The service load bending moments acting on a rectangular beam 307 mm wide and 607 mm deep are 147 kN-m for dead load and 195 kN-m for live load. Use the following properties: f'c = 28 MPa fy = 414 MPa db = 20 mm (bar diameter) ds = 10 mm (stirrups diameter) Consider that the stirrups used are spiral stirrups

The service load bending moments acting on a rectangular beam 307 mm wide and 607 mm deep are 147 kN-m for dead load and 195 kN-m for live load. Use the following properties: f'c = 28 MPa fy = 414 MPa db = 20 mm (bar diameter) ds = 10 mm (stirrups diameter) Consider that the stirrups used are spiral stirrups

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

ANALY 2 THE STATICALLY METHOD DRAW SHEAP AND BENDING MOMENT INDETERMINATE BEAM SHOWN BY TA E DIAGRAM A @= 10tt W=2k/# ~! P=30k ↓ L=18pt b=8ft SLOPE DEFLECTION
04. Analyze completely the compound truss loaded as shwon below. D 24 kN 3 m E 2 m F 3 @ 4 m 12 PROBLEM 4 TABLE OF ANSWERS: МЕМBER NUMBER FORCE (k) AB 1 CD 2 EF 3 BC CF 5 BF 6 DE 7 AD 8 AE 9 4.
A simply supported beam has a maximum moment equal to 20 KN.m. The beam has a width of 200 mm and a depth of 250 mm and is made up of 80% grade Narra. What is the actual bending stress of the beam in MPa? Blank 1 (use two decimal places, no need for unit)
Safe Support Spacing and S-1-bd Table below shows formulas for shear, bending moment and deflection for three-span continuous uniform load beam. R₁ = R₁ = 0.4wl R₁ = R₁ = 1.1wl V V Vmax=0.6wl Vdes = 0.6wl-w M M w(² Mmax= 10 A AMEE we Amax 145E'I If the bending and shear stress equation, and allowable deflection values are given as follow: • fb = M/S • fv = 3V/2bd • A = 1/16" a- Calculate safe support spacing for this beam (express the safe support spacing as a function of several variable). b- Express the equations for section modulus, moment of inertia and bd. Three-span, continuous, uniform l | = -w (d+ 12 )
The service load bending moments actong on a rectangular beam 310 mm wide and 624 mm deep are 58.35 kN-m for dead load and 69.11 kN-m for live load. Use the following properties: f'c = 30.01 MPa fy = 414 MPa p= 0.81 pbal db = 20 mm (bar diameter) ds = 10 mm (strirrups diameter) Consider the stirrups used are spiral stirrups. Calculate the D/C ratio in percentage (%) for the particular beam.
H.W Draw S.f,B.M and elastic dliagrams for the beam Shown below by method of equntions ? 6W/m 20 KN 6m 2m 2m
H.wi. Draw the Sher stress elistribution for the beam Cross- Section shown. Knowng that, Q= 150 kN. --I A 13 150 mm loomm
The service load bending moments acting on a rectangular beam 337 mm wide and 630 mm deep are 77 kN-m for dead load and 151 kN-m for live load. Use the following properties: f'c - 28 MPa fy = 414 MPa d₁ 20 mm (bar diameter) d, Bri 10 mm (stirrups diameter) Consider that the stirrups used are spiral stirrups. Calculate the D/C ratio in percentage (%) for the particular beam considering the strain in steel is 0.018. NOTE: USE STORED VALUES IN YOUR CALCULATION
Question #2: Calculate the maximum load P that the beam shown can carry based on plastic moment Mp. Assume a yield stress Fy of 40,000 psi. Neglect the weight of the beam. P 3" Diam. hole 6" |3'–0"|3'-0"|3'–0"| 7. SIMPLE BEAM: TWO EQUAL CONCENTRATED LOADS SYMMETRICALLY PLACED |P |P R= V R M max (between loads) M. (when x a and < (I - Pa (3/x - 3x2- a2) 6EI Moment
Problem 1. A W30 x 116 beam hasthe top flange restrained againstlateral displacement and rotationand has an unstiffened web. A pointload is applied to the top flange at2m. From the left support. The beamspans 12 m. Use A36 steel with yieldstrength of Fy = 248 Mpa. a. Compute the maximum value of point loadthat may be applied without web yieldingoccurring. b. Compute the maximum value of point loadthat may be applied without web cripplingoccurring. Properties of W30 x 116d = 762 mmK = 41.28 mmbf = 266.7 mmtf = 21.59 mmtw = 14.33 mm
12. A % in. -diameter rod made of the same material as rods AC and AD in the truss shown was tested to failure and an ultimate load of 29 kips was recorded. Using a factor of safety of 3.0, determine the required diameter (a) of rod AC, (b) of rod AD. 5 ft |B 10 ft 10 ft 10 kips 10 kips
Determine the design strength of the beam(properties given below) and the safe service live load if the service dead load is 320 kN.m.Use f'c =22 Mpa and fy=415 MPa. given properties: b=400 mm d'=70 mm d=620 mm As=10-28 mm A's=3-25 mm bars hello please answer this thanks ☺️