A First Course in the Finite Element Method 6th edition Daryl L. Logan - Solutions

2000 Ib ft (2 8 ft 2000 lb 6 ft 45° 3) 2.

4 m 8. 10 kN 3 m 6. 20 kN E = 200 GPa A = 2 x 10* m? 1 = 4 x 10-4 m* 3 m 20 kN 3 O 3 m 3.

20 kip 25 ft 25 ft 3 3 40 ft (2 в - - 20 ft - 20 ft 30 ft-

12 kN/m Н 5 m 12 kN/m D 10 m 10 m за

к 25 kN 4 m 50 kN 6 m A -6 m - 8 m 8 m

G Н 30 kN 3 m 60 kN 3 m -5m -5m

м 7.5 kip - 15 ft 15 kip - 15 ft 15 kip - 15 ft A - 30 ft - 20 ft --30 ft -

90

ro I 7- L-

T (Ib-in./in.)

5000 ib 10 ft 2.

y 4 m 2 20 kN · m 25 kN · m 30 kN

3 3000 lb 5 10 ft 10 ft 10 ft-

c= 29 in. d=58 in.- e= 86 in. b=58 in. 6000 lb a = 100 in. |D

-30 30- -72- 32- Dimensions are in inches 30 600 lb 800 lb 600 lb 24 30 45° B.

5000 lb 8

-9.0

8 in. 72 in.- 5000 lb 60 in. 20°

Wind |1}}}}}} 16 ft 3.00 psf 11 ft L= 40 ft (a) (b) 7.50 psf

1500 Pa 3 m (4 m spacing of frames) 4 m 6 m –

8 ft- 12 ft 2 ft 15 ft 3 ft 10 ton

- 25 ft 25 ft 50 ft 15 ft 10 ft 0.2 W 0.8 W 14 ft - W = total weight of truck and load 32 k H20 - 44 8 k H truck loading

3 s = 5000 lb /in. 10 in. 4 20 in.

20 kN 250 mm 100 mm 20 kN 30° 500 mm -100 mm (b) 40 kN Pr= 10 MPa 400 mm 400 mm 400 mm 30 kN 400 mm (d) (c)

8. 4 (a) (b) 4) 2. 2. 3.

fbix fbiy Хь | fbjx {fb} = fbiy Хь | At |Y 3 Xp fbmx fbmy YB

(b — х)(h — у) N2 4bh (b + x)(h – y) 4bh N1 (b + x)(h + y) 4bh (b – x)(h + y) N3 = N4 4bh ||

uj = 0 in. v2 = 0.0025 in. u4 = 0 in. vI = 0 in. uz = 0.0025 in. %3D u2 = 0.005 in. v3 = -0.0025 in. V4 = 0 in. V4 =

For the finite element mesh shown in Figure P7-1, comment

-1.4 m- 0.6 m

5000 lb 10

25 mm radius 10 kPa 500 mm 10 kPa 500 mm

P = 1000 lb КUИ o, - bt ШШ dy bt -Ь aв

in. radius 0.5-in. radius in. Axis of 0.3125 in. symmetry in. 40* 2 in. 0.3125 in. | 0.75 in. -2 in.- l00

Axis of symmetry -100 mm- 100 mm 100 mm -1.5 N/mm2 50 mm 1.5 N/mm? 38 mm 88 mm 6 mm radius

10 in. 1000 lb/in. 10 in. 5 in. 0.25 in. rad. 10 in. 20 in.

15 lb 10 Ib 0.1 in. rad. (typ.) in. 16 in. 16 6

A D B

40,000-lb total shear load parabolically distributed 8 in. 48 in.

50 kN/m Beam 2 m 4 m -4 m 10 m

I mm rad. each corner 500 mm 25 mm 25-mm radius | KN/m? I KN/m 1 kN/m| 1 kN/m? 500 mm 500 mm

2 k/ft 18 ft 11.5-ft radius 8 ft 1wat-10n-105- 10 ft 10 ft 10 ft-

+0.3 m+0.4 m-0.3 m- l0.75 m 2 kN/m? t = 10 mm 75-mm radius

25-mm radius 0.1 m 5 kN (total force) 0.3 m 0.15 m ! = 10 mm 0.25 m -0.25 m -0.25 m

700 N spread over 50 mm length 50 mm 70 mm 15 mm 40 mm 160 mm 40 mm 20 mm 40 mm 20 mm 50 mm 20 mm 40 mm - -75 mm 40 mm

S000 Ib RO.10

2 in. 12 in. P= 500 lb 2 in. 12 in.

120 30 60 E 24 50 D Dimensions in millimeters Released position

8.0000 RO.1409 0.6000 R0.6000 3.6541 (a) Crimper arm with dimensions (inches) RO.1409 - F = 60 lb 1.5 in. 2.3 in. 5.8 in

The sides of the middle hexagon are 9 mm long. R= 15 mm 1 mm rad. (typ.) 20 mm 40 mm 1 mm rad. (typ.) The sides of the c

E 10 in.- 1000 Ib/in. 10 in. rad = 1 in. 5 in. 10 in. 20 in.

(0, h)3 (0. 4). 6. (0. 0) (b, 0) 5.0)

6. 2.

(6, 6) (3, 3) 4. (6, 3) (0, 0) (3, 0) (6, 0)

(60, 60) х (0, 0) (60, 0)

6000 Ib total force ft Soil mass

Model of a nuclear reactor Concrete Steel liner 13 in. -3-in-radius hole -7.5 in. 16 in. 50 in. 25 in. 2 in. -30 in.-

Concrete Steel liner 1250 mm 400 mm 325 mm- 750 mm

3 1 -in. radius 4 2 in. in. in. I in 10° I in. 1000 lb 1000 ib Axis of symmetry in. 7, in.

Po 2.

25 18.75 30° 60° 25 300 225 Dimensions in millimeters -200 dia. 250 dia.- 310 dia.-

+в- (a) (b)

45 N Plunger 15 mm 25 mm- 20 mm Liquid 90 mm Glass syringe 45 8 mm 8 mm + 4 mm mm 12 mm