Question

Problem 1

The beam shown below is subjected to only two types of loads: live load w.= 0.7 k/ft, and dead load wp=

1.5 k/ft. Based on the ASCE 7-16 basic load combinations, determine the following:

(1) maximum factored negative moment Mu at support B

(2) maximum factored negative moment Mu at support D

(3) maximum factored positive moment M. at section E (mid-span of DF)

(4) maximum factored shear V. at the immediately right side of section D

F

TRB

Load combination

At C

B

16 ft

4 MB =

T

32 ft

с

RD = 327.04 K

RD

= 146 k

+

16 ft

1.2 №

+ 1.6 WL

= 1.2 (1.5) + 1.6 (0.7)

PB + RD +RF = 2.92 (1125+) = 327.04 k.

19

=) RB (32) = 3363.84

B = 105.12 K

FRD

+ +

16 ft

2.92 (48) (24) R8 (32)

16 ft

= 75.92 K

-

Moment at D: 2.92 (64) (32) - RB (48) = (2.92 x 48×24) - RF (32)

=) RF

F

105. 12k 75.92k

- 2.92 (16) (8) = √-373.76 k-ft

16 ft

= 2.92 k/ft

-2

MD = 75.92 (16)- 2.92 (48) (24)

37 Maximum positive My at E =0 because there in

(2149.126-11)

no positive

moment at DF

estion with Student's solution | Need to match with Professor's solution | Handwritten solution is needed/nVD =

V0 =

VD= 64.24 kips

2.92 (48)

RF

2-92 (48) -75.92/nops.tutorbin.com/assignment X

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For the HW1,

Please consider the following two fundamental load

combinations (same in ASCE 7-16 and ASCE 7-22):

(1) 1.4D

(2) 1,2D + 1.6L

The following shows the answers of the first

problem:

Mu for negative moment at B: 373.8 k-ft

Mu for negative moment at D: 1006 k-ft

Mu for positive moment at E: 6.4 k-ft

Vu for shear at right side of D: 70.96 kips

Professor's answer

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