2. EXAMPLE CHAPTER ONE
Determine the horizontal and vertical components of reaction on the beam caused by the pin at B and the roller
at A as shown in Fig. a. below. Neglect the weight of the beam.
3. Solution:
Equations of equilibrium: Summing forces in the x direction yields
+ → 𝛴𝐹𝑥 = 0; 600cos45°𝑁 − 𝐵𝑥 = 0
𝐵𝑥 = 424𝑁
A direct solution for Ay can be obtained by applying the moment
equation 𝜮𝑴𝐵 = 0 about point B
+↺ 𝜮𝑴𝐵 = 0; 100N 2m + (600 sin 45 °N) 5m − 600cos45°𝑁
0.2𝑚) − 𝐀y(7m = 0 𝐀y = 319N
Summing forces in the Y direction , using this result , gives
+ ↑ 𝛴𝐹𝑥 = 0; 319N − (600 sin 45 °N) − 100N − 200N + 𝐁y = 0
0.2𝑚) − 𝐀y(7m = 0 𝐀y = 319N
𝐁y = 405N
𝑨𝒏𝒔
𝑨𝒏𝒔
𝑨𝒏𝒔
5. Solution:
There are four separate areas of glue. Each area must transmit half of the
24 KN load the force , 𝐹 = 12 𝐾𝑁 = 12 ∗ 103
N
Shearing stress in glue 𝜏 = 800 ∗ 103
pa
𝜏 =
𝐹
𝐴
→ 𝐴 =
𝐹
𝜏
→ 𝐴 =
12∗103𝑁
800∗103 𝑝𝑎
= 15 ∗ 10−3
𝑚2
Let 𝑙 = length of glue area, and w = width = 100mm = 0.1m
A = 𝑙 ∗ w 𝑙 =
𝐴
𝑊
𝑙 =
15∗103 𝑚2
0.1m
= 150 ∗ 10−3
m = 150 mm
L = 2𝑙 + gap → L = 2 150 + 8 = 308mm
𝑨𝒏𝒔
𝑨𝒏𝒔
𝑨𝒏𝒔
6. EXAMPLE CHAPTER THREE
The steel frame (E = 200 GPa) shown has a diagonal brace BD with an area of 1920 𝑚𝑚2
. Determine the largest
allowable load P if the change in length of member BD is not to exceed 1.6 mm.
7. Solution:
SBC = 1.6 ∗ 10−3
, ABD = 1920𝑚𝑚2
→ ABD = 1920 ∗ 10−6
𝑚2
𝐿𝐵𝐶 = 52 + 62 = 7.810𝑚 𝐸𝐵𝐶 = 200 ∗ 104
𝑝𝑎
𝛿𝐵𝐶 =
𝐹𝐵𝐶∗𝐿𝐵𝐶
𝐸𝐵𝑐∗𝐴𝐵𝑐
→ 𝐹𝐵𝐶=
𝛿𝐵𝐶∗𝐸𝐵𝑐∗𝐴𝐵𝑐
𝐿𝐵𝐶
→
(200∗109)(1920∗10
−6)(1.6∗10−3)
7⋅81
𝐹𝐵𝐶= 78.76 ∗ 103
N
Use joint B as a free body: + → 𝛴𝐹𝑥 = 0;
5
7.810
𝐹𝐵𝐶 − P = 0 → P =
5
7.810
𝐹𝐵𝐶 → =
5∗(78.76∗103N )
7.810
50.4 ∗ 103
N → 50.4KN
𝑨𝒏𝒔
𝑨𝒏𝒔
8. EXAMPLE CHAPTER FOUR
SFy = 0:
SFy = 0:
Draw the shear and
moment diagrams
for the beam.
The distributed load is replaced by its resultant
1– w0 x x
—
—
w x
force and the reactions have been determined as shown
The reactions have been
determined and are shown on the
free-body diagram.
9. Solution:
Shear Diagram: The end points V = 1.5,
and x = 4.5, V =-3
From the behavior of the distributed load,
the slope of the shear diagram will vary
from zero at x =0 to -2 at x= 4.5.As a
result, the shear diagram is a having the
shape shown. The point of zero shear can
be found by using the method of sections
for a beam segment of length x
+ ↑ 𝜮𝑭𝒙 = 𝟎; 𝟏. 𝟓𝐤𝐍 −
𝟏
𝟐
[𝟐𝐊𝐍/𝐦(
𝐱
𝟒.𝟓𝒎
)]
x=0; 𝒙 = 𝟐. 𝟔𝒎
𝑨𝒏𝒔
𝑨𝒏𝒔
𝑨𝒏𝒔
𝑨𝒏𝒔
Shear Diagram
10. Solution:
Moment Diagram : The end points 𝑥 =
0, 𝑀 = 0 𝑎𝑛𝑑 𝑥 = 45, 𝑀 = 0 are
plotted first From the behavior of the
shear diagram, the slope of the moment
diagram will begin at +1.5 then it
becomes decreasingly positive until it
reaches zero at 2.6 m. It then becomes
increasingly negative reaching -3 at x =
4.5 m. Here the moment diagram is a
cubic function of x. Why? Notice that the
maximum moment is at x = 2.6
Since V =
𝑑𝑀
𝑑𝑥
= 0 at this point. From the
free-body diagram
+↺ 𝜮𝑴𝐵 = 0; −1.5𝑘𝑁 2.6𝑚 +
1
2
[2𝑘𝑁/m(
2.6𝑚
4.5𝑚
)](2.6)(
2.6𝑚
3
) + 𝑀 = 0
𝑀 = 2.6𝑘𝑁. 𝑚 𝑨𝒏𝒔
free−body diagram
Moment Diagram
11. EXAMPLE CHAPTER FIVE
Knowing that 𝜎 = 24 ksi for the steel strip AB, determine
(a) the largest couple M that can be applied,
(b) the corresponding radius of curvature. Use E = 29 x 106
psi.
