study material_2011-12 se sqs errata

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ERRATAforStructural Engineering Sample Questions and Solutions

ISBN 978-1-932613-58-2

Copyright 2011 (November 2011, second printing)

Errata posted 04-10-12

Page 1 of 2

1 2 3 4A

B

C

D

THIS BEAM

PLAN

Solution 603, p. 100

Revisions indicated in red type

603. (a) Determine maximum factored shear and moment in floor beam

Calculate factored loads at beam at floor:

Floor DL:Superimposed 130 psf 30 ft 3,900 plf

Floor LL 70 psf

LL

Per ASCE-7 Section 4.8

15Reduced live load, L 70 0.25 70 0.60 42 psf with K 2

2 30 30

U 1.2 D 1.6 L IBC Eq.16-2

1.2 3.9 1.6 42 psf 30 ft/1, 000

4.68 2.02 6.70 kips/ft

n

n

2u n

2

Calculate moments and shears: ACI 8.3.3

2.0End span, L 30.0 2.0 27.0 ft

2

2.0 2.0Interior span, L 30.0 2

(discontinuous end integral with support

8.0 ft2 2

w LEnd span M

6.7 27ft-k

)14

349 ips

M at ext

14

erior fa

2u n

n

2

2u n

u n

w Lce of first interior support (more than two spans)

10

27 28L 27.5

2

6.7 27.5M 507 ft-kips

10w L

M at exterior face at exterior column 305 ft-kips16

Shear at end member 1.15 w L /2 (1.15)(6.7)(27/2)

u n

104.0 kips

Shear at all others w L /2 6.7 27/2 90.5 kips


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ISBN 978-1-932613-58-2

Copyright 2011 (November 2011, second printing)

Errata posted 03-27-2012

Page 2 of 2

3 4

507 ft-kips

SHEAR AND MOMENT DIAGRAMS

SHEAR

MOMENT

104.0 kips

90.5 kips

305 ft-kips

349 ft-kips

BEAM

603. (Continued)

(b) Sketch the shear and moment diagrams for the beam in Requirement (a).

(c) Design the beam for maximum positive moment in the span and negative moment and

shear at column grid.

u

Positive moment (design as T-beam):

30M ft-kips; b top flange T-beam width span/4 ft 12 in./ft 90 in. Controls

4

or 8 9 72 in.

or (30/2) 12 180 in.

d 24 0.5 1.5 0.5 21.5 in.From ACI Flexural Strength Tables (or o

349

2u c 2

ther design aids):

12Flexural strength coefficient = M / f bd

0.9 5.0 90 21.5

(Note: since the thickness of the slab flanges > a, design as rectangular section with

tension reinforcing onl

3490.02

y

2

c y

2s

s min w y

2w y

c

)

From the table,

5Ratio of reinforcement f / f

60

A bd 9

0.0223

0.02

0 21.5 in

3 5,000 90 21.5Min steel A 3 f b d/f ACI 318 Sec.10.5

60,000

200 90 21.56.84 in 200 b d/f

60,

230.00186

0.00186 3.6

2

2s required

6.45 in000

A 6.84 in Use (6) #10s


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ISBN 978-1-932613-58-2

Copyright 2011 (November 2010 printing)Errata posted 10-20-2011

Page 1 of 2

Vertical Forces PM Buildings Solutions

Solution 601, p. 91

Replace page 91 with the following:

601. (Continued)2

2 51(0.25)(2,000)(91.5 in /ft) 1 39,678 plf 140

a

a actual

P

P P

Check flexure:2 2

m m

s

b

2s

m

12 in.0.31 in /ft 0.116 in

32 in.

E 900 f 1,800,000 ACI 530-05, Sec.1.8.2.2.1

E 29,000,000 ACI 530-05, Sec.1.8.2.1

2,000F 667 psi ACI 530-05, Sec. 2.3.3.2.2

3

E n29 (0.116 in )(1n 16.11 n

E 1.8 b d

st

st

A

A

2

6.11)

(12 in.)(5.25 in.)

0.0297

k 2 n ( n) n

0.216

kj 1 0.928

3

allow

2

2 2b

allow

M (steel) jd

in.-lb ft-lb(0.116 in /ft)(24 ksi)(0.928)(5.25 in.) 13,565 1,130

ft ft

F jkbd (667)(0.928)(0.216)(12)(5.25)M (CMU)

2 2

22,110 in.-lb/ft

1,843 ft-lb/ft

st sA F

Steel stress controls 1,130 ft-lb > 691 ft-lb OK


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ISBN 978-1-932613-58-2

Copyright 2011 (November 2010 printing)Errata posted 10-20-2011

Page 2 of 2

Solution 601, p. 91 (continued)

Check combined stresses:

b 2 2

a

a b m

a b

2M (2)(691 ft-lb 12 in./ft)f 250 psi

jkbd (0.928)(0.216)(12 in.)(5.25)

(1,372 plf 600 plf )f 22 psi

(12 in./ft)(7.625 in.)

1 2,000per 2.3.3.2.2 maximum f f f 667 psi

3 3

actual f f 250 22 272 psi

study material_2011-12 se sqs errata

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