PROBLEM 7: The shear diagram due to distributed load of a beam is given in the figure. The reactions at columns C1 and C2 are uniformly distributed loads. Given: L₁ = 2.6 m; L₂ = 5.8 m; L3 = 3.6 m; C₁ = 0.4 m; C₂ = 0.42 m VB = 43.68 KN; Vc = 39.77 kN; VD = 57.67 KN; VE = 60.48 KN (a) Calculate the loads on the beam and the reactions on the columns and illustrate the load diagram. (b) Draw the moment diagram and identify the maximum positive and bending moments. (c) If the beam is an I-section with the properties below, calculate the maximum bending stress and the maximum flexural shear stress. Beam Properties: W220X26.6 L1 VB Vc w = 26.6 kg/m³ tw = 5.84 mm rx = 87.1 mm Zy76.4 x 103³ mm³ A = 3390 mm² tf = 8.38 mm lx 25.8 x 106 mm² Sy d = 207 mm bf = 133 mm k = 16 mm Zx 279 x 103 mm³ ly 49.8 x 103 mm³ 3.32 x 106 mm² Sx 249 x 103 mm³ ho 198 mm VE VD L₂ ry = 31.2 mm J = 71.6 x 103 mm Cw=32.8 x 109 mm6 rts = 36.3 mm

Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter6: Stresses In Beams (advanced Topics)
Section: Chapter Questions
Problem 6.2.17P: Repeat Problem 6.2-1 but now assume that the steel plate is smaller (0.5 in. × 5 in.) and is aligned...
icon
Related questions
Question

CORRECT AND DETAILED HANDWRITTEN SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED.

Answers:

P1 = 208.625 KN/M

P2 = 281.310 KN/M

P = 15.491 KN/M

FB = 463.402 MPA

FV = 55.034 MPA

PROBLEM 7: The shear diagram due to distributed load of a
beam is given in the figure. The reactions at columns C1 and
C2 are uniformly distributed loads.
Given:
L₁ = 2.6 m; L₂ = 5.8 m; L3 = 3.6 m; C₁ = 0.4 m; C₂ = 0.42 m
VB = 43.68 KN; Vc = 39.77 kN; VD = 57.67 KN; VE = 60.48 KN
(a) Calculate the loads on the beam and the reactions on the
columns and illustrate the load diagram. (b) Draw the
moment diagram and identify the maximum positive and
bending moments. (c) If the beam is an I-section with the
properties below, calculate the maximum bending stress
and the maximum flexural shear stress.
Beam Properties: W220X26.6
L1
VB
Vc
w = 26.6 kg/m³
tw = 5.84 mm
rx = 87.1 mm
Zy76.4 x 103³ mm³
A = 3390 mm²
tf = 8.38 mm
lx 25.8 x 106 mm²
Sy
d = 207 mm
bf = 133 mm
k = 16 mm
Zx
279 x 103 mm³
ly
49.8 x 103 mm³
3.32 x 106 mm²
Sx
249 x 103 mm³
ho
198 mm
VE
VD
L₂
ry = 31.2 mm
J = 71.6 x 103 mm
Cw=32.8 x 109 mm6
rts = 36.3 mm
Transcribed Image Text:PROBLEM 7: The shear diagram due to distributed load of a beam is given in the figure. The reactions at columns C1 and C2 are uniformly distributed loads. Given: L₁ = 2.6 m; L₂ = 5.8 m; L3 = 3.6 m; C₁ = 0.4 m; C₂ = 0.42 m VB = 43.68 KN; Vc = 39.77 kN; VD = 57.67 KN; VE = 60.48 KN (a) Calculate the loads on the beam and the reactions on the columns and illustrate the load diagram. (b) Draw the moment diagram and identify the maximum positive and bending moments. (c) If the beam is an I-section with the properties below, calculate the maximum bending stress and the maximum flexural shear stress. Beam Properties: W220X26.6 L1 VB Vc w = 26.6 kg/m³ tw = 5.84 mm rx = 87.1 mm Zy76.4 x 103³ mm³ A = 3390 mm² tf = 8.38 mm lx 25.8 x 106 mm² Sy d = 207 mm bf = 133 mm k = 16 mm Zx 279 x 103 mm³ ly 49.8 x 103 mm³ 3.32 x 106 mm² Sx 249 x 103 mm³ ho 198 mm VE VD L₂ ry = 31.2 mm J = 71.6 x 103 mm Cw=32.8 x 109 mm6 rts = 36.3 mm
Expert Solution
steps

Step by step

Solved in 2 steps with 4 images

Blurred answer
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:
9781337093347
Author:
Barry J. Goodno, James M. Gere
Publisher:
Cengage Learning