π
<-

## RESISTENCIA245

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Catégorie :Category: mViewer GX Creator App HP-Prime
Auteur Author: jeanpaul2809
Type : Application
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### Description

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8–10. Determine the slope at B and the maximum 15 k
displacement of the beam. Use the moment-area theorems.
Take E = 29(103) ksi, I = 500 in4. A
C

B

6 ft 6 ft

M
Using the diagram and the elastic curve shown in Fig. a and b, respectively,
EI
Theorem 1 and 2 give

1 90 k # ft
uB = ƒ uB>A ƒ = a b (6 ft)
2 EI

270 k # ft2 270 (144) k # in2
= = = 0.00268 rad Ans.
c 29(10 ) 2 d (500 in4)
EI k
3
in
1 90 k # ft
¢ max = ¢ C = ƒ tB>A ƒ = c a b (6 ft) d c 6 ft + (6 ft) d
2
2 EI 3
2700 k # ft3
=
EI
2700 (1728) k # in3
=
c 29(103) d (500 in4)
k
in2
= 0.322 in T Ans.

278
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8–11. Solve Prob. 8–10 using the conjugate-beam method. 15 k

A
C

B

6 ft 6 ft

The real beam and conjugate beam are shown in Fig. b and c, respectively.
Referring to Fig. c,
1 90 k # ft
+ c a Fy = 0; - V¿B - a b (6 ft) = 0
2 EI
270 k # ft2
uB = V¿B = –
EI
270 (122) k # in2
c 29(103) d (500 in4)
k
in2
Referring to Fig. d,
1 90 k # ft
M¿C + c a b (6 ft) d c6 ft + (6 ft) d = 0
2
a+ a MC = 0;
2 EI 3
2700 k # ft3
¢ max = ¢ C = MC¿ = -
EI
2700 (123) k # in3
= = 0.322 in T Ans.
c 29(10 ) 2 d (500 in )
3 k 4
in

279
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*8–12. Determine the slope and displacement at C. EI is 15 k
constant. Use the moment-area theorems.

A C
B
30 ft 15 ft

M
Using the diagram and the elastic curve shown in Fig. a and b, respectively,
EI
Theorem 1 and 2 give

225 k # ft 5062.5 k # ft2 5062.5 k # ft2
a- b (45 ft) = -
1
uC>A = =
2 EI EI EI

1 225 k # ft 33750 k # ft3
ƒ tB>A ƒ = c a b (30 ft) d c (30 ft) d =
1
2 EI 3 EI
1 225 k # ft 1 225 k # ft
ƒ tC>A ƒ = c a b (30 ft) d c 15 ft + (30 ft) d + c a b(15 ft) d c (15 ft) d
1 2
2 EI 3 2 EI 3

101250 k # ft3
=
EI
Then,

45 33750 k # ft3 50625 k # ft3
a b =
45
¢¿ = (tB>A) =
30 30 EI EI

ƒ tB>A ƒ 33750 k # ft3>EI 1125 k # ft2
uA = = =
LAB 30 ft EI
+b uC = uA + uC>A

–1125 k # ft2 5062.5 k # ft2 3937.5 k # ft2
uC = + = Ans.
EI EI EI

101250 k # ft3 50625 k # ft3
¢ C =  tC>A  – ¢ ¿ = -
EI EI
50625 k # ft3
= T Ans.
EI

280
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8–13. Solve Prob. 8–12 using the conjugate-beam method. 15 k

A C
B
30 ft 15 ft

The real beam and conjugate beam are shown in Fig. a and b, respectively.
Referring to Fig. c,

1 225 k # ft
a+ a MA = 0; B¿y(30 ft) - c a b (30 ft) d (20 ft)
2 EI
2250 k # ft2
B¿y =
EI
Referring to Fig. d,

1 225 k # ft 2250 k # ft
+ c a Fy = 0; - V¿C - a b (15 ft) -
2 EI EI
3937.5 k # ft2 3937.5 k # ft2
uC = V¿C = - = Ans.
EI EI

1 225 k # ft 2250 k # ft2
a+ a MC = 0; M¿C + c a b (15 ft) d (10 ft) + a b(15 ft)
2 EI EI

50625 k # ft3 50625 k # ft3
¢ C = M¿C = = T Ans.
EI EI

281
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8–14. Determine the value of a so that the slope at A is P ...

### Archive contentsContenu de l'archive

Action(s) SizeTaille FileFichier
3.65 Ko KB lisezmoi.txt
1.09 Ko KB RESISTEN.hpprgm
1.09 Mo MB RESISTEN 01_20.hpappdir.zip
110.12 Ko KB RESISTEN 21_22.hpappdir.zip
167 octets bytes appslist.txt

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