(0.125) 2.825 100-mm cinder fill (17.0)(0.100) 1.700 metal lath & plaster 0.48 Total dead load 5.0 kN/m2
*1–8. 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.
900 mm
150 mm
200 mm
900 mm A = 900(150) + 200(500) + 300(250) = 310 000 mm2 DL = (310 000 10–6 m2)(19.6 kN/m3) = 6.08 kN/m From Table 1 – 4,
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¥ 900 mm ´ = (11.97 kN/m2) § 1000 mm/m ¶ = 10.77 kN/m
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LL
250 mm
300 mm
1–9. 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 roof’s 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.
Weight per square meter
From
Table 1 – 3 Shingles Total
p
¥ 38 mm ´ µ = (7.86 kN/m3) ¦ § 1000 mm/m ¶ 2 = 0.30 kN/m = 0.10 kN/m2 = 0.40 kN/m2 r = 0.40 kN/m2
rx = (0.40) sin 30o = 0.20 kN/m2 ry = (0.40) cos 30o = 0.35 kN/m2
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Ans Ans
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© 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.
1–10. A two-storey school has interior columns that are spaced 4.5 m apart in two perpendicular directions. If the loading on the flat roof is estimated to be 1 kN/m2, determine the reduced live load supported by a typical interior column at (a) the ground-floor level, and (b) the second-floor level. Tributary area AT = (4.5)(4.5) = 20.25 m2 FR = 1(20.25) = 20.25 kN Since KLLAT = 4(20.25) > 37.2 Live load for second floor can be reduced.
4.57 K LL At )
L = LO(0.25 + L = 1.92(0.25 + a)
b)
4.57 2 ( 4)(20.25) = 1.45 kN/m
For ground floor column: L = 1.45 > 0.5 LO = 0.96 FF = 1.45(20.25) = 29.46 kN Fg = FF + FR = 29.46 + 20.25 = 49.71 kN
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For second floor column: F = FR = 20.25 kN
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1.11. *1–12. A four-storey office building has interior columns spaced 9 m apart in two perpendicular directions. If the flat-roof loading is estimated to be 1.5 kN/m2, determine the reduced live load supported by a typical interior column located at ground level.
Floor load: LO = 2.40 kN/m2 At = (9)(9) = 81 m2 L
= LO(0.25 + = 2.40(0.25 +
4.57 K LL At ) 4.57 2 4(81) ) = 1.21 kN/m
1.21 = 50% > 40% (OK) 2.40 Fg = 3[1.21 kN/m2)(9 m)(9 m)] + 1.5 kN/m2(9 m)(9 m) = 415.5 kN % reduction =
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1–11. *1.12. A three-storey hotel has interior columns that are spaced 6 m apart in two perpendicular directions. If the loading on the flat roof is estimated to be 1.5 kN/m2, determine the live load supported by a typical interior column at (a) the ground-floor level, and (b) the second-floor level. At = (6)(6) = 36 m2 LO = 1.92 kN/m2 L
a) b)
= LO (0.25 +
4.57 K LL At )
=1.92(0.25 +
4.57 2 4(36) ) = 1.21 kN/m
Fg = 2[(36 m2)(1.21 kN/m2)] + (36 m2)(1.5 kN/m2) = 141.2 kN F2F = (36 m2)(1.21 kN/m2) + (36 m2)(1.5 kN/m2) = 97.6 kN
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© 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.
z
1.8 m 2.4 m 1.35 m
3m
y
x
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© 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.
B A 4.5 m 30 m wind
C D
15 m
0.613KzKz1KdV 2I 0.613Kz(1)(1)(40)2(0.87) 0.613(0.85)(1)(1)(40)2(0.87) 725 N/m2 0.613(0.90)(1)(1)(40)2(0.87) 768 N/m2 1 h 15 + (7.5 tan 10°) 5.16 m 2 qh − 725 768 − 725 = 5.16 − 4.6 6.1 − 4.6 qh 741 N/m2 q1 q15 q20
External pressure on windward side of roof p qhGCp h 5.16 = = 0.344 L 15 [(−0.9 ) − (−0.7 )] (−0.9 − C p ) = (0.5 − 0.344 ) (0.5 − 0.25 ) Cp –0.7752 p 741(0.85)(–0.7752) –488 N/m2
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External pressure on leeward side of roof [ −0.5 − (−0.3)] (−0.5 − C p ) = (0.5 − 0.25 ) (0.5 − 0.344 ) Cp –0.3752 p qhGCp 741(0.85)(0.3752) = –236 N/m2
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Internal pressure p –qh(GCp1) –741(0.18) 134 N/m2
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© 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.
B A 4.5 m 30 m wind
C D
15 m
0.613KzKz1KdV 2I 0.613Kz(1)(1)(40)2(0.87) 0.613(0.85)(1)(1)(40)2(0.87) 725 N/m2 0.613(0.90)(1)(1)(40)2(0.87) 768 N/m2 1 q 4.5 + (7.5 tan 10°) 5.16 m 2 qh − 725 768 − 725 = 5.16 − 4.6 6.1 − 4.6 qh 741 N/m2 q1 q15 q20
External pressure on windward wall p qzG Cp 725(0.85)(0.8) 493 N/m2
External pressure on leeward wall
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L 15 = = 0.5 B 30
p qhGCp 741(0.85)(–0.5) –315 N/m2
External pressure on side walls p qhGCp 741(0.85)(–0.7) –441 N/m2
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Internal pressure p –qh(GCp1) –741(0.18) 134 N/m2
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© 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.
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© 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.
1–19. 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 0.7CICIIpg pf 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
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