10.3. The two-span beam with equal spans I is subjected to force P as shown in Fig. 10.14a. The beam is divided into ten equal portions; the number of joints is presented in Fig. 10.1le. Find the reaction at the middle support B. Solve this problem by three different ways (a) Use the influence lines for Mg and Qk: (b) use the influence lines for primary unknown of the displacement method; (c) use the influence lines for primary unknown of the force method. 0.4/ RA 0.4/ 0.6 R 0.096P b Influence line for M P=1 LoadP=1 0.0384 PI 0.4/ in the left span 0.096P M depends of position P the left or right at k 0.41 0.6/ Load P=1 in the right span 8. M = 0 for any position Pt ul=0.41 ul=0.6! R RC 8. 0.6/ X;{1+0.6P Fig. 10.14 (a) Design diagram of the beam and calculation of reaction at support B using the influ- ence line for internal forces at section k. (b) Calculation of reaction at support B using the influence line Z, of the displacement method. (e) Calculation of reaction at support B using the influence line X, of the force method 902 0.0336 19180'0 8001'0 0.0192

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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1. Influence lines for Mo and Q° for the force method are bounded by the straight
lines, because the primary system is a set of statically determinate simply sup-
ported beams. The same influence lines in the displacement method are bounded
by the curved lines, because the primary system is a set of statically indetermi-
nate beams.
2. If a load is located in the right span, then M° and 0° are zero. In the force
method it happens because the primary system presents two separate beams;
therefore, the load from one beam cannot be transmitted to another. In the dis-
placement method it happens because the introduced support in the primary
system does not allow transmitting of internal forces from the right span to the
left one.
10.3. The two-span beam with equal spans I is subjected to force P as
shown in Fig. 10.14a. The beam is divided into ten equal portions; the number of
joints is presented in Fig. 10.11e. Find the reaction at the middle support B. Solve
this problem by three different ways (a) Use the influence lines for Mk and Qk; (b)
use the influence lines for primary unknown of the displacement method; (c) use the
influence lines for primary unknown of the force method.
P=1
4
6.
708
9.
10 11
a 4
k
B
k
8
0.4/
RA
0.41.
0.61 R
-1.24 IL(Z,)
RC
a
(All ordinates
must be multiplied
by factor /)
0.096P
6
b Influence line for M
P=1
8
P=1
-
LoadP=1
in the left span
4 05
0,0384 PI
.3
8 9
10
0.41
0.096P
M depends of position P
the left or right at k
0.41
0.6/
ul
ul
b
Z,
P
ul
P=1
8 89
10 11
Load P=1
in the right span
**** -
B
M = 0 for any position P
ul=0.41
RB
ul=0.6/
R
RC
Inf. line M
(factorl /)
P
11
X +
Final
0.6/
Inf. line M.
(factor /)
X,/1+0.6P
Fig. 10.14 (a) Design diagram of the beam and calculation of reaction at support B using the influ-
ence line for internal forces at section k. (b) Calculation of reaction at support B using the influence
line Zj of the displacement method. (c) Calculation of reaction at support B using the influence
line X, of the force method
8001'0
0.1216E
0.2064
Azısoo
---
7610'0
0.0384
A 9EED'0
A 8870 0
0.0288
0.0384
0.0336
7610
Transcribed Image Text:1. Influence lines for Mo and Q° for the force method are bounded by the straight lines, because the primary system is a set of statically determinate simply sup- ported beams. The same influence lines in the displacement method are bounded by the curved lines, because the primary system is a set of statically indetermi- nate beams. 2. If a load is located in the right span, then M° and 0° are zero. In the force method it happens because the primary system presents two separate beams; therefore, the load from one beam cannot be transmitted to another. In the dis- placement method it happens because the introduced support in the primary system does not allow transmitting of internal forces from the right span to the left one. 10.3. The two-span beam with equal spans I is subjected to force P as shown in Fig. 10.14a. The beam is divided into ten equal portions; the number of joints is presented in Fig. 10.11e. Find the reaction at the middle support B. Solve this problem by three different ways (a) Use the influence lines for Mk and Qk; (b) use the influence lines for primary unknown of the displacement method; (c) use the influence lines for primary unknown of the force method. P=1 4 6. 708 9. 10 11 a 4 k B k 8 0.4/ RA 0.41. 0.61 R -1.24 IL(Z,) RC a (All ordinates must be multiplied by factor /) 0.096P 6 b Influence line for M P=1 8 P=1 - LoadP=1 in the left span 4 05 0,0384 PI .3 8 9 10 0.41 0.096P M depends of position P the left or right at k 0.41 0.6/ ul ul b Z, P ul P=1 8 89 10 11 Load P=1 in the right span **** - B M = 0 for any position P ul=0.41 RB ul=0.6/ R RC Inf. line M (factorl /) P 11 X + Final 0.6/ Inf. line M. (factor /) X,/1+0.6P Fig. 10.14 (a) Design diagram of the beam and calculation of reaction at support B using the influ- ence line for internal forces at section k. (b) Calculation of reaction at support B using the influence line Zj of the displacement method. (c) Calculation of reaction at support B using the influence line X, of the force method 8001'0 0.1216E 0.2064 Azısoo --- 7610'0 0.0384 A 9EED'0 A 8870 0 0.0288 0.0384 0.0336 7610
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