PROBLEM 567
Layer 1 (20 mm from top and bottom layers)
answer
Layer 2 (40 mm from top and bottom layers)
answer
Layer 3 (60 from top and bottom layers)
answer
Layer 4 (The Neutral Axis, NA)
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Guidelines and tips
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
1 / 37
This page cannot be seen from the preview
Don't miss anything!
Related documents
Partial preview of the text
Download solution manuals for student and more Summaries Design in PDF only on Docsity!
PROBLEM 567
Layer 1 (20 mm from top and bottom layers)
answer
Layer 2 (40 mm from top and bottom layers)
answer
Layer 3 (60 from top and bottom layers)
answer
Layer 4 (The Neutral Axis, NA)
answer
Checking: For rectangular section
( okay! )
PROBLEM 568
Where
Thus,
( okay! )
PROBLEM 569
Where
Checking: at the neutral axis
( okay! )
PROBLEM 572
( okay! )
By transfer formula for moment of inertia
Thus,
( okay! )
(a) At the Neutral Axis
answer
Maximum shearing stress occurs at the mid-height of the altitude, h
answer
PROBLEM 574
Where
Location of maximum horizontal shearing stress:
answer
PROBLEM 575
Where
Maximum horizontal shear stress occurs at the neutral axis
Thus,
answer
Minimum horizontal shear stress in the web occurs at the junction of flange and web
answer
The horizontal shearing stresses vary parabolically from the top to the bottom of the web. Recall that the average height of parabolic segment is 2/3 of its altitude measured from its base. Thus,
Shear force carried by web alone Force = Stress × Area
Shear force carried by web alone Force = Stress × Area
Percentage of shear force carried by web alone
answer
PROBLEM 577
Where:
Thus,
answer
PROBLEM 580
From the figure:
From flexure formula:
From shear stress formula:
answer
PROBLEM 581
Maximum moment for simple beam
Maximum shear for simple beam
For bending stress of wood
For shear stress of wood
PROBLEM 583
Based on allowable bending stress:
Based on allowable shearing stress:
For safe value of W, use W = 2 222.22 lb. answer
PROBLEM 584
Bending stress:
where:
Thus,
Shear stress:
where:
(see computation above)
Thus,
Ratio (flexural stress : shear stress)
answer
PROBLEM 585
Flexural Stress
Where fb = 16 MPa M = 1/8 woL^2 = 1/8 (6000)L^2 = 750L^2 N·m c = ½(250) = 125 mm I = 300(250^3 )/12 - 200(150^3 )/12 = 334 375 000 mm^4
Thus,
answer
Shearing Stress
Where V = ½ woL = ½(6000)(7.55) = 22 650 N Q = 10 000(100) + 2(6 250)(62.5) Q = 1 781 250 mm^3 b = 2(50) = 100 mm
Thus,
answer
PROBLEM 587
To draw the Shear Diagram
- VA = -P lb
- VB = VA + Area in load diagram VB = -P + 0 = -P lb VB2 = VB + R 1 = -P + 2P = P lb
- VC = VB2 + Area in load diagram VC = P - ½(12)(½P) = -2P lb VC2 = VC + R 2 = -2P + 3P = P lb
- VD = VC2 + Area in load diagram VD = P + 0 = P VD2 = VD - P = P - P = 0
- Shear at AB and CD are rectangular.
- Shear at BC is parabolic (2nd degree curve).
- Location of zero shear: By squared property of parabola x^2 / P = 122 / 3P x = 6.93 ft 12 - x = 5.07 ft
To draw the Moment Diagram
- MA = 0
- MB = MA + Area in shear diagram MB = 0 - 4P = -4P lb·ft
- ME = MB + Area in shear diagram ME = -4P + 2/3 (6.93)(P) = 0.62P lb·ft
- MC = ME + Area in shear diagram MC = 0.62P - [ 1/3 (12)3P - 2.31P - 5.07P ] MC = -4P lb·ft
- MD = MC + Area in shear diagram MD = -4P + 4P = 0
- The moment diagram at AB and CD are straight lines (1st degree curves) while at BC is 3rd degree curve.
Based on allowable bending stress
Where (From Solution 577) c = 6 in I = 350.67 in^4
Thus,
Based on allowable shear stress
Where (From Solution 577) Q = 35.5 in^3 I = 350.67 in^4 b = 0.75 in
Thus,
For safe value of P, use P = 740.85 lb. answer
PROBLEM 588
From the shear diagram
Maximum moment = sum of area in Shear Diagram at the left of point of zero shear
( okay! )
By transfer formula for moment of inertia
( okay! )
For M = -2W lb·ft Top fiber in tension
Bottom fiber in compression
For M = W lb·ft Top fiber in compression
Bottom fiber in tension
Based on allowable shear stress:
Where
Thus,
For safe value of W, use W = 4045 lb answer
PROBLEM 590
Check MC from the overhang segment
( okay! )
Based on allowable bending stress
Where M = 2.5P + 1700 lb·ft