influence lines for trusses
DESCRIPTION
the variation of forces in truss members due to moving loads is of great importance. Influence lines may be drawn and used for making force calculations, or they may be sketched roughly without computing the values of the ordinates and used only for placing the moving loads to cause maximum or minimum forces.TRANSCRIPT
THEORY1 Structural Theory 1 Chapter 7
Influence Lines for Trusses;
Introduction; the variation of forces in truss members due to moving loads is of great
importance. Influence lines may be drawn and used for making force calculations, or they may be sketched roughly without computing the values of the ordinates and used only for placing the moving loads to cause maximum or minimum forces. The procedure used for preparing influence lines for trusses is closely related to the one used for beams.
EX. For the Howe bridge truss shown, draw the influence lines for the stresses in members , and .
for ,
(+) = 0;
(24) – 1(24 – x) = 0; =
for ,
when 0 x 12,
using the left segment,
(+) = 0;
(8) – 1(8 - x) - 4 = 0
(8) - 8 + x - 4 = 0
= 0; =
if, x = 0; = 0
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0L 6L
1L 2L 3L 4L
5U4U3U2U1U
5L
4.0 m
6 @ 4.0 m = 24.0 m
1.50
IL for
IL for
0.471
0.707
IL for
1.50
THEORY1 Structural Theory 1 Chapter 7
x = 8; = 1.33when 12 x 24,
using the left segment,
(+) = 0;
(12) – 4 = 0
(12) - 4 = 0
= 0; =
if, x = 12; = 1.50
x = 24; = 0
for ,
when 0 x 8,
using the left segment,
(+) = 0;
– 1 + = 0
- 1 + = 0
+ = 0
- + = 0; =
if, x = 0; = 0
x = 8; = 0.471
when 12 x 24,
using the left segment,
(+) = 0;
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THEORY1 Structural Theory 1 Chapter 7
+ = 0
+ = 0; =
if, x = 12; = -0.707
x = 24; = 0
for ,
when 0 x 8,
using the left segment,
(+) = 0;
(8) – 1(8 - x) + 4 = 0
(8) - 8 + x - 4 = 0
= 0; = -
if, x = 0; = 0
x = 8; = -1.33
when 12 x 24,
using the left segment,
(+) = 0;
(12) + 4 = 0
(12) + 4 = 0
= 0; =
if, x = 12; = -1.50
x = 24; = 0
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THEORY1 Structural Theory 1 Chapter 7
EX. For the bridge truss shown, draw the influence lines for the stresses in members , and .
for ,
(+) = 0;
(18) – 1(18 – x) = 0; =
for ,
when 0 x 6,
using the left segment,
(+) = 0;
(6) – 1(6 - x) – 2.5 = 0
(6) - 6 + x – 2.5 = 0
= 0; =
if, x = 0; = 0
x = 6; = 1.60
when 9 x 18,
using the left segment,
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0L 6L1L 2L 3L 4L
5U4U3U
1U
5L6 @ 3.0 m = 18.0 m
1.5 m
2.5 m
3.0 m
IL for
1.86
1.60
IL for
IL for
1.30
THEORY1 Structural Theory 1 Chapter 7
(+) = 0;
(6) – 2.5 = 0
(6) – 2.5 = 0
= 0; =
if, x = 9; = 1.20
x = 18; = 0
for ,
when 0 x 9,
using the left segment,
(+) = 0;
(9) – 1(9 - x) + 2.5 + 3 = 0
(9) – 9 + x + = 0
= 0; = -
if, x = 0; = 0
x = 9; = -1.30
when 9 x 18,
using the left segment,
(+) = 0;
(9) + 2.5 + 3 = 0
(9) + = 0
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THEORY1 Structural Theory 1 Chapter 7
= 0; = -
if, x = 9; = -1.30
x = 18; = 0
for ,
when 0 x 15,
using the left segment,
(+) = 0;
(15) – 1(15 - x) + 1.5 = 0
(15) - 15 + x + 1.5 = 0
= 0; = -
if, x = 0; = 0
x = 15; = -1.86
when 15 x 18,
using the left segment,
(+) = 0;
(15) + 1.5 = 0
(15) + 1.5 = 0
= 0; = -
if, x = 15; = -1.86
x = 18; = 0
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THEORY1 Structural Theory 1 Chapter 7
Engr. I.R. Bonzon 7
THEORY1 Structural Theory 1 Chapter 7
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