![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](https://image.slidesharecdn.com/strengthexample1-5-210501235512/85/Strength-example-1-5-9-320.jpg)
(
2.6𝑚
3
) + 𝑀 = 0
𝑀 = 2.6𝑘𝑁. 𝑚 𝑨𝒏𝒔
free−body diagram
Moment Diagram](https://image.slidesharecdn.com/strengthexample1-5-210501235512/85/Strength-example-1-5-10-320.jpg)
Strength example 1 5
- 1. Prepared : Majeed azad Supervisor: D.r Mahmoud Strength of material
- 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
- 4. EXAMPLE CHAPTER TWO If (maximum shear stress) τ max=800 KPa, calculate L.
- 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.
- 12. Solution: 𝐼 = 1 12 𝑏ℎ3 → 𝐼 = 1 12 3 4 3 16 3 = 412 ∗ 10−6 𝑖𝑛4 𝜎 = Mc 𝐼 → 𝐶 = 1 2 3 16 = 0.09375 in A- M = 𝜎𝐼 𝐶 → 𝑀 = (24∗103)(412∗10−6) 0.09375 𝑀 = 105.5 𝐼𝑏-in B- 𝐶 𝑃 = 𝜎 𝐸 → 𝑝 = 𝐸𝐶 𝜎 𝑝 = (29 x 106 )(0.09375) (24 ∗ 103 = 113.3 in 𝑨𝒏𝒔 𝑨𝒏𝒔 𝑨𝒏𝒔
- 13. Reference • http://mguler.etu.edu.tr/MAK205_Chapter6.pdf • https://construccion.uv.cl/docs/textos/TEXTO.20.pdf • https://www.chegg.com/homework-help/solid-steel- rod-diameter-d-supported-shown-knowing-steel-490- chapter-12-problem-28p-solution-9780073380155-exc
- 14. Any question